Chemistry Answer

ORGANIC CHEMISTRY ASSIGNMENTS

ASSIGNMENT-1

Q.1 Match the following: [4]Column I Column II(Wrong IUPAC name) (Reason)

(A)

cyanide3PentaneCN|

CHCHCHCHCH 3223

(P) Numbering of selected carbon chain is wrong

(B)butene3bromo3chloro2

CHCHCCH||

ClBr

32

(Q) Selection of carbon chain is wrong

(C)

OH

Bicyclo [4,2,0] octan-2-ol(R) Selection of principal F.G. is wrong

(D)

CH3

3,5-dimethyl octa-1,3-diene

CH3

(S) Alphabetic order is not considered for writting

the name

(T) Suffix of functional group is wrong

[Ans. (A) Q,T (B) P, S (C) P (D) P]

Q.2 Match the following: [4]Column I Column II (DOU)

(A) Azulene (P) 4

(B) Napthalene (Q) 5

(C) Tetralin (R) 6

(D) 1,2,3-tri cyclo propyl cyclo propane (S) 7

(T) 8

[Ans. (A) S, (B) S, (C) Q, (D) P]

Question No. 3 to 4 ( 2 questions)Questions below consist of an “Assertion” in column I and “Reason” in column II. Use the followingkey to choose the appropriate answer(A) If both “Assertion” and “Reason” are correct and “Reason” is the correct explanation of the “Assertion”.(B) If “Assertion” and “Reason” are correct, but “Reason” is not the correct explanation of the “Assertion”.(C) If “Assertion” is correct, but “Reason” is incorrect.(D) If “Assertion” is incorrect, but “Reason” is correct.

Q.3 Assertion : Effective molecular mass of acetic acid is 120. [3]Reason : It form dimer due to intramolecular H-bonding. [Ans. C]

Q.4 Assertion : Cyclohexanol is less soluble in water than 1-hexanol. [3]Reason : 1-hexanol can form intermolecular H-bond with H2O. [Ans.D]

Q.5 and are [3]

(A) Chain isomers (B*) Metamers (C) Positional isomers (D) Conformers

Q.6 The boiling points of isomeric alcohols follows the order [3](A*) primary > secondary > tertiary (B) tertiary > secondary > primary(C) secondary > tertiary > primary (D) does not follow any order

Q.7 Explain why 2-methylpyrrolidine boils at a higher temperature than its isomers N- methylpyrrolidine.[3]

Q.8 Explain why p-dichlorobenzene is more soluble in n-propyl alcohol than methyl alcohol, whileo-dichlorobenzene is less soluble in n-propyl alcohol than methanol. [3]

Q.9(a) How many metamers (aromatic) are possible for the structural formula of O – Et .

(b) Write bond line structural formula for all the possible isomers having the molecular formula C4H6.[4][Ans. (a) 4 , (b) 9]

Q.10 Fill up the blank space showing minimum number of C-atoms required to exhibit the given isomerism.[24]

Aldehyde)f(Ether)e(Alcohol)d(Alkyne)c(Alkene)b(Alkane)a(

MetamerismFunctionalPositionChain

[Ans.

354Aldehyde)f(4245Ether)e(

234Alcohol)d(345Alkyne)c(

44Alkene)b(64Alkane)a(

MetamerismFunctionalPositionChain

]

“Well begin is half done”

ASSIGNMENT-2

Q.1 Match the column: [4]Column I Column II

(A) A pair of metamers (P)

ClCl

Br

Br

&

(B) A pair of functional isomer (Q) Acetaldoxime & acetamide

(C) A apir of position isomer (R)

OH

CH3

CH2–OH

&

(D) A pair of chain isomer (S)

CN|

CHCHCH 223 &

CN|

CHCHCH 33

(T) Cl C–O–C

Cl

O O

&

C–O–C

Cl Cl

O O

[Ans. (A) T, (B) Q,R (C) P (D) S]

Q.2 Match the column : [4] Column I Column II(Compound) (Double bond equivalent)

(A) S

N

C—N

OO

COOH

HCH3

(penicillin)

(P) 4

(B)

(Hexahelicene)

(Q) 5

(C)(Prismane)

(R) 7

(D)C

CC

OH

OH

O

O(Ninhydrin)

(S) 19

(T) 8

[Ans. (A) Q; (B) S; (C) P; (D) R]

Q.3 Statement-1 : Homologoues are having similar physical properties & different chemical properties.Statement-2 : They are having different molecular masses & having same F.G. [3](A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

Q.4 Statement-1 : IUPAC name of H2C = CH–CH2–CCH is pent-4-en-1-yne. [3]Statement-2 : Alkene is preferred over alkyne when they are placed at same position.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.5 Which of the following is/are a bicyclo compound(s): [3]

(I) (II) (III) (IV)

(A) III only (B*) III and IV only (C) II, III and IV (D) I, II, III and IV

Q.6 Compare water solubility and boiling point of iso-pentyl alcohol and iso pentyl amine. [3][Ans. (boiling point i) (H2O solubility ii)]

Q.7 Boiling point of ethanol (78°C) is more than that of ethanamine (16.6°C), but boiling point of ethanoicacid (117°C) is less than that of ethanamide (221°C). Explain with reasons. [3]

Q.8 Compare boiling point: [4]

(I) (a) 2223

3

NHCHCHCHCH|

CH

(b) 323

3

CHNHCHCHCH|CH

(c) 3223

3

CHCHNCHCH|

CH

[Ans. a > b > c]

(II) (CH2OH)2 (CH2OMe)2 CH2OH – CH2OMe A B C [Ans. A > C > B]

(III)

3

33

3

CH|

CHCCH|CH

neopantane n-hexane 2,3-dimethylbutane

1-pentanol 2-methyl-2-butanol[Ans.neopantane < 2,3-dimethylbutane < n-hexane < 2-methyl-2-butanol < 1-pentanol ]

(IV) 3-methylbutanol, n-Pentane, 2, 2-dimethylpropanol and n-pentanol [Ans. 4 > 1 > 3 > 2]

Q.9NH

NH2 NH2

O

O

OH

OH

[9]

(a) Find out the total number of 1° hydrogen.(b) Find out the total number of 2° hydrogen.(c) Find out the total number of 3° hydrogen.(d) Find out the total number of 1° carbon.(e) Find out the total number of 2° carbon.(f) Find out the total number of 3° carbon.(g) Find out the total number of 4° carbon.(h) Find out the total number of functional groups.(i) Find out the number of degree of unsaturation.

[Ans. a-9, b-11, c-3, d-5, e-9, f-3, g-0, h-6, i-5]

Q.10 What isomerism is exhibited by the following pairs of compound. [10](a) Me4C & Me2CHCH2CH3

(b) CH3COCH2OH &CH3 – C – CH2

OOH(c) o-xylene & p-xylene(d) Butanal & 2-buten-2-ol(e) Bicyclo [1,1,0] & 2-methyl cyclopropene(f) Acrylic acid & Vinyl formate

(g)O

& O

(h)

H|

CCCCNCC &

C|

CCCNHCC

(i)

O||

MeNHCEt &

O||

EtNHCMe

(j) C—O

O

& C—O

O

[Ans. (a) chain (b) func.(c) posi. (d) func. (e) func. (f) func. (g) posi. (h) chain (i) meta (j) meta ]

A diamond is merely a lump of coal that did well under pressure.

ASSIGNMENT-3

Q.1 Match the column:Column-I and column-II contains four entries each. Entry of column-I are to be uniquely matched withonly one entry of column-II. [4]

Column I Column II (Compound) (Boiling point in °C)

(A) H3C–CH2–CH2–CH2–CH2–OH (P) 290°C

(B)

OH|

CHCHCH 33 (Q) 138°C

(C)

OHOHOH|||CH—CH—CH 22 (R) 102°C

(D)

OH|

CHCHCCH|CH

323

3

(S) 82.4°C

[Ans. (A) Q; (B) S; (C) P; (D) R]

Q.2 Match the Column : [4]Column I Column II

(A) C6H14 (P) Tautomerism

(B) C4H10O (Q) Positional isomerism

(C) C4H8O (R) Functional isomerism

(D) C2H5N (S) Geometrical isomerism

[Ans. (A) Q (B) Q,R (C) P,Q,R,S (D) P,R,S]

Q.3 Statement 1 :

H5C2

CH3

Cl

BrC=C and

H5C2

CH3

Cl

BrC=C are structural isomers. [3]

Statement 2 : The above mentioned compounds can show geometrical isomerism.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement-1.(C) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

Q.4 Statement 1 : Resonating structures of NH

O

can show geometrical isomerism at room temperature.

Statement 2 : Lone pair of nitrogen will participate in resonance but only E form is stable in it’s resonatingstructure. [3]

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

Q.5 Conformational changes in a ethane leads to change in [3](A*) torsional angle (B) bond angle (C) bond length (D) all of the above

Q.6 Which of the following isomeric structure have lowest energy? [3]

(A) (B)

(C) (D*)

Q.7 The terpene ocimene has the IUPAC name (3E)-3,7-dimethyl-1,3,6-octatriene, what is the structuralformula of this compound? [3]

(A) (B) (C) (D*)

Q.8 2223 CHCHCHCH|

NC

and

3

33

CH|

CHCCH|NC

are [3]

(A*) chain isomers (B) positional isomers (C) both (D) none

Q.9 Circle the member of each pair that is more soluble in water. [6](a) CH3CH2OCH2CH3 or CH3CH2CH2CH2CH3(b) CH3CH2NHCH3 or CH3CH2CH2CH3(c) CH3CH2OH or CH3CH2CH2CH2OH

(d) or

(e) C3H8, C2H5OH, (CH3)2O and CH2OH·CH2OH(f) 3-methylbutanol, n-Pentane, 2, 2-dimethylpropanol and n-pentanol

[Ans. (a)CH3CH2OCH2CH3, (b)CH3CH2NHCH3, (c) CH3CH2OH , (d) ,

(e) 4>2>3>1, (f) 3>1>4>2 ]

Q.10 For each pair of compounds, circle the compound you expect to have the higher boiling point. Explainyour reasoning. [12]

(a) (CH3)3C–C(CH3)3 and (CH3)2CH–CH2CH2–CH(CH3)2(b) CH3(CH2)6CH3 and CH3(CH2)5CH2OH(c) HOCH2 – (CH2)4–CH2OH and (CH3)3CCH(OH)CH3(d) (CH3CH2CH2)2NH and (CH3CH2)3N

(e) and

(f) and

N-methylpyrrolidine piperidine

(g) and

tetrahydropyran cyclopentanol(h) CH3CH2OCH3 or CH3CH(OH)CH3(i) CH3CH2CH2CH3 or CH3CH2CH2CH2CH3(j) CH3CH2CH2CH2CH3 or (CH3)2CHCH2CH3(k) CH3CH2CH2CH2CH3 or CH3CH2CH2CH2CH2Cl(l) CH3CH2–O–CH2CH3 and CH3CH2CH2CH2–OH

[Ans. (a) (CH3)2CH–CH2CH2–CH(CH3)2, (b) CH3(CH2)5CH2OH, (c) HOCH2 – (CH2)4–CH2OH,

(d)(CH3CH2CH2)2NH, (e) , (f) piperidine, (g) cyclopentanol, (h)

CH3CH(OH)CH3, ( i) CH3CH2CH2CH2CH3 , (j) CH3CH2CH2CH2CH3,(k)CH3CH2CH2CH2CH2Cl, (l) CH3CH2CH2CH2–OH ]

Discipline is the bridge between goal and accomplishment.

ASSIGNMENT-4

Q.1 Match the column : [4]Column-I and column-II contains four entries each. Entry of column-I are to be uniquely matched withonly one entry of column-II.

Column I Column II (Compound) (Solubility in g / 100 g H2O)

(A) CH3CH2–OH (P) 0.05(B) (C6H5)2CH–OH (Q) 12.5(C) CH3(CH2)5CH2OH (R)

(D)

OH|

CHCHCHCH 323 (S) 0.2

[Ans. (A) R; (B) P; (C) S; (D) Q]

Q.2 Match the column: [4]Column I Column II

(A)

OH

OH

(P) Conformation of maximum torsional strain

(B) (Q) Conformation with strongest intramolecular hydrogen bond

(C) (R) Highest boiling point out of ABCD

(D) (S) Conformation of minimum Vander Waal strain

(T) Maximum heat of combustion out of ABCD[Ans. (A) R,T (B) Q, (C) P, (D) S ]


Q.3 Statement-1 : Boat form is the least stable conformation of cyclohexane. [3]Statement-2 : Boat form is eclipsed form. [Ans. D]

Q.4 Statement-1 : Gauche form of butane is less stable than anti-form. [3]Statement-2 : Anti-form is more symmetrical the gauche. [Ans. B]

Q.5 Potential energy barrier for the direct interconversion of the two gauche forms of butane is 3.6 kcal/mol.What should be the expected potential energy barrier for direct interconversion of the two gauche formof 1,2-dichloroethane Vander Waal’s radius of a covalently bond chlorine atom is about the same as thatof a methyl group. [3](A) 2.8 kcal/mol (B) 3.6 kcal/mol (C) 3.8 kcal/mol (D*) 9.3 kcal/mol

Q.6 Select correct statement(s) about (a) & (b). [3]

(A*) a and b have equal torsional strain (B*) a is more stable(C*) a is having higher dipole moment (D) b is the least stable form of this compound.

Q.7 Find out the correct option(s) ? [3]

(A*)

CH3C=C

H

NH

NH Orientation is E

(B*)

DC=C

H Orientation is Z

(C) CH3 CH3

H HOrientation is Z

(D*)

CH3

CH3

geometrical isomers are not possible

Q.8 Which compound(s) will show the geometrical isomerism? [3]

(A*)

CH3N–H

(B) C = C = CCl

H

Cl

H

(C*) CH3

H3C (D)

HN

NH

O

O

Q.9(a) Arrange the groups or atoms accroding to their priority order using CIP sequence rules ? [5]

(1) , , –CH2–CH3 , –CH=CH2 , –CCH

(2) –COOH, –COCl, –CONH2 ,–CHO, –COCH3

(3) –CH2–NH2, –CH2–Cl, –CH2–Br,–CH2–CN, –CH2–CH2–Cl, –CH2–OH

(4) –CN, –OH, Cl, –D, –H, –SO3H, –SH, –F, Br, I, –NH2

(5) O, O

, O

, O

[Ans. (1) 2 >5 > 1 > 4 > 3 (2) 2 > 1 > 3 > 5 > 4(3) 3 > 2 > 6 > 1 > 5 > 4 (4) 10 > 9 > 3 > 6 > 7 > 8 > 2 > 11 > 1 > 4 > 5(5) 2 > 4 > 1 > 3 ]

(b) Which of the following compounds should have the larger energy barrier to rotation about the indicatedbond? [3]

(a) Me C – CMe3 3 (b) Me Si – SiMe3 3

(c) CH = CH – CH = CH2 2

[Ans. c > a > b ]

Q.1084goc-1 Find total number of Geometrical isomerism of following compounds. [6]

(a) CH3–CH=CH–CH=N–OH (b)

(c) (d) CH3–(CH=CH)3 – Ph

(e) (f) CH3–CC–CH=CH–CH3

[Ans. (a) 4, (b) 2, (c) 2, (d) 8, (e) 2, (f) 2]

I’ll run till I can, I’ll walk till I can, when I can only crawl, I’ll crawl but by the grace of god,I’ll always be moving forward.

ASSIGNMENT-5

Q.1 Match the following: [4] Column I Column II(Compound) (Double bond equivalent)

(A) (P) 12

(B)

HH

N

NN

N

Porphyrin (C H N )20 14 4

(Q) 13

(C)

Ph

(R) 14

(D) (S) 16

[Ans. (A) Q, (B) S, (C) R, (D) P]

Q.2 Match the following structural formulae with their possible geometrical isomers? [4]Column I Column II

(Structural formula) (Total geometrical isomers)

(A) CH3–CH = CH CH2–CH3 (P) 8

(B) CH3–CH =CCH=CH–CH3

CH=CH–CH3

(Q) 6

(C) CH –CH=CH3 CH=CH–CH3 (R) 4

(D) Cl – CH = CH – CH = CH – CH = CH – CH3 (S) 2

[Ans. (A) S, (B) R (C) Q (D) P ]

Q.3 Statement-1 : Van-der-wall’s forces are present only in non polar compounds [3]Statement-2 : In polar compounds stronger dipolar forces are present.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.4 Statement-1 :

O

tautomer of this compound can show geometrical isomer.. [3]

Statement-2 :

OH

across the double bond both the terminals having different groups.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.5

CH3 CD2 CF3

CH–CH3

CH2–ClCH3 CH2

C=CC=C

Me

H

14 [3]

Configuration of both the double bond in this compound respectively are(A) 2E 4E (B*) 2E 4Z (C) 2Z 4E (D) 2Z 4Z

Q.6 What is the correct stereochemical assignment for each of the following alkenes? [3]

F(I)

F(II) (III) (IV)

(A*) I = E; II = Z; III = E; IV = Z (B) I = Z; II = E; III = E; IV = E(C) I = E; II = E; III = Z; IV = Z (D) I = E; II = E; III = Z; IV = E

Q.7 Correct statement about given Ketoxime is [3]

CH3 C = NOH

..

(A)(A*) A is named as syn-p-tolylphenyl ketoxime(B*) A is named as anti-phenyl-p-tolyl ketoxime(C*) Compound (A) is Z isomer(D) Compound (A) is E isomer

Q.8 What is correct about (B) & C) [3]

H

D

CH3

D

CH3

+180° rear carbon

(B)

H–180° front carbon

(C)

(A*) Both are achiral molecules (B*) Both contains chiral carbon(C*) Both are meso (D*) Both having same bond length.

Q.9 How many chiral carbon atoms are present in the following compounds? [4]

(i) (ii)

(iii) (iv)

[Ans. (i) 2, (ii) 3, (iii) 4, (iv) 10]

Q.10 Assign priority number to the following groups as per Cahn, Ingold, Prelog sequence rule [21](a) –CH2OH, –CH3, –CH2CH2OH, – H(b) –Cl, –Br, –CH=CH2, –CH3

(c) HC||

O

–OH, –CH3, –CH2OH(d) –CH(CH3)2, –CH2CH2Br, –Cl, –CH2CH2CH2Br

(e) –CH=CH2, –CH2CH3 , –CH3

(f) –CH=CH2, –CCH

33

2

CHCH||CHCH

33

33

CHCH||

CH—C||

CHCH

(g) –CH2CH2CH2I

Br|

CHCHCH 3

Cl|

CHCHCH 32 –F

[Sol. (a) 1 > 3 > 2 > 4 (b) 2 > 1 > 3 > 4 (c) 2 > 1 > 4 > 3 (d) 3 > 1 > 2 > 4(e) 3 > 1 > 2 > 4 (f) 4 > 2 > 3 > 1 (g) 4 > 3 > 2 > 1(see the Cahn, Ingold, Prelog rule) ]

Dreams ....Are not the ones you see when you get in to the sleep, But are those which will never let you sleep...Till they are achieved.....

Q.1 Indicate whether each of the following structure has the R configuration or the S-configuration.[21]

(a) (b) (c) (d)

(e) (f) (g) (h)

(i) (j) (k) (l)

(m) (n) (o) (p)

(q) (r) (s) (t) H

Br

CH CH2 3

CH3C CH3H2

CH3

(u)

[Ans.. (a) R (b) R (c) S (d) S (e) R (f) S (g) R (h) S (i) S (j) S (k) S (l) R(m) S (n) R (o) S (p) S (q) 2R 3R (r) 2R 3R (s) 2S 3R (t) 3S 4R(u) 2R 3S ]

Q.2 Indicate whether each of the following structure has the R configuration or the S-configuration. [26]

(a) (b) (c) (d)

(e) (f) (g) (h)

(i) (j) (k) (l)

ASSIGNMENT-6

In time of trouble, don't say "God I have a big problem." say "Hey Problem. I have a big God.

(m) (n) (o) (p)

(q) (r) (s) (t)

(u) (v) (w) (x)

(y) (z)

[Ans. (a) R, (b) R, (c) S, (d) S, (e) R, (f) S, (g) R, (h) S, (i) S, (j) S, (k) S, (l) R, (m) S, (n) R, (o) S, (p) S(q) 2R 3R, (r) 3S 4R, (s) 2S 3R, (t) 3S 4R, (u) 1R 2S, (v) 5R 6S, (w) 3S 4R, (x) 4S 5R,(y) 2R3R4R, (z) R ]

Q.1 Match the column : [4]Column I Column II(Property) (Compound)

(A) Plane of symmetry (P) C CH

HMe

Me

(B) Centre of symmetry (Q)

Me Me

(C) Meso Compound (R) CH

Me

Me

(D) Chiral atom is / are present (S) H

COOH

H

Ph

H

COOH

H

Ph

(T) C = C = C = C HCl

Cl H

[Ans. (A) P,S,T (B) P,T (C) S (D) Q,R,S]

ASSIGNMENT-12

Q.2 Column I Column II [4]

(A)Me

Me

Cl

Cl

O

(P) Total number of stereo isomers are odd.

(B)Br

Br

(Q) Total number of stereocenter are even

(C)C

C

H

H

Cl

Cl(R) Compounds having plane of symmetry

or axis of symmetry

(D)

COOH

COOH

HHOH

HO (S) Compounds have zero dipole in given form.

(T) Compounds having centre of symmetry.

[Ans. (A) P,Q,R (B) Q,R,S,T (C) Q,R,S,T (D) P,Q,R]

Q.3 Statement-1 :

F

F(I)

F

F(II)

[3]

(I) and (II) are optically inactive molecules.

Statement-2 : Molecules containing plane of symmetry or centre of symmetry are opticallyinactive.

(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.4 Select the correct statement: [5]

(A*) N

H EtCH3

is optically inactive while N

is optically active

(B*) )I(

CHCHCHCH 3223

)II(CHCHCH 323

About C() – C() bonds; (I) has more energy barrier for rotation as compared to (II).(C*) The boat form of cyclohexane is less stable than chair form due to torsional strain as well as flagpole interaction.(D*) In t-butyl substituted cyclohexane, most stable chair conformer has – CMe3 group at equitorialposition.

Q.5 Most stable form of meso-2,3- diflouro-2,3-butandiol is: [3]

(A) (B) (C*) (D)

Q.6 True statement(s) about this compound is/are: [3]

Cl

Cl

Cl

Cl

(A) It is E isomer (B) It is Z isomer(C*) It is optically active (D) It has Centre of Symmetry

Q.7 Consider the following statements regarding the given projection and select the correct statement(s)?[3]

(W) (X) (Y) (Z)

(A*) W & Y are diastereoisomers (B*) Z is the newmann projection of X(C*) W, X, Y and Z are optically active (D) Y & Z are meso.

Q.8 30.9 g of (+) 2-butanol was taken in a 309 mL solution and was mixed with 15.45 g of (–) 2-butanolplaced in 154.5 mL and the final solution is passed through 444 mm tube. What will be observed rotation( in degree) for the final solution if specific rotation of 2-butanol is 13.5°. [5]

[Ans. 0002.00 or 0001.99]

Q.9 From the following set of compounds, select: [5](a) enantiomer pairs(b) distereomer pairs in which both are optically active(c) label them as D or L sugar

(I) (II) (III)

(IV) (V)

[Ans.

Also II D-sugar V L-sugarSugar I, III and IV are all L-sugar; Any pair of two structure other than II and V are diastereomer pair ]

Q.10 How many stereoisomers are possible for following compounds: [7]

(a)

(b) CH3 – CH = CH – CH = CH – CH = C = CH2

(c) COOHCHCHHOOC||

OHOH

(d) 1,2-dichlorocyclopropane(e) 1,3- dimethylcyclobutane(f) 2-bromo-3- chlorobutane(g) 1,3-dimethyl cyclohexane

[Ans. (a) 8 (b) 4 (c) 3 (d) 3 (e) 2 (f) 4 (g) 3 ]

Tough Times never last, tough people do.

ASSIGNMENT-13


(A) (P) Chiral

(B) (Q) Plane of symmetry

(C) (R) S

(D) (S) E (Entegen)

(T) Even number of stereoisomer possible

[Ans. (A) P,R,S,T (B) P, R,T (C) P, S, (D) P,T]

Q.2 Match the column : [4]Column I Column IICompound Number of Geometrical isomerism

(A)BrI

Cl

(P) 2

(B*) (Q) 3

(C) (R) 4

(D) (S) 5

(T) 6

Q.3 Statement-1 : Resonance energy of 22 NHCCHCH||

O

is very high as compaired to

223 NHCCHCH||

O

.Statement-2 : Another double bond is introduced in conjugation to C = O.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true. [3]

Q.4 The configuration of chiral carbon in the molecule. [3]CHO

H

CH2OHHO

(A*) D, R (B) D, S (C) L, R (D) L, S

Paragraph for question nos. 5 to 7Observe following structures and answer questions given below: [9]

(I)

HO

HO H

HMe

Me

(II)

HO

H

H

OHMe

Me

(III)H

H OHMe

MeHO

(IV)

H

H

OH

OH

Q.5 What is the relationship between II & III.(A) Enantiomers (B) Diastereomers (C*) Identical (D) Positional isomers

Q.6 Which of the following is pair of diastereomers(A*) I & III (B) II & III (C) I & IV (D) None

Q.7 Compound IV is a vicinal diol, total how many vicinal diol's are possible with the same molecular formulaas IV.(A) 3 (B) 4 (C) 5 (D*) 6

Q.8 Select correct statement: [5]

(A*)

CH – OH2

and

CH3

OH

are functional isomers

(B*)

Cl

has eight stereoisomers

(C)

ClClCl|||

MeCHCHCHMe has four optically active stereoisomers

(D) H CH3

CH3H

Me

Me

is meso compound

Q.9 Find relationship between following pairs. [6]

(a)

HO OH

Cl

Cl

Me Me

CHO CHOCH OH2

CH OH2CH3 CH3

&

(b)

CH3

CH3 Et

&H

H

H MeMeEt HEt Et

(c) CH3 – O – CH2 – CH2CH3 &

3

33

CH|

CHCHOCH

(d) &

Me

Me

OH OH

Br

Br

(e) & Cl

OH

OH Cl

(f) &

Br Br BrBr D D HH

O O[Sol. (a) Diastereoisomers (b) Identical (c) Positional isomer (d) Enantiomer

(e) Geometrical isomer (f) Identical ]

Q.10 Among following pairs which will require lower heat of activation for cis to trans interconversion: [8](a) Me–CH=CH–Me & Me3C–CH=CH–CMe3(b) Me–CH=CH–Me & Me–CH=CH–OMe

(c) CH=CH & CH=CH NO2

(d) CH=CH & CH=CH OMe

(e) O2N CH=CH NO2 & O2N CH=CH OMe

(f)

OO||||

MeOCCHCHCOMe &

OO||||

MeCOCHCHCOMe

(g) & (h) &

[Sol. a to h in all case right one]

Confidence begets confidence. The man who wins is the man who thinks he can

Q.1 Match the column: [4]Group Effect

(A) –NO (P) + I

(B)3HN

(Q) – I

(C) – NC (R) + M(D) –NH2 (S) – M

[Ans. (A) Q,R,S ; (B) Q ; (C) Q ; (D) Q, R]

Q.2 Match the column-I with column-II. Note that column-I may have more than one matching options incolumn-II. [4]

Column–I (stability) Column-II (Reason)

(A) > 3CH (P) Inductive effect

(B) < –3CH (Q) Resonance

(C) > H3C– 2CH (R) Hyperconjugation

(D) > (S) steric hindrance

[Ans. (A) P,Q,R; (B) P; (C) P, R; (D) Q]

Q.3 Statement-1 : Correct order of acidic strength is CH3CH3 > CH2=CH2 > HCCH [3]Statement-2 : C–H bond energy in ethene is more than ethane but less than ethyne.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

ASSIGNMENT-14

Page No.# 1

Q.4 Among the following pair of compounds geometrical isomers are? [3]

(A) O–C–Me O–C–Me

O–C–Me C–O–Me

O O

O O&

(B) & H H H

Me Me Me

MeN

(C) & C = C = C C = C = C H HH

HMe MeMe

Me(D*) None of these

Q.5 Identify total number of 2° amines in given compound? [3]

O

O

NHN

N

N

Me

(A) 1 (B*) 2 (C) 3 (D) 4

Q.6 Identify total number of geometrical isomers in given compound? [3]H Me

H N–N2 N–NH2(A) 2 (B*) 4 (C) 6 (D) 8

Q.7 Which order regarding basicity is correct: [4]

(A*) > (B*) <

(C*) > (D*) >

Page No.# 2

Q.8 Which of the following carboxylic acids could be resolved by reaction with an enantiomerically purechiral amine? [3]

(A) (B)

(C*) (D)

Q.9 Identify relationship between following pairs : [4]

(a) (b)

(c) (d)

If they are enantiomer answer will be 1, if they are diastereomers answer will be 2, if they are constitutionalisomers answer will be 3 and if they are identical present 4 as the answer. Present sum of answer of eachpart a + b + c + d in OMR sheet.(for example if a, b, c, d are enantiomer then final answer is a + b + c + d = 4). [Ans. 13]

Q.10 S(+) Mono Sodium Glutamate (MSG) is a flavour enhancer used in many foods. Fast foods oftencontain substantial amount of MSG and is widely used in Chinese food. If one mole of above MSG wasplaced in 845 ml solution and passed through 200 mm tube, the observed rotation was found to be +9.6°. [9]

¯OOC—CH —CH —C22

COO¯Na+

NH3

H

MSG(a) Find out the specific rotation of (–) MSG:

(A) + 24° (B) + 56.8° (C) – 48° (D*) None of these

(b) Find out the approximate percentage composition of (–) MSG in a mixture containing (+) MSG and(–)MSG whose specific optical rotation is –20°.(A) 83.3 % (B) 16.7 % (C*) 91.6 % (D) 74 %

(c) If 33.8 g of (+) MSG was put in 338 ml solution and was mixed with 16.9 g of (–) MSG put in 169 mlsolution and the final solution was passed through 400 mm tube. Find out observed rotation of the finalsolution(A) + 1.6° (B) + 4.8° (C*) + 3.2° (D) None of these

Page No.# 3

Q.11 [6]

(a) is dimethyl derivative of a compound 'A' and is fairly stable but most of the molecules of 'A'

gets converted into another compound 'B' on keeping. What are the structures of 'A' and 'B'. Explain thereason of conversion A into B.

(b) Write increasing order of C–N bond length in following compounds.

(i) (ii) (iii) (iv)(c) Write increasing order of heat of hydrogenation

CH3

CH3

(i) (ii) (iii)

[Sol.(a) A is and it tautomerise into B as it is more stable because it is aromatic compound

B is

(b) (i) < (ii) < (iii) < (iv)(c) (ii) > (i) > (iii) ]

Challenges of the life should be accepted strongly, Don’t say “Why me?”, instead tell God “Tryme!!!

Page No.# 4

Q.1 (a) Column-I (Carbanions) Column -II (Half lives) [4]

(A) (P) 4.3 × 10–8 s

(B) (Q) 2.5 × 10–5 s

(C) (R) 8.7 × 10–7 s

(D) (S) 1.4 × 10–4 s

[Ans. (A) R, (B) S, (C) Q, (D) P ]

(b) Match the column : [4]Column I Column II

(Compounds) (Magnitude of heat of hydrogenation)

(A) (P) 52.5 Kcal / mole

(B) (Q) 45.9 Kcal / mole

(C) (R) 26.6 Kcal / mole

(D) (S) 30.7 Kcal / mole[Ans. (A) S (B) R (C) Q (D) P ]

ASSIGNMENT-15

Page No.# 5


(A)

CH3

CH3

Cl

ClH

HOH

H &

CH3

CH3

Cl

ClHO

HH

H(P) Identical

(B)

Et

HHHS

Me

OH &

OH

EtMe

SHH

H (Q) Diastereomers

(C)

H2C

MeH

C CH

Et

&

H2C Me

H

C CH

Et(R) Conformer

(D)

HH

H HH H

ClCl

&

HH

H

H

H H

Cl

Cl

(S) Positional

[Ans. (A) P, (B) S, (C) R, (D) Q ]

Q.3 Questions below consist of an “Assertion” in column I and “Reason” in column II. Use the followingkey to choose the appropriate answer [3](A) If both “Assertion” and “Reason” are correct and “Reason” is the correct explanation of the “Assertion”.(B) If “Assertion” and “Reason” are correct, but “Reason” is not the correct explanation of the “Assertion”.(C) If “Assertion” is correct, but “Reason” is incorrect.(D) If “Assertion” is incorrect, but “Reason” is correct.Assertion : n-propyl carbocation is more stable than ethyl carbocation.Reason : n-propyl carbocation rearrange to give 2° carbocation. [Ans. D]

Q.4 Which of following pair is Diastereomers: [3]

(A)

CO H2

CO H2 CO H2

CO H2

HO HHH

H OH OHOH

(B*) CH3

CH3

CH3 CH3

C=C C=CH

HC C

H H

H H

Br Br

(C)

CH3

CH3Et

Et

H HSH HOOH H

SH H(D)

Cl Cl

Cl Cl

Cl Cl

Page No.# 6

Q.5 What is the correct order of basicity values of the following compounds? [3] CH2 – NH2 CH2– NH2(A) | > | > CH3NH2 > NH3 +SMe2 Br(B) NH3 > CH3NH2 > CH2=CH-NH2 > NH2-CH2-CH2-N+R3

(C) > > > >

(D*) None

Q.6 Consider the following compound [4]

(A*) (B*) (C*) CCOOHCH||

O

3(D*)

Which of the above compounds reacts with NaHCO3 giving CO2

Q.7 Out of following the most basic isomers of the formula C4H7N is [3]

(A*) CC–C–C–NH2 (B) C–CC–C–NH2 (C) C–C–CC–NH2 (D) N|

H

Q.8 Select incorrect statement from the following. [4]

(A*) O

O is having very high resonance energy than

O

O

(B) CH3

COOH

is more acidic than COOH

(C*) CH2

is more stable than OHr.

(D*) O

is more stable than OCMe||

O

Page No.# 7

Q.9 Following figure give Potential energy of different types of isomers of unsaturated compound CnH2n–2.There are three isomers shown is the following figure with there o

fH (standard heat of formation),which involves two alkynes and one alkadiene, with 5 carbon atoms. Usually alkynes are less stable thanisomeric alkadienes as sp-hybridisation state is relatively less stable than the sp2-hybridisation state.Based on it answer the following questions. [9]

A B C

Pot e

ntia

l Ene

rgy

(a) Which of the following will react with NaNH2 most readily.(A*) A (B) B (C) C (D) All of these

(b) Which is most stable alkyne?(A) A (B*) B(C) C (D) Additional data os required to decide relative stability

(c) Which are of the following is not the isomer of the compounds shown is figure:

(A) (B*) (C) (D)

Q.10 Observed the following compound and answer the following questions. [9]

H – OaO

N

MeNH2

NMe

S

O – H

b c

d e

(a) Arrange the H-atom in decreasing order of their acidic strength.(A) a > b > e (B*) a > e > b (C) b > a > c (D) e > a > b

(b) Arrange the N-atom in decreasing order of their basicity(A*) b > d > c (B) c > b > d (C) d > c > b (D) c > d > b

(c) What will be the degree of unsaturation(A) 5 (B) 6 (C*) 7 (D) 8

Success is waiting for you. Difficulties are blocking the path. Take the weapon of confidence andreach your success every day!.....

Page No.# 8

Q. 1 Match the following : [4]

Column I Column II(Number of planes of symmetry)

(A)

Me

MeMe

(P) 6

(B) (Q) 0

(C) CCl4 (R) 7

(D)

CH3CH3

H C3 H

HH

(S) 1

[Ans. (A) S (B) R (C) P (D) Q]


Q.2 Assertion : Heat of hydrogenation of trans-2-butene is more than cis-2-butene. [3]Reason : Trans-2-butene is more stable cis-2-butene. [Ans. D]

Q.3 Assertion : Cyclohexene shows geometrical isomerism.Reason : Its two sp2 H-atoms are cis to each other. [And. D] [3]

ASSIGNMENT-16

Page No.# 9

Q.4 Which of the following is most stable carbocation. [3]

(A)

OEt

(B)

OEt

(C*)

OEt

(D)

OEt

Question No. 5 to 8 (4 questions)

Cortisone

Q.5 Cortisone contains which functional groups? [3](A) Ether, alkene, alcohol (B) Alcohol, ketone, amine(C*) Alcohol, ketone, alkene (D) Ether, amine, ketone

Q.6 Correct order of acidic strength in Cortisone [3](A) a > b > c > d (B*) b > a > c > d (C) c > b > a > d (D) b > c > a > d

Q.7 Total number of chiral center in Cortisone : [3](A) 4 (B) 5 (C*) 6 (D) 7

Q.8 Total stereoisomer of the compound Cortisone is [3](A) 32 (B*) 64 (C) 66 (D) 128

Q.9 Arrange the following in increasing order of their heat of combustion: [5]

I. II. III. [Ans. I < II < III ]

Q.10 For the following compounds, arrange the labelled proton in increasing order of their case of deprotonation:

(a) CH2 H2

H1

C C–H 3

(b) (c) [5]

[Ans. (a) 2 < 1 < 3 (b) 3 < 1 < 2 (c) 3 < 1 < 2 ]

There isn't any problem which is without a gift for you in its hand.

Page No.# 10

Q.1 Four compounds are given in column I, match these with their 1apK values given in column II. [4]

Column I Column II

(Compound) ( 1apK )

(A) CH3COOH (P) 4.87

(B)COOH|COOH

(Q) 4.76

(C) ClCH2COOH (R) 2.86

(D) CH3CH2COOH (S) 1.2[Ans. (A) Q, (B) S, (C) R, (D) P]

Q.2 Statement-I : A tertiary carbocation is more stable than a secondary carbocation which is morestable than a primary carbocation.

Statement-II : The inductive effect operates through -bonds and decrease rapidly with increasein distance. [3]

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

Q.3 Statement-1: p-amino benzoic acid do not exists as zwittor ion [3]Statement-2: Acidic and basic groups in this compound are too far from each other to react, so they

can't form salt.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is false, statement-2 is true.(D*)Statement-1 is true, statement-2 is false.

Q.4 Arrange the following in increasing order of acid strength. [3]

OH OH

COOH

CH –H2H—CH2

HO S3

(a)

(b)

(f)

(c) (d)

(e)

(A) c < f < d < a < b < e (B) f < d < c < b < e < a(C) f < c < d < b < e < a (D*) c < d < f < a < b < e

ASSIGNMENT-17

Page No.# 11

Q.5 Select true statement(s) : [3](A*) Resonance effects bond length(B*) cis-1-bromo-1,2-difluoro ethene and Z-1-bromo-1,2-difluoro ethene are geometrical isomers

(C*) In CH2

O

O

COH

COH most acidic H is connected directly to oxygen not on carbon.

(D*) Boiling point of OH

is less than

OHOH

Q.6 Which of the following is the correct order of acidic nature. [3]

(I)

O

NH H

(II) N

H H

(III)

H HN

(A*) III > I > II (B) I > II > III (C) II > I > III (D) III > II > I

Q.7 Which of the following is the correct order of basic nature. [3]

(P) NH

(Q) N

(R) NN

(A) Q > R > P (B*) R > Q > P (C) R > P > Q (D) Q > P > R

Q.8 Correct increasing order of acidic strength of following compounds is / are ? [3]

(A)

O OHHN ,

OOHH

N ,

OOH

NH

(B*) OH , OH , OH

(C*) Br

ClOHF

, Br

Cl OHF

,

BrCl OH

F

(D*) CH3–C C–H , H2O , HF

Page No.# 12

Q.9 Among the following compounds select those [6](i) Which will readily dimerise.(ii) Which are Quasi aromatic

(1) (2) 244HC (3)

(4) (5)

t–Bu t–Bu

t–Bu N(6) Azulene

[Ans. (1)Quasi (2) Quasi (3) Quasi(4)Quasi (5) It is antiaromatic but does not undergos dimerisedue to bulky t-Bu groups.(6) Quasi ]

Q.10 Resonance energy of is much lower than that of but resonance energies of and

are identical. Explain. [5]

Q.11 Arrange the following in increasing order of pKa values [6](a) HNO3 H2O HCO2H CH3CO2H PhOH(b) CH4 PhCH3 PhC CH

(c)PhCH3 CF3CO2H NH3

(d) PhCH3 HCN HCO2H CH3OH

(e) NH3 PhNH2

(f) CH2=CH–OH HC C–OH

[Ans. (a) 1 < 3 < 4 < 5 < 2 (b) 4 < 3 < 2 < 1 (c) 3 < 2 < 4 < 1 (d) 3 < 2 < 4 < 1(e) 2 < 4 < 3 < 1 (f) 4 < 1 < 3 < 2 ]

Don't repeat what you have already done, unless you can do it in a better way

Page No.# 13

Q.1 Match the pKa values: [4]

Column I Column II

(A) H3O+ (P) – 9

(B) CH3CO2H (Q) – 1

(C) 4NH (R) 4.76

(D) H2SO4 (S) 9.4

[Ans. (A) Q; (B) R; (C) S ; (D) P ]

Assertion & Reason. (Q.2 & Q.3) [3 + 3](A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.2 Statement-1 :

O

is more stable than

S

[Ans. D]

Statement-2 : Overlapping between orbitals of C and O is better than orbitals of C and S

Q.3 Statement-1 :

OO

having more resonance energy than

O

O

Statement-2 : Extended conjugation is more effective the cross conjugation. [Ans. D]

Q.4 [3]

Correct order of basicity is(A) 3>1>2>4 (B) 3<1<2<4 (C) 3<4<1<2 (D*) 3>4>1>2

ASSIGNMENT-18

Page No.# 14

Q.5 Which one of following represents different molecules? [3]

(A) and (B*) and

(C) and (D) and

Q.6 Arrange the following carbocations in the increasing order of their stability. [3]

(I) (II) (III)

(A*) I > II > III (B) I > II = III (C) I > III > II (D) III > I > II

Q.7 The most stable resonating structure of following compound is [3]

(A) (B)

(C) (D*)

Q.8 [5]

(i) Rank the following in order of decreasing acidity.

[Ans. 1 > 4 > 3 > 2](ii) Consider the following compound. Rank the labeled proton in increasing order of acidity

(a) (b) Write the stepwise ionization of acid

[Ans. (a) Carboxy H is more acidic than hydroxy or thioxy hydroxy, between —OH and —SH, latter His more acidic thus the order of acid strength is: c < b < a

(b) x > y > z ]

Q.9 [5](a) NaF is dissolved in (CH3)2SO and CH3OH. In which solution F– will be more reactive and why?(b) Explain why iso octane is less viscous than n-octane.

Page No.# 15

Q.10 [6](a) Write correct order of acidity of a to e marked in structure of vitamin B6.

[Ans. b > e > d > a > c]

(b) Write correct order of acidic strength as a to d.

[Ans. a > c > b > d ]

(c) Write the correct order of pKa of marked position C–H bond in following compound.

[Ans. R > T > P > Q > S ]

Q.11 [6](a) Ordinarily the barrier to rotation about a carbon-carbon double bond is quite high but compound A

have a rotational barrier of only about 20 K cal / mole

What is the reason for this ?

(A)

(b) Which is more basic & why ?

A B [Ans. A < B]

(c) Among & which is more acidic & why? Explain through canonical

forms.You can't get different fruit, if you keep doing the same things.

Page No.# 16

ASSIGNMENT-19

Q.1 Match the names of carboxylic acids in column I with pka value in column II. [4]Column I Column II

(A) Benzoic acid (P) 4.17

(B) Ethanoic acid (Q) 4.14

(C) o-methyl benzoic acid (R) 4.74

(D) p-flourobenzoic acid (S) 3.91

[Ans. (A) P, (B) R, (C) S, (D) Q]

Q.2

O N2 NO2

NO2

I

Br

x y

z[3]

Correct order of bond length is(A) z > y > x (B*) x > y > z (C) y > x > z (D) x > z > y

Q.3 Ph N

2 is more stable than R N

2 because [3]

(A) in PhN

2 , +ve charge undergoing delocalisation in the ring.(B) Ph having –I effect whereas R having +I effect.(C) in Ph group carbon is more electron defficient w.r.t. carbon in R group.(D*) all are incorrect

Q.4 Arrange the following in the increasing order of their basic strength: [3]

I. II. III. IV.

(A) I < II < III < IV (B) IV < III < II < I (C) IV < I < II < III (D*) III < II < IV < I

Q.5 Energies of three resonating structures of a compound are E1, E2 & E3 respectively and energy of realmolecule is E0 . If E1 > E2 > E3 then the resonance energy will be: [3](A) (E + E2 + E3) – E0 (B) E1 – E0

(C*) E3 – E0 (D) 3

EEE 321

Page No.# 17

Q.6 Which order of acid strength is wrong [3]

(A) > (B) >

(C*) > (D) >

Q.7 In which of the following the 2nd structure having more resonance energy. [4]

(A*) vs

(B*) vs

(C*) CH2=CH–CH=CH– 2HC—

vs

(D*) vs

Q.8 [3+3](a) In which of the following pairs, indicated bond is of greater strength.

(i) ClCCH

||O

3 &

ClCHCH 23 , (ii) & (iii) &

(b) Arrange the following compound, in order of C–N bond length.

(A) (B) (C) (D)

[Ans. D > C > A > B ]

Page No.# 18

Q.9 Arrange in order of C–H bond energy [4]

[Ans. d < f < b < c < a < eOn the basis of stability of free radical formed after removal of H ]

Q.10 Which of the following pairs is stronger acid. Suggest a suitable explanation in each case. [4]

(i) and

(ii) and

Q.11 [8]

(a) , ,

Arrange these compounds in basic strength order:[Ans. I > II > III ]

(b) Following compound contains three C=C, arrange these in stability order:

[Ans. b > a > c ]

(c) (i) Write the correct order of case of deprotonation of labelled H-atoms.

[Ans. 1 > 2 > 3 > 4 ]

Page No.# 19

(ii)

[Ans. II > I > III ]

Opportunities are better recognised not when they are coming but when they are going.

Page No.# 20

Q. 1 Match the following : [4]Column I Column II

(Number of planes of symmetry)

(A)

Me

MeMe

(P) 6

(B) (Q) 0

(C) CCl4 (R) 7

(D)

CH3CH3

H C3 H

HH

(S) 1


Q.2 Assertion : Heat of hydrogenation of trans-2-butene is more than cis-2-butene. [3]Reason : Trans-2-butene is more stable cis-2-butene.

Q.3 Assertion : Cyclohexene shows geometrical isomerism.Reason : Its two sp2 H-atoms are cis to each other. [3]

Q.4 Which of the following is most stable carbocation. [3]

(A)

OEt

(B)

OEt

(C)

OEt

(D)

OEt

ASSIGNMENT-16

Page No.# 1


Cortisone

Q.5 Cortisone contains which functional groups? [3](A) Ether, alkene, alcohol (B) Alcohol, ketone, amine(C) Alcohol, ketone, alkene (D) Ether, amine, ketone

Q.6 Correct order of acidic strength in Cortisone [3](A) a > b > c > d (B) b > a > c > d (C) c > b > a > d (D) b > c > a > d

Q.7 Total number of chiral center in Cortisone : [3](A) 4 (B) 5 (C) 6 (D) 7

Q.8 Total stereoisomer of the compound Cortisone is [3](A) 32 (B) 64 (C) 66 (D) 128

Q.9 Arrange the following in increasing order of their heat of combustion: [5]

I. II. III.

Q.10 For the following compounds, arrange the labelled proton in increasing order of their case of deprotonation:

(a) CH2 H2

H1

C C–H 3

(b) (c) [5]

There isn't any problem which is without a gift for you in its hand.

Page No.# 2

Q.1 Four compounds are given in column I, match these with their 1apK values given in column II. [4]

Column I Column II

(Compound) ( 1apK )

(A) CH3COOH (P) 4.87

(B)COOH|COOH

(Q) 4.76

(C) ClCH2COOH (R) 2.86

(D) CH3CH2COOH (S) 1.2

Q.2 Statement-I : A tertiary carbocation is more stable than a secondary carbocation which is morestable than a primary carbocation.

Statement-II : The inductive effect operates through -bonds and decrease rapidly with increasein distance. [3]

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

Q.3 Statement-1: p-amino benzoic acid do not exists as zwittor ion [3]Statement-2: Acidic and basic groups in this compound are too far from each other to react, so they

can't form salt.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is false, statement-2 is true.(D)Statement-1 is true, statement-2 is false.

Q.4 Arrange the following in increasing order of acid strength. [3]

OH OH

COOH

CH –H2H—CH2

HO S3

(a)

(b)

(f)

(c) (d)

(e)

(A) c < f < d < a < b < e (B) f < d < c < b < e < a(C) f < c < d < b < e < a (D) c < d < f < a < b < e

CLASS : XIII (VXYZ) MARKS:42 DATE : 02-03/09/2009 TIME :40 MIN. DPP. NO.-17

Page No.# 3

Q.5 Select true statement(s) : [3](A) Resonance effects bond length(B) cis-1-bromo-1,2-difluoro ethene and Z-1-bromo-1,2-difluoro ethene are geometrical isomers

(C) In CH2

O

O

COH

COH most acidic H is connected directly to oxygen not on carbon.

(D) Boiling point of OH

is less than

OHOH

Q.6 Which of the following is the correct order of acidic nature. [3]

(I)

O

NH H

(II) N

H H

(III)

H HN

(A) III > I > II (B) I > II > III (C) II > I > III (D) III > II > I

Q.7 Which of the following is the correct order of basic nature. [3]

(P) NH

(Q) N

(R) NN

(A) Q > R > P (B) R > Q > P (C) R > P > Q (D) Q > P > R

Q.8 Correct increasing order of acidic strength of following compounds is / are ? [3]

(A)

O OHHN ,

OOHH

N ,

OOH

NH

(B) OH , OH , OH

(C) Br

ClOHF

, Br

Cl OHF

,

BrCl OH

F

(D) CH3–C C–H , H2O , HF

Page No.# 4

Q.9 Among the following compounds select those [6](i) Which will readily dimerise.(ii) Which are Quasi aromatic

(1) (2) 244HC (3)

(4) (5)

t–Bu t–Bu

t–Bu N(6) Azulene

Q.10 Resonance energy of is much lower than that of but resonance energies of and

are identical. Explain. [5]

Q.11 Arrange the following in increasing order of pKa values [6](a) HNO3 H2O HCO2H CH3CO2H PhOH(b) CH4 PhCH3 PhC CH

(c)PhCH3 CF3CO2H NH3

(d) PhCH3 HCN HCO2H CH3OH

(e) NH3 PhNH2

(f) CH2=CH–OH HC C–OH

Don't repeat what you have already done, unless you can do it in a better way

Page No.# 5

Q.1 Match the pKa values: [4]

Column I Column II

(A) H3O+ (P) – 9

(B) CH3CO2H (Q) – 1

(C) 4NH (R) 4.76

(D) H2SO4 (S) 9.4

Assertion & Reason. (Q.2 & Q.3) [3 + 3](A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.2 Statement-1 :

O

is more stable than

S

Statement-2 : Overlapping between orbitals of C and O is better than orbitals of C and S

Q.3 Statement-1 :

OO

having more resonance energy than

O

O

Statement-2 : Extended conjugation is more effective the cross conjugation.

Q.4 [3]

Correct order of basicity is(A) 3>1>2>4 (B) 3<1<2<4 (C) 3<4<1<2 (D) 3>4>1>2

ASSIGNMENT-18

Page No.# 6

Q.5 Which one of following represents different molecules? [3]

(A) and (B) and

(C) and (D) and

Q.6 Arrange the following carbocations in the increasing order of their stability. [3]

(I) (II) (III)

(A) I > II > III (B) I > II = III (C) I > III > II (D) III > I > II

Q.7 The most stable resonating structure of following compound is [3]

(A) (B)

(C) (D)

Q.8 [5]

(i) Rank the following in order of decreasing acidity.

(ii) Consider the following compound. Rank the labeled proton in increasing order of acidity

(a) (b) Write the stepwise ionization of acid

Q.9 [5](a) NaF is dissolved in (CH3)2SO and CH3OH. In which solution F– will be more reactive and why?(b) Explain why iso octane is less viscous than n-octane.

Page No.# 7

Q.10 [6](a) Write correct order of acidity of a to e marked in structure of vitamin B6.

(b) Write correct order of acidic strength as a to d.

(c) Write the correct order of pKa of marked position C–H bond in following compound.

Q.11 [6](a) Ordinarily the barrier to rotation about a carbon-carbon double bond is quite high but compound A

have a rotational barrier of only about 20 K cal / mole

What is the reason for this ?

(A)

(b) Which is more basic & why ?

A B

(c) Among & which is more acidic & why? Explain through canonical

forms.You can't get different fruit, if you keep doing the same things.

Page No.# 8

CLASS : XIII (VXYZ) MARKS:46 DATE : 07-08/08/2009 TIME :40 MIN. DPP. NO.-19Q.1 Match the names of carboxylic acids in column I with pka value in column II. [4]

Column I Column II

(A) Benzoic acid (P) 4.17

(B) Ethanoic acid (Q) 4.14

(C) o-methyl benzoic acid (R) 4.74

(D) p-flourobenzoic acid (S) 3.91

Q.2

O N2 NO2

NO2

I

Br

x y

z[3]

Correct order of bond length is(A) z > y > x (B) x > y > z (C) y > x > z (D) x > z > y

Q.3 Ph N

2 is more stable than R N

2 because [3]

(A) in PhN

2 , +ve charge undergoing delocalisation in the ring.(B) Ph having –I effect whereas R having +I effect.(C) in Ph group carbon is more electron defficient w.r.t. carbon in R group.(D) all are incorrect

Q.4 Arrange the following in the increasing order of their basic strength: [3]

I. II. III. IV.

(A) I < II < III < IV (B) IV < III < II < I (C) IV < I < II < III (D) III < II < IV < I

Q.5 Energies of three resonating structures of a compound are E1, E2 & E3 respectively and energy of realmolecule is E0 . If E1 > E2 > E3 then the resonance energy will be: [3](A) (E + E2 + E3) – E0 (B) E1 – E0

(C) E3 – E0 (D) 3

EEE 321

Page No.# 9

Q.6 Which order of acid strength is wrong [3]

(A) > (B) >

(C) > (D) >

Q.7 In which of the following the 2nd structure having more resonance energy. [4]

(A) vs

(B) vs

(C) CH2=CH–CH=CH– 2HC—

vs

(D) vs

Q.8 [3+3](a) In which of the following pairs, indicated bond is of greater strength.

(i) ClCCH

||O

3 &

ClCHCH 23 , (ii) & (iii) &

(b) Arrange the following compound, in order of C–N bond length.

(A) (B) (C) (D)

Page No.# 10

Q.9 Arrange in order of C–H bond energy [4]

Q.10 Which of the following pairs is stronger acid. Suggest a suitable explanation in each case. [4]

(i) and

(ii) and

Q.11 [8]

(a) , ,

Arrange these compounds in basic strength order:

(b) Following compound contains three C=C, arrange these in stability order:

(c) (i) Write the correct order of case of deprotonation of labelled H-atoms.

(ii)

Page No.# 11

Opportunities are better recognised not when they are coming but when they are going.


(A) Cl

ClCl

ClCl

Cl

(P) Compound can show optical isomerism

(B)

Cl

Cl

(Q) Compound can show geometrical isomerism

(R) Compound can show positional isomerism

(C) (S) Structure may have centre of symmetry in oneof its form.

[Ans. (A) P,Q,R,S (B) R,S(C) P,Q,R]Q.2 Assertion & Reason. [3]

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.Statement-1 : LDA is a very strong nucleophile but it is not a very good base.Statement-2 : Nucleophilicity is a kinetic property but basicity is thermodynamic property.

[Ans. D]Q.3 Select the incorrect statement. [3]

(A) Cyclopropane is more acidic than propane

(B*) Br

Br

FFis a meso compound with (2S,3S) configuration

(C) N-methylaniline is a non-resolvable compound

(D)

NO2

COOH

is more acidic than

NO2

COOH

ASSIGNMENT-20

Q.4 Give number of the products including stereoisomers formed in the following reaction. [3]

CH2

h

NBS

(A) 5 (B*) 6 (C) 7 (D) 4

Q.5 Percentage of hydrate [3]

+ H2O l

H|

OHCH|

OH

X

+ H2O l

3

3

CH|

OHCCH|

OH

Y

X & Y respectively is(A) X = 0.2 %, Y = 99.95 (B*) X = 99.9%, Y = 0.2%(C) X = 0.2%, Y = 0.2% (D) X = 99.9%, Y = 99.9%

Q.6 Which is the most likely site of protonation. [3]

(A) a (B) b (C) c (D) d

Q.7 322 HNCHCHOCMe||

O

NaOH Q [3]

Q is:

(A) 222 NHCHCHOCMe||

O

(B*) OHCHCHNHCMe||

O

22

(C) (D) MeCOONa + HOCH2CH2NH2

Q.8 Which form of given compound is more stable.

EtEt||

MeCHMeCH [3](a) meso(b) optically active

Q.9 o-nitrobenzoic acid (Pka = 2.21) is stronger acid than 3,5-dinitrobenzoic acid (Pka = 2.80) in waterwhereas in ethanolic solution the opposite is true (their respective Pka in ethanol are 8.82 and 8.09).Suggest a possible reason for this. [5]

Q.10 Give product(s) in each of the following reactions [5]

(a) CH3 –

3CH|CH – CH2 – CH2– CH3 hv/Br2 (A)

(b) + NBS O)COHC( 256 (B)

(c) CH3 – CH2 – CH = CH2 + Me3COCl hv (C) + (D)

(d) C6H5 – CH2 – CH2 – CH3

3

3

3

CH|

/ClOCCH|

CH

(E)

“Remember you are born to live not living because you are born.”

ASSIGNMENT-21

Q.1 Column I Column II [4](Molecular formula) (Theoritically possible isomers)

(A) C2H5N (P) Tautomers

(B) C3H6O (Q) Functional isomers

(C) C2H4O2 (R) Geometrical isomers

(S) Optical isomers[Ans. (A) P,Q,R (B) P,Q,R,S (C) P,Q,R,S ]

Q.2 Statement-1 : Yield of R – Br follows the order 1° < 2° < 3° in Hundsdicher reaction [3]Statement-2 : Hundsdicher reaction follows free redical mechanism.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.3 Which of the following is/are the chain termination step during photochlorination of ethane? [3]

(1)

ClCHCH 23 CH3–CH2–Cl

(2)

2323 CHCHCHCH CH3CH2CH2CH3

(3) ClCl Cl2

(4) HCl HCl

(A) 1 and 2 (B) 1 and 3 (C*) 1, 2 and 3 (D) 1, 2, 3 and 4

Q.4 Pick the correct statement for monochlorination of R-secbutyl chloride. [3]

C300

Cl2

(A) There are four possible products ; three are optically active one is optically inactive(B) There are five possible products ; three are optically inactive & two are optically active(C*) There are five possible products ; two are optically inactive & three are optically active(D) There are four possible products ; two are optically active & two are optically inactive

Q.5

OH

OHO

O

+ 2RCOONa l

ONa

ONaO

O

+ 2RCOOH [3]

above reaction is favoured is which direction.(A*) backward direction (B) forward direction(C) equal in both direction (D) can not be predicted without pKa

Q.6 Select stable molecule (s): [3]

(A*) (B*) CCl3CH(OH)2 (C) (D)

Q.7 When an aqueous solution containing MeCOOK and Me3C–COOK is electrolysed, possible product(s)is/are: [5](A*) Me – Me (B*) Me4C(C*) Me3C–COOCMe3 (D) CH2 = CH2

Q.8 CH3NO2 when dissolved in NaOH it forms salt. When the Na salt is acidified at low temperature thereis not always an immediate separation of oily drops (of nitro form). On standing however the acidifiedsolution slowly deposits oily drops. How you explain the time factor in this observation. [3]

Q.9 Hydrate of following compound are stable explain. [5]

(a) (b) HCCCl||

O

3 (c)

(d) 33 CFCCF||

O

(e) PhCCCPh||||||

OOO

Q.10 Identify unknown: [6]

(i)

,SOH)ii(

)Hg(Mg)i(

42

(ii) etherdry

Na

)CC(14*

(iii)

I

Cu

Q.11 Each of the following carbocations has the potential to rearrange to a more stable one. Write the structureof the rearranged carbocation. [8]

(a) CH3CH2CH2+ (b) 323 CHCHCH)CH(

(c) 333 CHCHC)CH(

(d) (CH3CH2)3CCH2+

(e) (f)

(g)

OCHCH|CHCHHC

23

222

(h)

“Every bad situation will have something positive. even a stopped clockshows correct time twice a day... so be positive in life.”

OZONOLYSIS : The reaction of alkene or Alkyne with ozone (O3) followed by hydrolysis is known asozonolysis.It is two types : (I) Reductive ozonolysis In presence of reducing agent

(II) Oxidative ozonolysis In presence of oxidising agentReducing agents : Zn, H2O or Zn, CH3COOH or (CH3)2S or (Ph)3P etc.Oxidising agents : H2O2 or

O||

HOOCR or Ag2O etc.

Example 1 :C=C C=OC C

R RRR’ R’O

O OR RRH H

O3 H3O/Zn

H2O2

step I–70°C

Reductiveozonolysis

Oxidativeozonolysis

+ R’–C–H

R–C–R R’–C–OH

O

O O

+

[SCT- Cut the double bondand paste two oxgen atomsand vice versa]

Mechanism :

C=OR’

H

R–C–R R’–C–H

O O

+

C

C

R

R’

R

H

+ +O

O

O

C

C

R R

H

H H

O

OO

OO

R

R’R C

CO

OR

R

R’O=C

R’

R—C—O

R

OC

O

Note : In case of oxidative ozonolysis aldehyde (not ketone) further undergoes oxidation whichgives acid as product.

Q.1 Give the product of the following reaction. [7 × 2 =14]

(i) H2C = CH2 OH/Zn)ii(

O)i(

2

3

[Sol. OH

H +

H

HO=C ]

ASSIGNMENT-28

(ii) CH3–CH = CH2 OH/Zn)ii(

O)i(

2

3

[Sol. HCCH||

O

3 + H

H C=O ]

(iii) 23

3

CHCCH|

CH

OH/Zn)ii(

O)i(

2

3

[Sol. CH3

CH3

O +

H

H C=O]

(iv) OH/Zn)ii(

O)i(

2

3

[Sol.

O O

H H ]

(v) OH/Zn)ii(

O)i(

2

3

[Sol.O O

H]

(vi)H

OH/Zn)ii(

O)i(

2

3

[Sol.O O

+

O O

H ]

(vii) H2C=CH–CH2–CH=CH–CH3 OH/Zn)ii(

O)i(

2

3

[Sol.H

H C=O +

HH||

OCCHCO 2 +

H|

CHCO 3 ]

Q.2 Find out the structure of reactant. [11 × 2 =22]

(i) X OH/Zn)ii(

O)i(

2

3

O||

HCCHCH 23

[Sol.H H

]

(ii) X OH/Zn)ii(

O)i(

2

3 O + OH

H

[Sol.H

H]

(iii) X OH/Zn)ii(

O)i(

2

3 O +

O

H

[Sol.

H

]

(iv) X OH/Zn)ii(

O)i(

2

3

O

H +

H

O

[Sol.

H

H ]

(v) X OH/Zn)ii(

O)i(

2

3 O

[Sol. ]

(vi) X OH/Zn)ii(

O)i(

2

3

O

O

+ CH3

H3CC=O +

H

H C=O

[Sol. ]

(vii) X OH/Zn)ii(

O)i(

2

3

O

O

[Sol. ]

(viii)1210HC

X OH/Zn)ii(

O)i(

2

3 OO

O + HCHO

[Sol. ]

(ix)1812HC

X OH/Zn)ii(

O)i(

2

3

O

+ O=C=O + OO

+ HCHO

[Sol.C=C—C=CH2

CH3

CH3

]

(x))X(

HC 46OH/Zn)ii(

O)i(

2

3 C3H2O3

[Sol. ]

(xi)1812HC

X OH/Zn)ii(

O)i(

2

3 O O

O

+ HCHO

[Sol. ]

Q.3 Give the ozonolysis product of the following. [3 × 2 =6]

(i) X Zn

)(O3 3 H

HCO

O

[Sol. ]

(ii) OH/Zn

O

2

3 ?

How many species.

[Sol. Only one

O

O

H

]

(iii) OH/Zn

/O

2

3

How many species are found.

[Sol.C

C

CH3

CH3

O

O +

C

C

CH3

H

O

O

+ C

CH

H O

O

]

When it rains all birds occupy shelter, but eagles, the only bird, avoids rain by flying above thecloud....Problem is same but ATTITUDE makes a DIFFERENCE.

OXYMERCURATION-DEMERCURATION (OMDM)OMDM is a hydration process of alkene according to Markawnikoff’s rule with no rearrangement ofcyclic mercuinium ion. In oxymercuration, the alkene is treated with mercuric acetate in aqueoustetrahydrofuran (THF). When reaction with that reagent is complete, sodium borohydride and hydroxideion are added to the reaction mixture.

R–CH = CH2 ¯HO,NaBH)ii(

THF,OH,)OAC(Hg)i(

4

22

OH|

CHCHR 3

(i)Hg(OAc) , MeOH, THF 2

(ii) NaBH , HO¯4

OMe|

CHCHR 3

AcO¯ = CH3COO¯

Mechanism for oxymercuration:

CH CH = CH + Hg — OAc 3 2 CH CH — CH3 2 CH CHCH3 2—Hg—OAc

CH CHCH3 2—Hg—OAc

H2O

OAcOAc

¯OAc

Hg+

OH

OH

+

H

+ AcOH

+ AcO¯

Sodium borohydride (NaBH4) converts the carbon-mercury bond into a carbon-hydrogen bond. Becausethe reaction results in the loss of mercury, it is called demercuration.

OH|CHCHCH 23 —Hg—OAc

¯HO

NaBH4

OH|CHCHCH 33 + Hg + AcO¯

Q.1 OH,NaBH)ii(

OHCH)OCOCH(Hg)i(

4

33 [Ans.OCH3

] [2]

Q.2 OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22 [Ans.

OH

] [2]

Q.3

CH3

¯HO,NaBH)ii(

OHCH,)COOCF(Hg)i(

4

323 [Ans.

H C3 OCH3

] [2]

ASSIGNMENT-29

Q.4 OH,NaBH)ii(

HOCH,)OAc(Hg)i(

4

32 [Ans.

OCH3

] [2]

Q.5OH

OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22 [Ans. O ] [2]

Q.6

C CH

OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22 [Ans.

COCH3

] [2]

Q.7 How could each of the following compounds be synthesized from an alkene by OMDM?

(i)

OH

(ii)

OCH CH2 3

(iii)

OH|

CHCCHCH|CH

323

3

(iv)

3

323

3

OCH|

CHCCHCH|CH

[Sol. (i) OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22

OH

(ii) OH,NaBH)ii(

OHCHCH,)OAc(Hg)i(

4

232

OCH CH2 3

(iii) OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22

OH|

CHCCHCH|CH

323

3

(iv) OH,NaBH)ii(

OHCH,)OAc(Hg)i(

4

32

3

323

3

OCH|

CHCCHCH|CH

]

HYDROBORATION-OXIDATIONHydroboration has been developed by brown as a reaction of tremendous synthetic utility because alkylboranes are able to undergo a variety of transformation. Hydroboration is a one step, four centre, cisaddition process in accordance with M. rule but after oxidation it seems to be appear to violate M.rule.

CH3CH = CH2 THFBH)i( 3

(CH3CH2CH2)3 B

OH¯ ,H O , H O2 2 2

CH COOH3

AgNO3

NH2Cl

Cl2

Hydroboration oxidation

Hydroboration reduction

Dimerisation

CH CH CH OH23 2

CH CH CH23 2

CH CH CH33 2

CH CH CH23 2 –NH2

CH CH CH23 2 –Cl

CH CH CH2 2 3

Mechanism of Hydroboration:

CH CH=CH3 2

H—BH2

CH3—C—C—H

H

H

H

BH2

+

—

More stable transition state

ealkylborananBH|

CHCHCH

2

223 2 (CH3CH2CH2)3 B

Mechanism of oxidation :HOOH + HO¯ l HOO¯ + H2O

R–B R—B—OR RO—B—OR

RO—B¯— OR

RO—BROH + RO—B

R—B—O—OH

R R OR

OR

OROR

OH

O—HO¯

R

R R

+ O—OH

repeat the twopreceding stepstwo times

repeat the twopreceding stepstwo times

+ OH¯ HO¯

RO¯ +3 ROH + BO33–

Q.1 CH2 = CH2 OH,OH¯,HO)ii(

BH)i(

222

3 [Ans.CH3CH2OH] [2]

Q.2 OH/OH

THF,BH

2

3 [Ans. HH

OHH] [2]

When God solves your problems, U have faith in his abilities. When he doesn’t solve ur problems,It means he has faith in ur abilities.

Q.3 OH,OH¯,HO)ii(

BH)i(

222

3 [Ans.

CH3

H

H

OH , CH3

H

H

OH

] [2]

Q.4 OH/OH

THF,BH

2

3 [Ans. CH3CH3

H

OH

,

CH3

CH3

HOH

] [2]

Q.5BD ,THF3

BH ,THF3

BT ,THF3

CH COOH3

CH COOD3

CH COOD3

[2]

[Ans. (i)

HD||CHCHCH 23 , (ii)

DH||CHCHCH 23 , (iii)

CH D2 CH D2

CH3 CH3

H HT T ]

Q.6

C CH

HO/OH)ii(

THF,BH)i(

22

3 [Ans.

CH2CHO

] [2]

ASSIGNMENT-30

Q.1 Match the column : [4]Column I Column II

(A)

H Me

42 CCl/Br (P) Optically active

(B)H Me

42 CCl/Br (Q) Diastereoisomers

(C)H

H

CH 3

OH /H (R) Optically inactive

(D)

Me

Me

42 CCl/Br (S) Meso

(T) Enantiomers

[Ans. (A) P,Q (B) R,T (C) R,T (D) R,S]

Q.2 Statement 1 : 1,1,1-trideutero-2-propanol reacts with conc. H2SO4 at high temperature to give onlyone alkene, 3,3,3-trideutero propene. [3]

becauseStatement 2 : C–D bond is stronger than C–H bond.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.3 Statement 1 : HCl in presence of peroxide will not undergo Antimarkonikov addition. [3]becauseStatement 2 : Attack of chlorine radical on alkene is endothermic and attack of radical formed on

HCl is exothermic.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.4

Me

Me

Et

Et

Peroxide

HBr Product. [3]

How many product will be formed in above reaction.(A) 2 (B*) 4 (C) 6 (D) 5

Q.5

42SOH.ConcA. [3]

Product A is:

(A) (B) (C) (D*)

Q.6 Hydrolysis of ether proceed via carbocation intermediate as shown below : [3]

H

stepgdetermininRate + CH3OH

Based on the above information, rationalize the relative rate of hydrolysis of compounds A to D below:

(I) (II) (III) (IV)

[Sol. I > II > III > IV]

Q.7 In the given reaction [5]

C7H12 (A) HBr

(A) can be

(A*) (B*)

(C*) (D)

Q.8 Which one of the following compounds gives acetone (CH3)2C=O as one of the products of its ozonolysis?[3]

(A) (B) (C) (D*)

Q.9 Explain mechanism of following reaction. [4]

HO 14

H

14

Q.10 The Corey-House Synthesis a way for coupling the alkyl groups of two alkyl halides to produce analkane: [9]

R – X + R’–X )X2(

steps

several

R – R’

R–X + 2 Li etherdiethyl

umAlkyllithi

RLi + LiX

umAlkyllithiRLi2 + CuI

ratedialkylcupLithium

CuLiR2 + LiI

ratedialkylcupLithium

CuLiR2 + halideAlkylX'R

AlkaneR'R + RCu + LiX

For the last step to give a good yield of the alkane, the alkyl halide R’–X must be either a methyl halide,a primary alkyl halide, or a secondary cycloalkyl halide.

(i) Complete the following reaction sequence:

CH3– I OEt

Li

2

(A) CuI (B)

ICHCHCHCHCH 22223 (C)

CH3CH2CH2CH2BrOEt

Li

2

(D) CuI

(E)

BrCHCHCHCHCH 22223 (F)

[Ans. (A) CH3Li, (B) (CH3)2CuLi, (C) CH3–CH2CH2CH2CH3, (D)CH3CH2CH2CH2Li,(E) (CH3CH2CH2CH2)2CuLi , (F) CH3CH2CH2CH2– CH2CH2CH2CH2CH3 ]

(ii) How will you prepare following using Corey-House Synthesis.

(a) Ph–CMe3 (b) (c)

[Sol. (a) CuLi + Me–Cl (b) CuLi + Me–Cl

(c) CuLi + Me–Cl ]

(iii) Identify the product of following reactions:

(a) 14

CuLi)CH(ClCHCCH||

CH

2323

2

(b)

[Sol. (a)

214

323

HC||

CHCHCCH (b)

OLi|

CHCHCHCCHCH|CH

323

3

]

U may get Delayed 2 reach ur Targets. , But Every step towards target is equal 2 victory.

ASSIGNMENT-31

Q.1 Match the column:

Column I Column II [4](Product)

(A)

Br

Mg, 35°CProductEt O2

CH3

(P) Racemic mixture

(B) Mg, 35°C

ProductEt O2

HH H

Br

CH3

CH3

(Q) Diastereomers

(C) Mg, 35°C

ProductEt O2H

Br

C H52

CH3CH3

CH3(R) Optically inactive product

(D) Mg(1 eq.)

Product35°C, Et O2

HH

H

CH – Br2

CH – Br2

CH3CH3

CH3

(S) Optically active product

(T) Most reactive G.R.[Ans. (A) Q,S (B) P,R (C) S,T (D) R]

Q.2 Statement-1 : Aldehyde is more reactive than ketone toward nucleophilic attack. [3]Statement-2 : Due to less +I effect and less steric hinderance in aldehyde it is more reactive than ketone.(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.3 Consider the following reactions [3]

M ICH3 L K273

LDA K298

NaH X ICH3 Y (LDA = lithiumdiisopropyl amide)

The products M and Y respectively are

(A) and (B) and

(C*) and (D) and

Q.4

O||

CHCCHCH 323 HCl

NaNO2 ? [3]

Major product of this reaction is:

(A)

NO|

CHCCHCH||

O

33 (B) OHNCHCCHCH||

O

23

(C*)

OHN||

CHCCCH||

O

33

(D)

OHN||

CHCCHCH 323

Q.5 Among the following compounds which liberate two equivalent of CO2 gas on oxidative ozonolysisfollowed by heating ? [3]

(A) (B*) (C) (D)

Q.6 What will be the product of hydration of 3-chloro-1-fluoro- propyne? [3]

(A) (B) (C*) (D)

Q.7 HCCH )excess(

HOCl A [3]

CH3–CCH )excess(

HOCl B

A & B are respectively(A) Cl2CH–CHO & CH3–CCl2–CHO (B) Cl2CH–CHO &

O||

CHClCCH 23

(C) Cl2CH–COOH & CH3CCl2COOH (D*) Cl2CHCOOH &

O||

CHClCCH 23

Q.8 H X [3]

X is:

(A*) (B)

(C) (D)

Q.9 CHMeCHCHCHCHCMe 222

654321 react with H2SO4 to give five structural isomer, C9H16, that react

with one eq. of H2/Pd. Give their structural formulas and account for their formation. [5]

[Ans. ]

Q.10 Write major product in each case. [7]

(a) + HBr A (b) 42SOH B

(c) + C (d) HBr D

(e)

COOEt|COOEt

H)ii(

)excess(MgBrCH)i( 3 E (f) 2HNO F

(g) H G

[Sol. (a) (b) , (c) , (f) ]

If the 1st button of a shirt is wrongly put, all the rest are surely crooked. So always be careful onyour 1st step. Rest will come correct....

ASSIGNMENT-32

Q.1 Match the column : [4]Column I Column II(Reactions) (Type of reaction)

(A) CH3 – CH = CH2 + HCl (P) Regioselective

(B)

CH3

CH=CH2

Cl H + H–Br (Q) Stereoselective

(C)CH3CH3

H HC=C + Br2

4CCl (R) Stereospecific

(D) + H2 OH,AcOH)ii(

HB)i(

2

62 (S) Anti Addition

(T) Syn Addition

[Ans. (A) P, (B) P,Q,S (C) Q,R,S (D) Q,T]

Q.2 Statement-1 : If during a reaction is generated, it quickly rearranges into

by hydride shift. [3]

Statement-2 : Hydride is better migrator than methyl during carbocation rearrangements.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.3 Statement-1 : Propan-2,2-diol is unstable. [3]Statement-2 : Repulsion between lone pairs of electron of two OH groups makes it unstable.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.4 Which of following on ozonolysis give 2-different product ? [3]

(A*) (B)

(C) (D) 23

3

CHCHCHCCH|

CH

Q.5 Which of following hydrocarbon can react with maleic anhydride [3]

(A) (B*) (C) (D) none of these

Comprehension (Q.6 & Q.8) [9]Consider the following figure representing variation of heat of combustion of isomeric butenes.

Based on this answer the following questions:Q.6 Minimum dipole moment is of isomers.

(A) P (B) Q (C) R (D*) S

Q.7 Which gives fastest reaction with HBr?(A) P (B*) Q (C) R (D) S

Q.8 Which will give product showing no optical activity, when treated with HBr.(A) P (B) Q (C) R (D*) all of these

Q.9 Propose mechanism of following reaction: [4]

H

Q.10 Provide a detailed, step-by-step mechanism for the following : [8]

(a) AA

(b) equivalent1

HCl B

(c) CH2=CH–CHD–CH=CH2 C

(d)

Me|

Me)I(C)OH(CPh 3AgNO2

D

[Sol. (a) (b) (d) Ph C (Me2) CO Me ]

Challenges are high, dreams are new, the world out there is waiting for you dare too dream daretoo try, no goal is too distant, no star too high !!

ASSIGNMENT-33

Q.1 Column I Column II [4](Reaction) (Type of intermediate formed)

(A) Ph–CHCl2 HO (A) (P) Carbocation

(B) R–Br etherdry

Na (Q) Carbanion

(C) 33 CHCCH||

O

,SOH)ii(

OH/Mg)i(

42

2 (R) Free – radical

(D) H (S) Carbene

[Ans. (A) Q,S, (B) Q, R, (C) P,R (D) P]

Q.2 Statement-1 : Propane on photochlorination gives 2-chloro propane as major product.Statement-2 : 2° carbocation is more stable than 1° carbocation. [3](A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.3 Statement-1 : Order of rate of decarboxylation in soda lime process of following compoundsis FCH2COOH > Cl – CH2COOH [3]

Statement-2 : F is a better –I group than Cl.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.4 Statement-1 : Me MgBr give CH4 with acetone. [3]Statement-2 : – H of carbonyl compounds are active.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.5

||O

(A)

H(B) [3]

Give structure of (B)

(A)

O

(B*) O (C) (D)

O

Q.6Me

3O A + B ; [3]

Total two type of Molozonides are formed identify relation between them?(A) Diastereoisomers (B) Geometrical isomers (C*) Enantiomer (D) Identical

Q.7 OMDM [3]

(A) OH (B) OH

(C) OH

(D*) OH

Q.8 Sentene ozonolysis [3]

Which is the correct structure of Sentene.

(A) (B) (C*) (D)

Q.9 The following equilibrium is driven to the right if the reaction is carried out in the presence of maleicanhydride. What is the function of maleic anhydride? [3]

+

Q.10(a) Prepare mechanism for following conversion. [7]

OH)ii(

MgBrCH)i(

2

3

(b) + HNO2 + H2SO4 OH2

(1,2-dione)Explain the mechanism 1,2-cyclohexadionefor this reaction

When God leads you to the edge of the cliff, trust him fully. Either he will catch you when youfall or he will teach you to fly.

OXIDATION OF ALKENES, ALCOHOLS & CARBONYL COMPOUNDS

(I) OXIDATION OF ALKENES

R–CH=CR2 R–CH—CR2

R–CH—CR2

OsO4

Cold dil.alkalineKMnO4

H2OOH

OH

OH

OH* Cold dil. alkaline KMnO4 is called as Bayer’s reagent.* Overall syn addition* Given by alkenes & alkynes* Benzene & Cyclopropane can not give this reaction.If we use acidic KMnO4 or warm KMnO4 or too concentrated KMnO4 the oxidative cleavage ofGlycol occurs resulting in mixture of Carboxylic acids & Ketones.

R–CH = CR2

4KMnO,H RCOOH + R2C = O

Hot acidic KMnO4, Hot acidic K2Cr2O7 & hot acidic NaIO4 gives same result with alkene. The effectis similar to that of oxidative ozonolysis on alkenes.

Preilschaive reaction :Epoxidation of alkenes is reaction of alkenes with peroxyacids.

CH2=CH2 + HOOCCH||

O

3 CH –CH2 2

O + OHCCH

||O

3

With the decrease in nucleophilicity of double bond, rate of reaction decreases. With the decrease in er withdrawing substituents in leaving group, rate decreases.

R RRR

CH2 CH2–OHCH2–OH

–H+

H O2 –CHCH2

CH HO–CHCHO O–H

dil.H SO2 4

H+

+ +

H O2

RCO H3

RCO H3O

OH

OH

OH

OH

HCO H3

H O3

R

CH2

CHO

R

CH2

O212

AgCH +

ASSIGNMENT-38

(II) OXIDATION OF ALCOHOLSOxidising agents

(1) Cu / 300°C (or Red hot Cu tube) (2) H /KMnO , 4 (Strong oxidising agent)(3) H /K Cr O , 2 2 7 (Strong oxidising agent) (4) PCC (Pyridinium chloro chromate)

HN CrO3 Cld or N CrO3 + HCl

(5) Collin’s reagent (6) Sarett reagent (i.e. PCC in CH2Cl2)

(N

(2 mol) + CrO3 + CH2Cl2)N

+ CrO3 + HCl + CH2Cl2

(7))dichromatePyridinium(

PDC

2

Cr O2 7

N

(8) Jones reagent (H2CrO4 in Anhydrous acetone)

or CrO3 + H2SO4 in acetone.Sufficiently mild so that it oxidises alcoholswithout oxidising or rearranging double bonds(8 or 9)

(9) TsCl + DMSO + NaHCO3 (10) MnO2-Oxidises only allylic or benzylic–OH.i.e.

RCH2OH ClTs RCH2OTss3NaHCO

DMSO RCHO 1° Allylic or benzylic OH 2MnO Aldehyde

R2CHOH ClTs R2CH–OTss3NaHCO

DMSO R2CO 2° Allylic or benzylic OH 2MnO Ketone

R3COH ClTs R3C–OTss3NaHCO

DMSO × No effect on 3° ROH and on Carbon-carbon

multiple bond.(11) Periodic cleavage (12) NBS

A similar oxidation is obtained incase of HIO4known as periodic cleavage. (13) Openaur oxidation

R–CH–OH

R– CH — O – I = O

R–CH

R C–OH2

R C–O–H2

+ +

+

HO–I=O

O

O

O

O

O

HIO3

R C=O2

R – CH – R

(R CH–O) 2 2

R CH–OH + 2

R – C – R

CH –C – CH3 3OH

O

O

Al(OCMe )3 3

Al(OCMe )3 3

Me COH + Al(OCHR )3 2 3

Al–O

3R C=O + (Me C–O)2 2 3

3Me C2

O CR2H

AlBut reaction is only observed for Vic-diols. Oxidation of alcohol with aluminium tertiary

butoxide is Openaur oxidation.OH

Al (OCMe )3 3

O

acetone

Different oxidising agents are used to oxidise alcohols in corresponding carbonyl compounds and carboxylic acids.

e.g. (I)alcohol1

OHCHR 2

agent

oxidisingmild

O||

HCR (Aldehyde) eg. 1,4,5,6,7,8,9,12

(II)alcohol2

'RCHR|

OH

agent

oxidisingmild 'RCR||

O

(Ketone) eg.1,2,3,4,5,6,7,8,9,12,13

(III)alcohol1

OHCHR 2

agent

oxidisingstrong

O||

OHCR eg. 2,3

(IV)

3

3

3

CH|

OHCCH|

CH

C300Cu

CH3

CH2

CH3

C Dehydration takes place.

(V) Double bond or Tripple bond is not affected by 1,4,5,6,7,8,9,10(VI) No effect on 3° alcohol by 2,3,4,5,6,7,8,9,10,12,13

(III) OXIDATION OF CARBONYL COMPOUNDS.

1. RCHO + [Ag(NH3)2]OH RCOOH + NH3 + Ag Aldehyde acts as reducing agent, they can reduce mild oxidizing agents like Tollen’s Reagent. Tollen’stest Gentle Heating for 20 to 25 mins.

2. Fehling’s SolutionsFehling’s A Fehling’s B

COONa

COOK

OHOHH

Haq. CuSO4 Alk. solution of Roschelyesalt (sodium potassium tartrate)It act’s a carrier for Cu2+ as it make reversible complex with Cu2+

This test is also used is Blood and Urine test.

RCHO + Cu2+ OH2 ).pptred(OCuRCOO

CuRCOOH

2–

3. Benedict’s solutionSodium Citrate + NaOH + NaHCO3 + CuSO4

RCHO + Cu2+ OH2 ).pptred(OCuRCOO

CuRCOOH

2–

4. RCHO + HgCl2 + H2O RCOOH + 2HCl + Hg2Cl2

RCHO + Hg2Cl2 + H2O RCOOH + 2HCl + 2Hg greyish solution

5. Schiff’s ReagentSchiff’s Reagent is aq. solution of following base decolourised by passing SO2.Aldehyde restore pink colour of Schiff’s reagent.

NH2

C

NH2

SO2 RCHO

Colourless solution (Schiff’s Reagent)

p-Rosaniline HydrochlorideMagenta colour (Fuschin)

RCOOH + Pink colour

NH2Cl+

Ketons are not easy to oxidize so they do not give these 5 tests. These five tests can be used to distinguishaldehyde and ketones. Both gives 2,4 DNP test

KETONES ARE DIFFICULT TO OXIDIZEKetones can be oxidized from their enolic form at high temperature with very strong oxidizing agent.Oxidation of ketones is sometimes governed by Popoff’s rule. According to this rule carbonyl groupremains with the smaller alkyl group. More electron rich alkene will be easy to oxidized.

Me – C – Me

O

MeCHO + CO + H2 2O[O]

Allylic oxidationSeO2 is a selective oxidizing agent with converts –CH2– group agjacent to carbonyl group into carbonylgroup.The reagent, in general, oxidises active methylene and methyl groups to ketonic and aldehydicgroups respectively.

CCH||

O

2 2SeO

CC||||

OO

; 3CHC||

O

2SeO CHOC||

O

Double bonds, triple bonds and aromatic rings may also activate the methylene group.The methylene ormethyl group to the most highly substitued end of the double bond is hydroxylated according to theorder of preference of oxidation CH2 > CH3 > CH groups.

CH3= CH–CH3 2SeO 22 CHCHCH|

OH

2° C – H > 1° C – H > 3° C – HRate of reactivity orderCH3 – CH = CH – CH2–CH3 2SeO

OH|

CHCHCHCHCH 323

CH3 HO–CH2

CH – CH3CH – CH3

SeO2

CH3CH3

SeO2

HO

Q.1 (i) CH2 = CH2

4KMNO/H (ii) CH3–CH= CH2

4KMNO/H [11]

(iii)

4KMNO/H (iv)

4KMNO/H

(v)

4KMNO/H (vi)

4KMNO/H

(vii)

4KMNO/H (viii)

4KMNO/H

(ix)

4KMNO/H (x)

4KMNO/H

(xi)

4KMNO/H

Q.2 A to F alkenes with minimum possible carbon. [7]

(i) A

4KMNO/H MeCOOH as the only product (ii) B

4KMNO/H O

(iii) C

4KMNO/H MeCH2COOH as the only organic (iv) D

4KMNO/H

O

O

(v) E

4KMNO/H

O||

COOHCCCCCCCCHOOC||

O

(vi) C10H10

4KMNO/H

COOHC|

COOHCCCHOOC

(vii) F

4KMNO/H acetone + ethanoic acid

Q.3 (i)

mCPBA\hydrolysis

1% alkalineKMnO4

(ii) mCPBA

[6]

(iii) mCPBA

Mehydrolysis (iv)

mCPBAhydrolysis

Me Me

H HC = C

(v) Me

MeH

H mCPBAhydrolysis

C = C (vi) Me

PhH

HC = C

Ag2O or 2Ag + O2

12

Q.4 (i) CH3– CH2 – CH2 – OH

,H/KMnOor

¯,OH/KMnO

4

4 ? [2]

K Cr O / H , 2 2 7

or conc. HNO3, ?

(ii) 323 CHCHCHCH|OH

H,OCrKor

H,KMnO

722

4 ? [1]

(iii)

HO

HOOH

OH(1)

(10)

(2) or (3)

(4) or (5) or (6)

??

?

?

or (7) or (8) or (9)[4]

Q.5 (i) CH2 = CH – (CH2)3 – CH2 – OH PCC [9]

(ii) C6H5 – CH = CH – CH2 –OH PCC

(iii) OHCHCHCHCHCH|OH

2223 PCC (A)

)ii(

NaOHDil)i( (B)

(iv)

CH OH2

OHPCC (v) CH2 = CH – CH2–OH 2MnO ?

(vi) CH O3

CH O3

CH–CH –CH –OH2 2

OH

Acetone

MnO2 ?

(vii) OHCHCCHCCH|

CH

2

3

Acetone

MnO2 ? (viii) 356 CHCHHC|OH

4

2

CCl

MnO ?

(ix)

3

2256

CH|

OHCHCHCHCHCHHC 3NaHCOTsClDMSO ?

Q.6 (i) HO Acetone

butoxidetertAluminium ? [3]

(ii) 322 CHCHCHCHCH|

OH

butoxidetertAluminium

(iii)

CH OH2

nebenzoquinop

butoxidetertAluminium

?

Q.7 Which one of the following alcohols are oxidised by MnO2? [3]

(A) C6H5 – CH2 – CH2–OH (B) 322 CHCHCHCHCH|

OH

(C*) 33 CHCHCHCHCH|

OH

(D) CH3–CH2 – CH2 –OH

Q.8 [9]

(i)

OH|

OHCHCHMe 2 4HIO (ii)

OHOH||

Et—CH—CMe2 4HIO

(iii)OH

OH 4HIO (iv)

OH|

OHCHCHCHCHHO 222 4HIO

(v)

OHOH||

CHCHCH—CH 322 4HIO (vi)

OHOHOH|||

CHCHCH—CH 32 4HIO

(vii)

OHOHOHOH||||

CHCHCH—CH 22 4HIO (viii)

OHO|||

MeCHCMe 4HIO

(ix)

OO||||

MeCCMe 4HIO

Q.9 Which will give the Tollen test. [3]

(i) O OH

H (ii) O OMe

(iii) 2CHCR|||

OHO

(iv) HO

HO

Q.10 (a) C – C – C –C

O[O] [Ans. 2C–COOH

O[O] ] [4]

(b) Me CH–C–Me2

O[O]

[Ans.Me CH–C–Me Me CO + MeCOOH2

O]

(c) Me CH–C–Me3 Me C–COOH + CO + H O

O[O] [Ans.Me CH–C–Me Me C–COOH + CO + H O3 2 2

O

]

(d)O [O]

[Ans. O O

O

[O]]

Q.11 (a) CH –CHO3

SeO2

[Ans.CH –CHO CH – CHO

O

SeO ] [10]

(b) Me CO2

SeO2

[Ans. Me – C – C – H

O

SeO

O

]

(c) C – C – C – CSeO2C – C – C –C

O

P1 HCPBA P2 LAH P3 [Ans. C – C – C – C

O O

O

]

(d)O SeO2

[Ans. O O

O

SeO]

(e) SeO2

HO

(f) CH – CH = CH 3 2

H – C – CH = CH 2CH – CH = CH 2 2

Acrolein?

SeO2

MnO2

OOH

(g)

SeO2

MnO2

CH3 C

CH2

2 step(1) step etard

H

OH

O

(h) CH –C – H3 CH – C – C – H3

O OSeO2

conc. NaOH H /+ P1 P2

O

Q.12 How will you differentiate HCHO and PhCHO ? [Ans. F and B test] [2]

Q.13 How will you differentiate HCHO and MeCHO ? [Ans. Iodoform test] [1]

“Success in not permanent & failure is not final.” So, never stop working after success & neverstop trying after failure.

Give major products of following reactions:

Q.1

O||

ROHClCMe Py Q.2 MeCOCl + EtOK

Q.3 MeCOCl + NH3 Q.4 Ph – NH2 + MeCOCl

Q.5 Ph – NH2 + PhCOCl Q.6 PhOH + PhCOCl

Q.7 RNH2 + PhSO2Cl Q.8 R2NH + PhSO2Cl

Q.9 ROH + KCN Q.10 ROH + KCN

Q.11 EtCOCl + EtCOONa Q.12 EtCOOH + EtCOOH

42SOH.Conc

Q.13 Q.14

Q.15

OO||||

MeCOCMe OH3

Q.16 NaOH

Q.17

O||

EtONaOMeCMe Q.18 + MeNH2

Q.19

O||

NHMeNHCPh 22 Q.20

Q.21 Q.22

O||

NHCMe NaOH2

Q.23 3NH

History proves that winners win because they refuse to be discouraged by temporary defeats!

ASSIGNMENT-39

Q.1 Write only the major product in each case:

(a) + 3AlCl AA (b) + 3AlCl B

(c) + 3AlCl C (d) + 3AlCl D

(e) + 3AlCl E (f) + CH2Cl2 3AlCl F

(g) + PhCH2Cl 3AlCl G (h) + CH2Cl2 3AlCl H

(i) 3AlCl I (j) +

O||

ClCMe 3AlCl J

(k) +

OH|||

ClCCMe|

Me

3AlCl K (l) +

OMe|||

ClC—CMe|

Me

3AlCl L

(m) 3AlCl M (n) +

OO||||

MeCOCMe 3AlCl N

(o) + 3AlCl O (p) + 3AlCl P

(q) + HNO2 H Q (r) + C – C – C – OH 42SOH.conc

(s) + C – C = C 42SOH.conc (t) + PhCH2CH2CH2CH2OH 42SOH.conc

Oxidation of aromatic compounds.

(a) (b) C500

OV 52

ASSIGNMENT-40

(c) C500

OV 52

(d)

C500

OV 52

(e)

722 OCrKH (f)

722 OCrKH

(g)

722 OCrKH (h)

722 OCrKH

(i)

4KMnOH (j) Ph n2 )CH( CH3 , PhCHMe2, Ph–CH2OH, Ph–CH2Br,,

PhCHCl2, PhCHO, Ph & all the compounds having atleast one H give PhCOOH.

(k) Ph–CMe3

4KMnOH Me3C–COOH (l)

4KMnOH terephthalic acid

(m)

4KMnOH isophthalic acid

(n)

4KMnOH

OH2

(m)

4KMnOH(n)

4KMnO

H

(o) 4KMnO

H

Etard oxidation: Alb's persulphate oxidation:

AcOAc

ClCrO 22 822 OSK

Our attitude towards life determines life’s attitude towards us.

INORGANIC CHEMISTRY ASSIGNMENTS

ASSIGNMENT-2

Q.1 What is the ratio of average formal charge of oxygen and Bond order of in CO32– . [3]

(A*) –0.5 (B) +0.5 (C) 1.0 (D) None

Q.2 Which option is correct among the following [3](A*) AlCl3(aq) is not hypovalent where as AlCl3 anhydrous is hypovalent(B) AlCl3(aq) is hypovalent where as AlCl3 anhydrous is hyper valent(C) AlCl3(aq) and AlCl3 anhydrous both are hypovalent(D) None

Q.3 Compound ‘X’ has eight atoms which has only four ‘’ bond no -bond, no ionic bond, no-coordinatebond, no H-bond. [3](A) C2H6 (B*) B2H6 (C) C2(CN)6 (D) All the above

Q.4 Although the electronegativity of ‘C’ is more in H – C C – H but it cannot form H-bond. [3](A) ‘C’ is linear (B) ‘C’ is sp-hybridised(C*) ‘C’ does not have lone pair e–s (D) None

Q.5 Statement-1 : S2F2 exist-in two forms where as S2Cl2 exist only is one form. [3]Statement-2 : Due to steric repulsion between two Cl and also in between S and S their is double

bond.(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.6 Calculate the percentage ionic character for flouride molecule where electronegativity of electropositiveelement is 2.1. Write your answer in integer –1. [3]

[Ans. 6]

Q.7 Which compound has 14 bond, 2 bond four coordinate bond and four hydrogen bond as well asionic bond : [3](A*) Blue vitriol (B) Glauber salt (C) Candy fluid (D) Norwagian salt petere

Q.8 Identify the element of third period whose successive I.E. are given below : [3]I.E1 I.E2 I.E3

X 513 4562 6920Y 738 1451 7733Z 1521 2666 3931

Q.9 Which one has large 4th I.E. [3](A*) Ga (B) Ge (C) Both (D) None

Q.10 Element A has 3e– in the valence shell and its principal quantum number for last e– is three and elementB has 4e– in the valance shell and its principal Q.N. for last e– is 2 what is the number of atom A andnumber of atom B in compound of AB and what is this compound. [3]

[Ans. 43]

Q.11 The X – X bond length is 1.0 Å and C – C bond length is 1.54 Å if electronegatives of X and ‘C’ are 3.0and 2.0 respectively. C – X bond length is likely to be. [3]

Q.1 Find the species / molecule is having maximum number of lone pair on the central atom. [3](A) ClOF4¯ (B) ClOF2

+ (C) BH4¯ (D*) XeOF2

Q.2 If y-axis is the approaching axis between two atoms, then which of the set of orbitals can not form the bond between two atoms in general. [3](A) pz – pz (B) px– px (C*) px – py (D) None of these

Q.3 The maximum number of bond and -bond can be formed between two atoms are respectively.[3](A) 4, 3 (B*) 3, 2 (C) 2, 3 (D) 3, 1

Q.4 Which of the following set of overlap can not provide -bond formation. [3](A) 3d and 2p (B) 2p and 3p (C) 2p and 2p (D*) 3p and 1s

Q.5 The ratio of number of -bond to -bond in N2 and CO molecules are [3]

(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2

Q.6 Which of the following overlapping is involved in formation of only -bond. [5](A*) s – p overlapping (B) p – d overlapping(C*) s – s overlapping (D) p – p overlapping

Paragraph for question nos. 7 to 9Different types of bonds are formed in the chemical compounds. These bond have different strength andbond energies associated with them. These bonds are formed with atoms in different environments.

Q.7 Which of the following bond has highest bond energy? [9](A*) -bond (B) -bond (C) Hydrogen bond (D) Metallic bond

Q.8 Shape of the molecule is decided by(A*) -bond (B) -bond(C) both and -bond (D) Never Nor -bond

Q.9 Which of the following hydrides is thermally least stable?(A) H2O (B*) H2Te (C) H2S (D) H2Se

Q.10 Match the column : [4]Column I Column II(Orders) (Sequences)

(A) Increasing order of ionic (P) Si, Mg, Al, Namobility in water

(B) Decreasing order of size (Q) Li+, Na+, K+, Rb+, Cs+

(C) Decreasing order of ionisation (R) O, O+, O2–, O¯energy

(S) Se, S, O, F[Ans. (A) Q, (B) S, (C) P]

ASSIGNMENT-4

ASSIGNMENT-5

Q.1 Explain type of hybridisation, shape, bond angle and geometry of the following compounds.[18](1) XeF2

(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2

Q.2 What is hybridisation of central atom of anionic part of PBr5 in crystalline state. [3](A) sp2 (B) sp3 (C) sp (D*) not applicable

Q.3 What is the difference between bond angles in cationic species of PCl5 and PBr5 in solid state. [3](A) 60° (B) 109°28 (C*) 0° (D) 90°

Q.4 All possible bond angles in anionic part of PCl5 are. [3](A) 109° 28 only (B*) 90°, 180° (C) 90°, 120°, 180° (D) 72°, 90°, 180°

Q.5 The hybridisation and shape of XeO3F2 molecule is [3](A) sp3 and tetrahedral (B) sp3d and see-saw(C*) sp3d and TBP (D) sp3d2 and octahedral

Q.6 TeF5¯, XeF2 , I3+ , XeF4 , PCl3

Which of the following shape does not describe to any of the above species ? [3](A) Square pyramidal (B) Square planar(C*) Trigonal planar (D) Linear

Q.7 Which of the following species does not exist? [3]

(A*) 3XeF (B) XeF4 (C)

5XeF (D) XeF6

Q.8 Which of following pair of species is having different hybridisation but same shape. [3]

(A) BeCl2 and CO2 (B) CO2 and SO2 (C*) SO2 and 3I (D*)

2ICl and BeH2

Q.9 Statement-1 : CH4 and CH2F2 are having regular tetrahedron geometry. [3]Statement-2 : Both are having same hybridization.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Paragraph for question nos. 10 to 12Hybridisation is the mixing of atomic orbital of comparable energy and the number of hybrid orbitalsformed is equal to the number of pure atomic orbitals mixed up and hybrid orbitals are occupied by -bond pair and lone pair. [9]

Q.10 Which of the following geometry is most likely to not form from sp3d hybridisation of the central atom.(A) Linear (B*) Tetrahedral (C) T-Shaped (D) See-Saw

Q.11 The orbital is not participated in sp3d2 hybridisation.

(A) px (B*) dxy (C) 22 yxd

(D) pz

Q.12 "The hybrid orbitals are at angle of X° to one another" this statement is not valid for which of thefollowing hybridisation.(A) sp3 (B) sp2 (C*) sp3d2 (D) sp

ASSIGNMENT-6

One or more than one correct:Q.1 Select the correct statement for non polar molecule. [3]

(A*) On time average the molecule is non-polar but at the particular moment it act as a dipole which isequally probale in all directions.(B) On time average the molecule is polar but at the particular moment it does not act as a dipole.(C) On time average the molecule is non-polar and the particular moment it must not act as dipole.(D) All are incorrect

Q.2 Select the correct order of B.P. [3](A) BF3 > BMe3 (B*) BF3 < BMe3 (C) BF3 = BMe3 (D) None of these

Q.3 Which molecular geometry are most likely to result, from a octahedral electron geometry? [3](A*) square planar (B*) square pyramidal (C) linear (D) V–shaped

Q.4 The correct order of hybridization of the central atom in the following species [3]NH3, XeO2F2, SeF4, NO2

+

(A) sp3, sp3, sp3d, sp (B*) sp3, sp3d, sp3d, sp(C) sp3, sp3d2, sp3d, sp2 (D) sp2, sp3d, sp3d2, sp

Q.5 Halogens form compounds among themselves with the formula XX', XX & '7XX (where X is the heavier

halogen) which of the following pair(s) represent(s) correct geometry with polar and non-polar nature(theoretically) [3](A*) XX' — Linear — Polar (B) XX — Linear — Polar(C) XX' — Linear — Non-polar (D*) '

7XX — Pentagonal bipyramidal — Non-polar

Q.6 In molecules of the type AX2Ln (where L represents lone pairs and n is its number) there exists a bondbetween element A and X. The X A X bond angle. [3](A) Always decreases if n increases (B) Always increases if n increases(C*) Will be maximum for n = 3, 0 (D) Generally decreases if n decreases

Q.7 Which of the following pairs of species have identical shapes? [3](A)

2NO and 2NO (B) PCl5 and BrF5 (C*) XeF4 and

4ICl (D) TeCl4 and XeO4

Q.8 London force works in [3](A) Polar molecule (B) Non-polar molecule(C*) All polar and non-polar molecule (D) Ionic compounds

Q.9 Choose the correct on the Cl–O bond length in NaClO4. [3](A) All Cl–O bonds are of equal length.(B) Three Cl–O bonds are of equal of length one longer.(C) Two Cl–O bonds are of same length which are longer compound to other two Cl–O bond length.(D) All are different.

Q.10 Statement-1 : Experimentally 100 % covalent bond formation is not possible [3]Statement-2 : Non polar molecule has instantaneous dipole – induced dipole interaction(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.11 Spin pairing and overlapping theory can not explain the equal bond length as well as equal bond anglesin CH4 molecule. To explain the above facts we are in need of hybridization theory. Hybridization is themixing of atomic orbitals of comparable energy and the number of atomic orbitals involved is equal to thenumber of hybrid orbitals formed of equal energy. [3 × 3 = 9](i) According to hybridization theory, the %s character in sp3d hybrid orbitals is

(A) 25% (B) 33.33% (C*) 20% (D) 16.66%

(ii) The number of axial and equitorial positions in octahedral geometry having sp3d2 hybridization(A) 2, 4 (B) 4, 2 (C) 3, 3 (D*) 0, 0

(iii) Increasing order of the energy of hybrid orbitals is(A) sp3 < sp2 < sp (B) sp2 < sp3 < sp (C) sp < sp3 < sp2 (D*) sp < sp2 < sp3


(i) 4IOF (P) See-saw

(ii) –22FIO (Q) Trigonal bipyramidal

(iii) –46XeO (R) Linear

(iv) XeF2 (S) Square bipyramidal

(A) (i) – P, (ii) – S, (iii) – R, (iv) – Q (B*) (i) – Q, (ii) – P, (iii) – S, (iv) – R(C) (i) – Q, (ii) – P, (iii) – S, (iv) – R (D) (i) – Q, (ii) – P, (iii) – R, (iv) – S

Q.1 The correct order of d C–H in the following option is [3](A) CHF3 = CH2F2 = CH3F (B) CHF3 > CH2F2 > CH3F(C) CH2F2 > CH3F > CHF3 (D*) CH3F > CH2F2 > CHF3

Q.2 O2F2 is an unstable yellow orange solid and H2O2 is a colourless liquid, both have O–O bond.O–O bond length in H2O2 & O2F2 is respectively. [3](A) 1.22Å, 1.48Å (B*) 1.48Å, 1.22Å (C) 1.22Å, 1.22Å (D) 1.48Å, 1.48Å

Q.3 The structure of O3 and 3N are [3]

(A) both linear (B) Linear and bent respectively.(C) both bent (D*) Bent and linear respectively.

Q.4 Select the correct order of following property. [3](A) % s-character : sp3 > sp2 > sp(B) ONO^ bond angle : —

3NO > 2ON

> NO2(C*) Second ionisation energy : O > F > N > C(D) C – F bond length : CF4 > CH3F > CH2F2 > CF3H

Q.5 Which of the following compounds is/are polar as well as non planar? [3]

(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl

Q.6 The strongest P–O bond is found in the molecule [3](A*) F3PO (B) Cl3PO (C) Br3PO (D) (CH3)3PO

Q.7 F-As-F bond angle in AsF3Cl2 can be [3](A*) 90° & 180° only (B) 120° only (C) 90° & 120° only (D) 90° only

Q.8 The number of S–S bonds in sulphur trioxide trimer (S3O9) is [3](A) three (B) two (C) one (D*) zero

Q.9 Select the incorrect statement(s) about N2F4 and N2H4. [3](A*) In N2F4 , d-orbitals are contracted by electronegative fluorine atoms, but d-orbital contraction isnot possible by H-atom in N2H4.(B) The N–N bond energy in N2F4 is more than N–N bond energy in N2H4.(C*) The N–N bond length in N2F4 is more than that of in N2H4.(D) The N–N bond length in N2F4 is less than that of in N2H4.

Q.10 Nodal planes of bond(s) in CH2=C=C=CH2 are located in [3](A) All are in molecular plane(B*) Two in molecular plane and one in a plane perpendicular to molecular plane which contains C–C

-bond(C) One in molecular plane and two in plane perpendicular to molecular plane which contains C–C

-bond(D) Two in molecular plane and one in a plane perpendicular to molecular plane which bisects C–C

-bond at right angleQ.11 Which of the following shape can not be obtained from sp3d2 hybridisation. [3]

(A) Square planar (B) Square pyramidal (C*) Tetrahedral (D) OctahedralQ.12 Which of the following statement(s) is/are correct about P4O6 and P4O10 [3]

(A*) Both oxides have closed cage like structure(B*) Each oxide contains six equivalent P–O–P bonds(C) Both P4O6 and P4O10 molecules have p–d bonds.(D*) Both are the anhydrides of their respective acids.

ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7


+ (C) BH4¯ (D*) XeOF2





(A) 2.0 , 2.0 (B) 2, 21

(C*) 21

, 21

(D) 21

, 2











ASSIGNMENT-4

ASSIGNMENT-5


(2) XeF4

(3) XeF5¯

(4) XeOF4

(5) PCl3

(6) PCl5

(7) SF2

(8) SF6

(9) IF3

(10) IF5

(11) IF7

(12) OF2

(13) 3NO

(14) 4ClO

(15) SF4

(16) 3I

(17) 3ClO

(18) OCl2









5XeF (D) XeF6



2ICl and BeH2






(D) pz


ASSIGNMENT-6






+














(A) 25% (B) 33.33% (C*) 20% (D) 16.66%














3NO > 2ON



(A*) S2Cl2 (B) B2H6 (C*) PCl2F3 (D) H

ClC=C=C=C

H

Cl











ASSIGNMENT-7

ASSIGNMENT10

Q.1 Column-I Column-II [12]

(A) BH4¯ (P) Central atom(s) of species/molecule is having sp3 hybridisation(B) I2Cl6 (Q) The species / molecule is having no lone pair(C) AlCl4¯ (R) All bond angles are identical(D) BeCl2 (solid) (S) Species / molecule having co-ordinate bond.

[Ans. (A) P, Q, R, S, (B)S, (C) P, R, S, (D) P, S ]

Q.2 Choose the correct statements. [4](A) CH3NCS molecule is linear (B*) SiH3NCS molecule is linear(C*) GeH3NCS molecule is bent (D*) P(SiH3)3 molecule is pyramidal

Q.3 Statement-1 : SO42– and S5

2– have different structures. [3]Statement-2 : O and S are of same group but of different period element.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.4 Only one match is to be found out. [9]Column I Column II

(A) Dimer of BeCl2 (P) All atoms are sp3 hybridised.(B) Dimer of AlBr3 (Q) Non-planar and see-saw in shape

(C) 5XeF (R) Central atoms are electron deficient

(S) Bond angles are of 72°[Ans. (A) R, (B) P, (C) S]

Q.5 In which of the following silicate structure, the number of corner shared is minimum. [3](A*) Pyrosilicate (B) Amphibole chain silicate(C) 3D-silicate (D) Six membered cyclic silicate

Paragraph for Question Nos. 6 to 8Bridge bonding is a specific kind of bonding in pages of chemistry. In general -bond pair delocalisation isvery difficult. But electron deficiency of the central atom forces to delocalise and forms this kind of bond.

Q.6 The B2H6 molecule is dissolved in tetrahydrofuran. Which atom(s) is/are having changes of hybridisationwith respect to reactant and final product of the process given. [3](A) B only (B) B and O (C) B, O and C (D*) None of these

Q.7 In which of the dimerisation process, the achievement of the octet is not the driving force. [3](A) 2AlCl3 Al2Cl6 (B) BeCl2 BeCl2 (solid)(C*) 2ICl3 I2Cl6 (D) 2NO2 N2O4

Q.8 The molecule is not having 3c–2e bond. [3](A) BeH2 (dimer) (B) BeH2 (solid) (C*) C2H6 (D) B2H6

Q.9b Only one match is to be found out. [12]Column I Column II

(A) ICl3 (P) Hybridisation of central atom is similar in bothdimer and monomer form.

(B) AlCl3 (Q) Both Monomer and dimer forms are planar.(C) AlF3 (R) In dimer form all atoms are sp3 hybridised.(D) NO2 (S) Does not exist in dimer form.

[Ans. (A) Q, (B) R, (C) S, (D) P ]

ComprehensionSilicate are existing mainly in the polymeric form. Several categories are available with us which dependon the mode of sharing of corners of SiO4

4– tetrahedron.

Q.10 Which of the following pyroxene chain silicate is having same formula. [3]

(A) (B)

(C) (D*) All of these

Q.11 Which of the following silicates are having same general formula. [3](A) Sheet silicate and single chain silicate(B) Single chain silicate and double chain silicate(C*) 3-membered cyclic silicate and 5-membered cyclic silicate(D) 2D-silicate and 3D-silicate.

Q.12 Which of the following silicates is having 212 o-atoms are shared per tetrahedron. [3]

(A*) Amphibole chain silicate (B) Single chain silicate(C) Sheet silicate (D) Pyro silicate

Choose only one correct option.Q.1 Which of the following statement is incorrect regarding the complete hydrolysis of Marshall's acid.

(A) Caro's acid is an intermediate product. [3](B) Two molecules of H2SO4 and one molecule of H2O2 are the final product.(C) Hybridisation and oxidation state of central atom remain unchanged in the final product.(D*) Both final products can act as oxidising as well as reducing agent.

Q.2 Which of the following is an uncommon hydrolysis product of XeF2 and XeF4? [3](A) Xe (B*) XeO3 (C) HF (D) O2

Paragraph for question nos. 3 to 5Covalent compounds undergo in hydrolysis via SN1 (unimolecular nucleophilic substitution) or SN2(Bimolecular nucleophilic substitution) mechanism, for SN2 mechanism within the molecule atom shouldhave at least one vaccant orbital, if it is not there then hydrolysis takes place via SN1 mechanism (dissociativestep) in drastic condition.

Q.3 What are the hydrolysed products of BeCl2 formed during hydrolysis [3](I) [Be(OH)4]2– (II) Be(OH)2 (III) HCl (IV) BeH2(A) I, II, III (B*) II, III (C) I, III (D) II, III and IV

Q.4 Least probable product formed on hydrolysis of BCl3 [3](A*) [B(OH)4]¯ (B) HCl (C) B(OH)3 (D) None of these

Q.5 CCl4 is inert towards hydrolysis under ordinary condition because [3](I) No vaccant orbital on attacking site of C-atom.(II) H2O molecule can not approach to the anti-bonding M.O. of C–Cl bond due to steric crowding(III) Bond dissociation energy of C–Cl bond is very high(IV) CCl4 is non polar and does not react with polar H2O moleculeSelect correct code:(A) I, II and IV (B) I and IV (C*) I and II (D) I, II, III and IV

Q.6 Which of the following substance has the largest negative lattice energy ? [3](A) NaCl (B) CaBr2 (C) NaBr (D*) CaCl2

Q.7 Which of the following is/are incorrect about solubility trend in group I & II? [3] Least soluble in water Most soluble in water

(A) Hydroxides : LiOH CsOH(B*) Carbonates : Cs2CO3 Li2CO3(C) Nitrate : Ba(NO3)2 Be(NO3)2(D) Sulphates : BaSO4 BeSO4

Q.8 The stability of dihalides of Si, Ge, Sn and Pb increases in the sequence [3](A) GeX2 < SiX2 < SnX2 < PbX2 (B) SiX2 < GeX2 < PbX2 < SnX2(C*) SiX2 < GeX2 < SnX2 < PbX2 (D) PbX2 < SnX2 < GeX2 < SiX2

ASSIGNMENT-11

Paragraph for Question Nos. 9 to 11

Boric acid has three –OH group, when it is involved in intermolecular H-bonding, results 2D-sheet likestructure in solid state.

Q.9 The number of H-atoms replaced from boric acid when it is dissolved in water is [3](A) 2 (B) 1 (C) 3 (D*) zero

Q.10 It acts as what kind of acid in water. [3](A) Arhenius acid (B) Bronsted acid (C*) Lewis acid (D) Proton donar acid.

Q.11 Which of the following property is caused by 2D-layer structure of it. [3](A) Very hard (B*) Slippery nature (C) White colour (D) Spongy nature

Match the column with multiple options: [12]Q.12 Column I Column II

(A) XeF5+ (P) Two lone pairs

(B) ICl4– (Q) Planar

(C) TeCl4 (R) Non-planar

(D) I3+ (S) sp3d2 (Hybridization of central atom)

[Ans. (A) R, S; (B) P, Q, S; (C) R; (D) P, Q]

One or more than one option(s) is/are correct

Q.1 Choose the correct option(s) : [4](A) Order of increasing ionic size is O–2 < F– < Na+ < Mg+2

(B*) Order of increasing Lewis acid nature is BF3 < BCl3 < BBr3(C*) Order of decreasing magnetic moment is Cr+2 > Co+2 > Ni+2 > Ti+3

(D) Order of decreasing Boiling point is GeH4 > CH4 > SiH4

Q.2 Which of the following statement is incorrect? [3](A) Oxidizing power order : SiCl4 < SnCl4 < PbCl4(B) Ionic character order : CsBr > RbBr > KBr > NaBr > LiBr(C) The ionic character of lead (II) halides decreases with increase in atomic no. of halogen(D*) The oxidation state of Tl in Tl I3 is +3.

Q.3 It has been observed that % 's' character in Sb–H bond in SbH3 is 0.5%. Predict the % 's' character inthe orbital occupied by the lone pair is [3](A) 99.5 % (B) 99.0 % (C*) 98.5 % (D) 98.0 %

Q.4 Choose the option(s) regarding correct order of acidic nature : [3](A*) MgO < ZnO < P2O5 < SO3 (B) MgO < ZnO < SO3 < P2O5(C*) Li2O < NO < CO2 < SO2 (D) Li2O < BeO < CO2 < NO

Q.5 Polarisation may be called as the distortion of the shape of an anion by an adjacently placed cation.Which of the following statements is/are not correct? [4](A*) Minimum polarization is brought about by a cation of low radius(B*) A large cation is likely to bring about a large degree of polarisation(C) Maximum polarization is brought about by a cation of high charge(D*) A small anion is likely to undergo a large degree of polarisation

Q.6 Out of the following which one has the highest values of covalent character [3](A) ZnCl2 (B) CdCl2 (C*) HgCl2 (D) CuCl

Q.7 Compound having lowest Melting point. [3](A*) BeCl2 (B) MgCl2 (C) CaCl2 (D) SrCl2

Q.8 Which of the following properties is having incorrect order. [3](A*) CaSO3 < SrSO3 < BaSO3 ; solubility order(B) CaCO3 > SrCO3 > BaCO3 ; solubility order(C) NaF > KF > LiF > RbF > CsF ; m.p. order(D) CH3Cl > CH2Cl2 > CHCl3 > CCl4 ; dipole moment order

Q.9 For H3PO3 and H3PO4, the correct choice is [4](A*) H3PO3 is dibasic and reducing agent. (B) H3PO3 is dibasic and non reducing agent.(C) H3PO4 is tribasic and reducing agent (D*) H3PO4 is tribasic and non reducing agent.

ASSIGNMENT-12

Q.10 Statement-1 : The acidic strength order of hydra acids is : HF < HCl < HBr < HI [3]Statement-2 : The E.N. of halogens are F > Cl > Br > I.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.11 Choose the correct code for the following statements. [3]I. The (–)ve value of H for the dissolution of ionic compound is enough to predict the compound

is soluble in water at any temperature.II. For the alkali metals carbonate, solubility order decreases down the group.III. For the alkali metals ozonide, the thermal stability order increases down the group.IV. For the alkaline earth metals nitride, the thermal stability order increases down the group.(A) T T F F (B) T F F T (C*) T F T F (D) F T T F

Match the column with multiple options:

Q.12 Column I Column II [8](A) Dithionous acid (P) S–O–S bond is not present

(B) Thiosulphuric acid (Q) All S atom in the molecule has oxidation state +3

(C) Caro's acid (R) Acidic strength of –OH groups present in themolecule is different

(D) Pyrosulphurous acid (S) at least one S atom has oxidation state +5 in molecule

[Ans. (A) P,Q; (B)P; (C) P,R; (D) P,R,S]

One or more than one option(s) is/are correctQuestion No. 1 to 4 (4 questions)Schrondinger equation can be written for any molecule. However, it cannot be solved exactly for anysystem containing more than one electron, molecular orbitals having one electron wave functions formolecules are difficult to obtain directly from the solution of the Schrondinger wave equation. Thisdifficulty is overcome by resorting to an approximation method called linear combination of atomicorbitals (LCAO) method to form molecular orbitals.The molecular orbital formed by the addition of atomic orbitals is called the bonding molecular orbitaland the molecular orbital formed by the subtraction of atomic orbitals is called antibonding molecularorbital. Qualitatively, the formation of molecular orbitals can be understood in terms of the constructiveor destructive interference of the electron waves of the combining atoms. In the formation of bondingmolecular orbital, the two electron waves of the bonding atoms reinforce each other (constructiveinterference) while in the formation of antibonding molecular orbital, these electron waves cancel eachother (destructive interference). The result is that in a bonding molecular orbital most of the electrondensity is located between the nuclei of the bonded atoms and hence the repulsion between the nuclei isvery low while in an antibonding molecular orbital, most of the electron density is located away from thespace between the nuclei, as a matter of fact there is a nodal plane (i.e., plane in which the electrondensity is zero)

Q.1 How many nodal plane is/are present in *sp antibonding molecular orbital? [3]

(A) zero (B*) 1 (C) 2 (D) 3

Q.2 Which of the following combination of orbitals is correct? [3](A) +

(B) +

(C*) +

(D) —

Q.3 Which of the following statements is not correct regarding bonding molecular orbitals? [3](A) Bonding molecular orbitals possess less energy than the atomic orbitals from which they are formed(B*) Bonding molecular orbitals have low electron density between the two nuclei(C) Electron in bonding molecular orbitals contributes to the attraction between atoms(D) They are formed when the lobes of the combining atomic orbitals have the same sign

Q.4 If X-axis is the molecular axis, then -molecular orbitals are formed by the overlap of [3](A) s + pz (B) px + py (C*) pz + pz (D) px + px

Q.5 Which statement is correct [4](A*) Higher is the polarisation, higher will be relative solubility in non-polar solvent(B*) Higher is the polarisation, higher will be the intensity of colour(C*) Diamagnetic substances some times become coloured due to HOMO-LUMO transition(D) Higher is the polarisation in metal oxide, higher will be the basic character

ASSIGNMENT-13

Q.6 If NB is the number of bonding electrons and NA is the number of antibonding electrons of a molecule.Then choose the incorrect statement(s) for the given relationship, [4]

NB > NA(A) Molecule may be stable or unstable(B*) Molecule may have any integral, fractional or zero value of bond order(C*) Molecule is only paramagnetic species(D*) Molecule does not exist

Q.7 In which of the following ionization processes, the bond order has increased and the magnetic behaviourhas changed? [3]

(A) C2 2C (B*) NO NO+ (C) O2

2O (D) N2 2N

Q.8 Which of the following pair of species is/are having same bond order and same magnetic moment.

(A*) He–H and H2– (B*)

2O and NO [4](C*) He2

+ and Li2+ (D*) N2+ and N2

–

Q.9 According to MOT [3]

the process N2 (g) 2N (g) + e is more endothermic than the process O2(g)

2O (g) + ebecause(A) N has half filled electronic configuration which is more stable.(B) Nitrogen is less electronegative than oxygen.(C) Electron is removed from anti bonding molecular orbital in N2 where as electron is removed from

bonding molecular orbital in O2.(D*) Electron is removed from bonding molecular orbital in N2 where electron is removed from antibonding

molecular orbital in O2.

Q.10 Which of the following order is correct of the given property. [3](A*) LiCl >NaCl > KCl > RbCl > CsCl : Thermal stability order(B) BeF2 < MgF2 < CaF2 < SrF2 < BaF2 : solubility order(C) NO2– > NO— > NO = NO2+ > NO+ : bond length order(D) BaO > SrO > CaO > BeO > MgO : basic character order

Match the column with multiple options:


(A) O2 2O (P) An electron is removed from the * molecular

orbital.

(B) N2 2N (Q) An electron is removed from the molecular

orbital.

(C) C2 2C (R) Fractional bond order of molecular ion

(D) NO NO+ (S) Bond order decreases[Ans. (A) P, R (B) R,S (C) Q,R,S (D) P]

Q.1 Find the molecule which is planar and polar. [3](A) B3N3H6 (B) F2C = C = C = CF2(C*) BrF2Cl (D) F2C = C = CF2

Q.2 Find out the incorrect order of the dipole moment among the following pair of compound [3](A) NH3 > NF3 (B*) p-nitrophenol < o-nitrophenol(C) CH3Cl > CH2Cl2 (D) SiF4 < SF4

Q.3 For the molecule MA2Ln (where A is number of single bonded surrounding atoms, L indicates lone pairand ‘n’ is the number of lone pair and M is the central atom of ‘s’ or ‘p’ block element). the possiblerange of ‘n’ is [3](A) ‘1’ to ‘4’ (B) zero to ‘4’ (C) ‘1’ to ‘3’ (D*) ‘0’ to ‘3’

Q.4 The number of identical P–O bonds in 462OP is [3]

(A) Four (B*) Six (C) Five (D) Three

Q.5 All the following molecules are polar except [3](A) CH2Cl2 (B) NBr3 (C*) XeF4 (D) FCN

Q.6 Statement-1 : Dipole moment of NF3 is less than that of NH3. [3]Statement-2 : Polarity of N–F bond is less than that of N–H bond.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.7 Statement-1 : Dipole moment of H2O is more than that of OF2. [3]Statement-2 : In H2O, the resultant bond dipole of O – H bond and the resultant lone

pair moment are in opposite direction.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.8 Statement-1 :P – O bond length in POCl3 is 1.45 Å whereas the sum of single bond [3]covalent radii of phosphorus and oxygen is 1.83 Å.

Statement-2 : p – d bond is present in P – O bond.(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

ASSIGNMENT-14

Q.9 Statement-1 : Allene is a non polar molecule. [3]Statement-2 : Allene is non planar molecule.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.10 Column-I Column-II [10]

(A) 2 lone pair (P) XeF5–

(B) Zero dipole moment (Q) NF3

(C) Planar (R) ICl3

(D) All adjacent bond angles are equal (S) XeF4

[Ans. (A) P, R, S; (B) P, S; (C) P, R, S; (D) P, Q, R, S ]

Q.11 Prove that dipole moment of C6H5Cl and m-C6H4Cl2 are same. [5]

Q.12 Which of the following molecule will have the highest boiling point (in bracket D of the molecule is given)(A) Dimethyl ether (1.3) (B) Propane (0.0) [3](C*) Methyl cyanide (3.9) (D) Acetaldehyde (2.7)

Q.13 Match List-I with List-II and select the correct answer using the codes given below. [3]List–I List–II

(a) H2O < H2S < H2Se < H2Te 1. Bond angle(b) H2O > H2S > H2Se > H2Te 2. melting and boiling point(c) H2O >> H2S < H2Se < H2Te 3. acidic strength

(a) (b) (c) (a) (b) (c)(A) 1 3 2 (B) 3 2 1(C*) 3 1 2 (D) 2 3 1

Q.14 Which of the following pair of molecule have same shape but different in polarity (Polar or nonpolar)(A) H2O & NH3 (B) SnCl2 & SO2 (C*) CO2 & N2O (D) SO2 & SO3 [3]

Q.15 Statement-1 : The dipole moment of O2F2 is not zero. [3]Statement-2 : All atoms are lying in the same plane.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

ASSIGNMENT-15

Q.1 Explain the structure of Boric acid in solid state.

Q.2 Boiling point of o-Nitrophenol is less than meta and para nitrophenol. Why?

Q.3 Maleic acid is more acidic than fumeric acid. Why?

Q.4 H – F is only liquid among halogen acid. Why?

Q.5 Ammonia is more easily liquefied than HCl, explain.

Q.6 Why ice floats on water?

Q.7 Water shows maximum density at 4°C. Why?

Q.8 HI is the strongest halogen acid, whereas H–F is the weakest. Why?

Q.9 Wood pieces are used to hold ice-cream. Why?

Q.10 KHF2 is possible but not KHBr2 or KHI2. Why?

Q.11 O – Nitrophenol is less soluble in H2O than p – Nitrophenol. Why?

Q.12 o-Hydroxy benzaldehyde is a liquid at room temperature while p-hydroxy benzaldehyde is a high meltingsolid.

Q.13 Glycerol is more viscous than ethanol. Explain.

Q.14 CH4 and H2O have nearly same molecular weight. Yet CH4 has a boiling point 112 K and water 373 K.Explain.

Q.15 The experimental molecular weight of acetic acid in just double than theoretical molecular weight ofacetic acid. Why?

Q.16 Although chlorine has same electronegativity as nitrogen but the former does not form effective H-bonding.Explain.

Q.17 Molar entropy change of vapourization of acetic acid is less than that of water. Explain

Q.18 Heat of vapourization of water is higher than HF, however strength of H-bond in HF is higher than water.Explain

Q.19 Statement-1 : The H-bond present in NH3 dissolved in water is best represented by

HH||

HO····HNH and not by

HH||

O–H····NH|

H

.

Statement-2 : The O–H bond polarity is more compared to that of N–H bond.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.20 Statement -1 : B(OH)3 is acidic while In(OH)3 is basic in nature.Statement -2 : B(OH)3 has a highly H–bonding structure in solid state.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement-1.(C) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

Q.21 Compare the basicity of[Me4N]+ OH¯ and [Me3NH]+ OH¯

ASSIGNMENT-18

Q.1 In which of the following metal carbonyls the C – O bond order is lowest? [3](A) [V(CO)6]– (B) [Mn(CO)6]+ (C) [Cr(CO)6] (D*) [Ti(CO)6]2–

Q.2 Which of the following names is/are correct for the compound Na[CoCl2(NO2)(-C3H5) (NH3)2](A*) Sodium allyldiamminedichloridonitrito-N-cobaltate (III) [3](B) Sodium diamminedichloridoallylnitrito-N-cobaltate (III)(C*) Sodium diamminedichloridocyclopropylnitrito-N-cobaltate (III)(D) Sodium diamminecyclopropylnitrito-N-dichloridocobaltate (III)

Q.3 IUPAC name(s) of possible coordination positional isomer(s) of complex ion [3]

Cl (NH ) Pt3 3

OH

OHPt(NH ) Cl3 3

2+

is/are

(A*) Di--hydroxido bis {diamminedichloridoplatinum (IV)} ion(B) Bis{-hydroxidodiamminedichloridoplatinate (IV)} ion(C*) Tetrachloridoplatinum (IV) di--hydroxidotetraammineplatinum (IV) ion(D*) Diamminedichloridoplatinum (IV) di--hydroxidodiamminedichloridoplatinum (IV) ion

Q.4 Arrange the following compound according to dC–C order. [3](I) C2F4 (II) C2H4 (III) [PtCl3(C2H4)]¯(A) I > II > III (B*) III > II > I (C) II > I > III (D) II > III > I

Q.5 Statement-1 : In Zeise's salt C – C bond order regarding ethylene molecule is less than two.Statement-2 : Ethylene accepts electron pair from filled d-orbital of Pt2+ into its vacant bonding

M.O. [3](A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.6 Statement-1 : [V(CO)6] can not act as oxidising agent. [3]Statement-2 : It can not be reduced by reducing agent.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is false.

Paragraph for question nos. 7 to 9Sidwick EAN rule says that complex compound has the tendency to achieve the EAN of 36, 54 and 86for first, second and third transition series elements. [9]

Q.7 Which of the following complex acts as reducing agent based on Sidwick EAN rule.(A) Mn(CO)5 (B) Mn2(CO)10 (C*) Mn(CO)6 (D) [V(CO)6]¯

Q.8 Which of the following complex is following sidwick EAN rule.(A) [Ag(S2O3)2]3– (when only 'S' atom is the donor atom)(B*) [Cd(CN)4]2–

(C) [Pt(en)2]2+

(D) [Mo(– C3H5) Br(NH3)2]°

Q.9 Which of following statement is not correct regarding complex "Ferrocene".(A*) EAN of central atom in ferrocene is not equal to its nearest noble gas(B) Molecule is having aromatic character(C) It has sandwich like structure(D) Two rings act as -donor ligand.

Q.10 Column I Column II [12](A) K3[Fe(CN)5(CO)] (P) complex having lowest bond length of CO ligand(B) K[PtCl3(C2H4)] (Q) follow Sidgwick's rule of EAN(C) Na[Co(CO)4] (R) complex involved in synergic bonding(D) V(CO)6 (S) complex having highest bond length of CO ligand

[Ans. (A) P, Q, R; (B) R; (C) Q, R, S; (D) R]

Q.11 Write down formula of the following complex compounds : [10](i) Pentaammine isothiocyanato cobalt (III) chloride(ii) Sodium diaquobisoxalato manganate (III)(iii) -amido--nitro-octaammine dicobalt (III) ion(iv) Sodium bis (thiosulphato) argentate (I)(v) trans dichloridobis [ethylenediammine] cobalt (III) ion(vi) -amido-bis-[pentammine cobalt (III)] nitrate(vii) trans-Tetraamminedichloridoplatinum (IV) ion(viii) cis-Dichloridobis (ethylenediamine) ruthenium (II)(ix) Sodium tris (oxalato) aluminate (III)(x) Hexaamminecobalt (III) tris (oxalato) cobaltate (III)

(Choose the correct option, only one is correct)Q.1 In the compound CoCl3·5NH3 [3]

(A) all the Cl show primary valency (PV) only(B) two Cl show (PV) and one Cl secondary valency (SV)(C*) two Cl show (PV) and one Cl (PV) as well as (SV)(D) all the Cl show (SV)

Q.2 The two compounds pentaamminesulphatocobalt(III) bromide and [3]pentaamminesulphatocobalt(III) chloride represent(A) Linkage isomerism (B) Ionization isomerism(C) Coordination isomerism (D*) No isomerism

Q.3 Which of the following is an oxidizing agent? [3](A*) Mn(CO)5 (B) Fe(CO)5 (C) Mn2(CO)10 (D) Fe2(CO)9

Q.4 EAN of the elements (*) are equal in: [3]

(A*) iN*

(CO)4, [*

Fe (CN)6]4– (B) [ iN*

(en)2]2+, [*

Fe (H2O)6]2+

(C) [*

Co (CN)6]3–, [*

Fe (CN)6]3– (D) [ iN*

(en)2]2+, [*

Sc (H2O)6]3+

Q.5 Which of the following statements is incorrect? [3](A) Co-ordination compounds and complexes are synonymous terms.(B*) Complexes must give ions in the solution.(C) Complexes may give ions in the solution or may not give ions in the solution.(D) Generally complex ion does not dissociate into its component parts even in the solution.

Q.6 Which one of the following is an example of coordination isomerism? [3](A) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br(B) [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2(C) [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2 ·H2O(D*) [Cr(NH3)6][Co(CN)6] and [Co(NH3)6][Cr(CN)6]

Q.7 Select correct code about complex [Cr(NO2)(NH3)5][ZnCl4] [3](I) IUPAC name of the compound is Pentaamminenitrito-N-chromium (III) tetrachloridozincate (II)(II) It shows geometrical isomerism(III) It shows linkage isomerism(IV) It shows co-ordination isomerism(A) III, IV (B*) I, III & IV (C) II, III & IV (D) I, II, III & IV

Q.8 Consider the following complexes: [3](I) K2PtCl6 (II) PtCl4· 2NH3 (III) PtCl4· 3NH3 (IV) PtCl4·5NH3Their electrical conductances in an aqueous solutions are:(A*) 256, 0, 97, 404 (B) 404, 0, 97, 256 (C) 256, 97, 0, 404 (D) 404, 97, 256, 0

ASSIGNMENT-19

Q.9 Statement-1 : In Mn2(CO)10 molecule, there are total 70 electrons in both Mn atoms. [3]Statement-2 : Mn2(CO)10 molecule acts as oxidising agent.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Paragraph for question nos. 10 to 12Complex compounds which have same molecular formula but have different structural arrangements ofligands around central metal atom or ion are called structural isomers and phenomenon is named asstructural isomerism. [9]

Q.10 Which of the following compounds is/are polymerisation isomer of [Fe(NO2)3(NH3)3].(I) [Fe(NO2)(NH3)5] [Fe(NO2)5(NH3)] (II) [Fe(NO2)2(NH3)4]2 [Fe(NO2)5(NH3)](III) [Fe(NO2)(NH3)5] [Fe(NO2)4(NH3)2]2 (IV) [Fe(NO2)2(NH3)4] [Fe(NO2)4(NH3)2]Choose the correct code :(A) II, III (B) II, III, IV (C) I, IV (D*) I, II, III, IV

Q.11 How many more coordination isomers are possible of the following complex compound.[PtCl2 (NH3)4] [Pt (SCN)4]

(A) 5 (B) 6 (C*) 8 (D) 9

Q.12 Select incorrect match(A) [Co(NO2)(H2O)(en)2)]Cl2, [CoCl(NO2)(en)2] Cl . H2O – Hydrate isomerism

(B*) [Cu(NH3)4][PtCl4], [CuCl2(NH3)2] [PtCl2(NH3)2] – Coordination isomerism

(C) [Ni(CN)(H2O)(NH3)4]Cl, [NiCl(H2O)(NH3)4]CN – Ionization isomerism

(D) [Cr(NCS)(NH3)5] [ZnCl4], [Cr(SCN)(NH3)5] [ZnCl4] – Linkage isomerism

Q.1 Which of the following complexes can exist as diastereoisomers? [3](A*) [Cr(NH3)2Cl4]– (B) [Co(NH3)5Br]2+

(C) [FeCl2(NCS)2]2– (D*) [PtCl2Br2]2–

Q.2 Lead poisoning in the body can be removed by: [3](A*) EDTA in the form of calcium dihydrogen salt(B) cis-platin(C) Zeisse’s salt(D) DMG

Q.3 How many geometrical isomers are possible for [Pd2+ (NH2–CH(CH3)– 2CO )2]o [3]

(A) 2 (B) 3 (C*) 4 (D) 6

Q.4 Isomerisms exhibited by [Cr(NH3)2(H2O)2Cl2]+ are [3](A) ionisation, optical (B) hydrate, optical(C*) geometrical, optical (D) coordinate, geometrical

Q.5 Of the following complex ions, one exhibits isomerism. That is: [3](A) [Ag(NH3)2]+ (B*) [Co(NH3)5NO2]2+

(C) [Pt(en)Cl2] (D) [Co(NH3)5Cl]2+

Q.6 [Cr(NH3)5Br]Cl and [Cr(NH3)5Cl]Br can be distinguished by/and isomerism shown is: [3](A) BaCl2, ionisation (B*) AgNO3, ionisation(C) AgNO3, coordinate (D) BaCl2, linkage

Q.7 If excess of AgNO3 solution is added to 100 mL of a 0.024 M solution of dichlorobis (ethylene diamine)cobalt (III) chloride, how many mol of AgCl be precipitated: [3](A) 0.0012 (B) 0.0016 (C*) 0.0024 (D) 0.0048

Q.8 Which kind of isomerism is exhibited by [Co(EDTA)]— [3](A) Optical & geometrical isomerism (B) Geometrical isomerism(C*) Optical isomerism (D) No isomerism

Q.9 Which of the following is correct IUPAC name of any complex compound. [3](A) Tris (acetyl acetonato) iron (III) chloride(B) Hexachloroplatinum (IV) tetraammine dicyano platinate (IV)(C*) Ammine bromochloro methylamine platinum (II)(D) Cis dichloro (ethylenediamine) platinum (II)

Q.10 [FeIICl(CN)4(O2)]4– is named as: [3](A) chlorotetracyanodioxoferrate (II) ion (B) chlorotetracyanoperoxoferrate (II) ion(C*) chlorotetracyanosuperoxoferrate (II) ion (D) none is correct

ASSIGNMENT-20

Q.11 The total possible co-ordination isomers for the following compounds respectively are [3][Co(en)3] [Cr(C2O4)3][Cu(NH3)4] [CuCl4][Ni(en)3] [Co(NO2)6]

(A) 4, 4, 4 (B) 2, 2, 2 (C) 2, 2, 4 (D*) 4, 2, 4

Q.12 Match The column: [12]Column I Column II

(Co-ordination compound) (Type of isomerism shown)(A) Na2[Pt(SCN)2(Ox)2] (P) Ionization isomerism(B) [CrCl2(NH3)4]NO3 (Q) Linkage isomerism(C) [Pt(NO2)(Gly)(NH3)] (R) Geometrical isomerism(D) K3[Fe(OH)2(C2O4)2] (S) Optical isomerism

[Ans. (A) Q,R,S (B) P,R (C) Q,R (D) R,S]

Q.1 [Ni(CN)4]2– and [NiCl4]2– have similarity but not in [3](A) magnetic moment (B) C.N. and O.N. (C) structure (D*) both (A) and (C)

Q.2 Which is true for [Ni(en)2]2+, Z(Ni) = 28? [3](A) paramagnetism, dsp2, square planar, C.N. of Ni = 2(B*) diamagnetism, dsp2, square planar, C.N. of Ni = 4(C) diamagnetism, sp3, tetrahedral, C.N. of Ni = 4(D) paramagnetism, sp3, square planar, C.N. of Ni = 4

Q.3 Among Ni(CO)4, [Ni(CN)4]2– and 24NiCl , [3]

(A*) Ni(CO)4 and [Ni(CN)4]2– are diamagnetic and 24NiCl is paramagnetic

(B) Ni(CO)4 and 24NiCl are diamagnetic and [Ni(CN)4]2– is paramagnetic

(C) Ni(CO)4 is diamagnetic and [Ni(CN)4]2– and 24NiCl are paramagnetic

(D) 24NiCl and [Ni(CN)4]2– are diamagnetic and Ni(CO)4 is paramagnetic.

Q.4 Geometrical isomerism in coordination compunds is exhibited by [3](A) Square planar and tetrahedral complexes (B*) Square planar and octahedral complexes(C) Tetrahedral and octahedral complexes (D) Squre planar, tetrahedral and octahedral complexes

Q.5 Which of the following is not optically active? [3](A) [Co(en)3]3+ (B) [Cr(ox)3]3– (C) cis-[CoCl2(en)2]+ (D*) trans-[CoCl2(en)2]+

Q.6 Arrange the following in order of decreasing number of unpaired electrons: [3](I) [Fe(H2O)6]2+ (II) [Fe(CN)6]3– (III) [Fe(CN)6]4– (IV) [Fe(H2O)6]3+

(A*) IV, I, II, III (B) I, II, III, IV (C) III, II, I, IV (D) II, III, I, IV

Q.7 For which of the following dn configuration of octahedral complex (es), can not exist in both high spinand low spin forms. [3](I) d3 (II) d5 (III) d6 (IV) d8

(A) II & III (B) I & III (C*) I & IV (D) III & IV

Q.8 Which of the following statements is/are false [4](A*) In [PtCl2(NH3)4]2+ complex ion, the cis-form is optically active, while trans-form is optically inactive(B) In [Fe(C2O4)3]3–, geometrical isomerism does not exist, while optical isomerism exists(C*) [Mabcd]n square planar complexes exhibit both optical as well as geometrical isomerism(D*) In [Mabcd]n± tetrahedral complexes, optical isomerism cannot be observed

ASSIGNMENT-21

COMPREHENSIONWhen a transition metal ion (usually) is involved in octahedral complex formation, the five degenerated-orbitals split into two set of degenerate orbitals (3 + 2). Three degenerate orbitals of lower energy(dxy, dyz, dzx) and a set of degenerate orbitals of higher energy )dandd( 222 zyx

. The orbitals with

lower energy are called t2g orbitals and those with higher energy are called eg orbitals.In octahedral complexes, positive metal ion may be considered to be present at the centre and negativeligands at the corner of a regular octahedron. As lobes of 22 yx

d

and 2zd lie along the axes, i.e., along

the ligands the repulsions are more and so high is the energy. The lobes of the remaining three d-orbitalslie between the axes i.e., between the ligands. The repulsion between them are less, so lesser the energy.In the octahedral complexes, if metal ion has electrons more than 3 then for pairing them, the options are(i) Pairing may start with 4th electron in t2g orbitals.(ii) Pairing may start normally with 6th electrons when t2g and eg orbitals are singly filled.

Q.9 In which of the following configurations, hybridisation and magnetic moment of octahedral complexesare independent of nature of ligands. [9](I) d3 configuration of any metal cation(II) d6 configuration of IIIrd transition series metal cation(III) d8 configuration of Ist transition series metal cation(IV) d7 configuration of any metal cationSelect the correct code:(A) III, IV (B) I, III, IV (C) I, II, IV (D*) I, II, III

Q.10 Which of the following electronic arrangement is / are possible for inner orbital oct complex.

(I) 2g

3g2 et (II) 1

g6

g2 et (III) 0g

3g2 et (IV) 2

g4

g2 et

Select the correct code:(A) I, IV (B*) II, III (C) III only (D) III, IV

Q.11 Select incorrect match for the following complexes.(A) [IrF6]3– ( > P) (B*) [Co(H2O)6]3+ ( < P)

(C) Fe(CO)5 ( > P) (D) 222 ])SCN(PdCl[ ( > P)

Q.12 Column-I Column II [12](A) [Ma2bcde]n ± (P) 3 optically inactive isomers(B) [Ma2b2c2]n ± (Q) 4 geometrical isomers(C) [Ma3bcd]n ± (R) 6 stereo(space)isomers(D) [M(AB)c2d2]n ± (S) 2 optically active isomers(where AB Unsym. bidentate ligand, a,b,c,d & e monodentate ligands)

[Ans. (A) – P ; (B) – R,S ; (C) – P, Q,S ; (D) – Q, R]

One option is correct

Q.1 Which of the following complexes is diamagnetic? [3](A*) [Fe(CN)6]4– (B) [Cu(NH3)4]2+ (C) [Ti(H2O)6]3+ (D) [Fe(CN)6]3–

Q.2 In nitroprusside ion the iron and NO exist as FeII and NO+ rather than FeIII and NO. These forms canbe differentiated by [3](A) estimating the concentration of iron (B) measuring the concentration of CN–

(C*) measuring the solid state magnetic moment (D) thermally decomposing the compound

Q.3 The complex ions [Fe(CN)6]3– and [Fe(CN)6]4– [3](A) Are both octahedral and paramagnetic(B) Are both octahedral and diamagnetic(C*) Have same structure but opposite magnetic character(D) Have different structure but opposite magnetic character.

Q.4 Of the following complex ions, the one that probably has the largest overall formation constant, Kf , is(A) [Co(NH3)6]3+ (B) [Co(H2O)6]3+ [3](C) [Co(H2O)4(NH3)2]3+ (D*) [Co(en)3]3+

Q.5 Of the following complex ions, one is a Bronsted-Lowry acid. That one is [3](A) [Cu(NH3)4]2+ (B) [FeCl4]– (C*) [Fe(H2O)6]3+ (D) [Zn(OH)4]2–

Q.6 The number of unpaired electrons expected for the complex ion [Cr(NH3)6]2+ is: [3](A*) 2 (B) 3 (C) 4 (D) 5

Q.7 The CFSE for [CoCl6]4– complex is 18000 cm–1. The for [CoCl4]2– will be: [3](A) 18000 cm–1 (B) 16000 cm–1 (C*) 8000 cm–1 (D) 2000 cm–1

More than one correctQ.8 Which one of the following statement(s) is/are false? [4]

(A*) Weak ligands like F—, Cl— and OH— usually form low spin complexes.

(B*) Strong ligand like CN— and –2NO , generally form high spin complexes.

(C) [FeF6]3– is high spin complex.(D*) [Ni(CO)4] is high spin complex

Q.9 Which is correct statement(s)? [4](A*) [Ag(NH3)2]+ is linear with sp hybridised Ag+ ion

(B*) 24NiCl , 3

4VO and 4MnO have tetrahedral geometry

(C*) [Cu(NH3)4]2+ , [Pt(NH3)4]2+ & [Ni(CN)4]2– have dsp2 hybridisation of the metal ion

(D*) Fe(CO)5 has trigonal bipyramidal structure with 3z

spd 2 hybridised iron.

ASSIGNMENT-22

Match the columneQ.10 Match the column-I with column-II. [12]

Note that column-I may have more than one matching options in column-II.Column-I Column-II

(A) [Ni(H2O)6]Cl2 (P) d2sp3 hybridisation(B) [Co(CN)2(NH3)4]OC2H5 (Q) Ionisation isomerism(C) [IrCl6]3– (R) = 2.83 B.M.(D) [PtCl2(NH3)4]Br2 (S) 0 < P

[Ans. (A) R, S; (B) P, Q; (C) P; (D) P, Q]

Q.11 Match the column-I with column-II. [12]Note that column-I may have more than one matching options in column-II.

Column-I Column-II(A) [Cr(gly)3]° (P) Low spin complex(B) [CoBr2Cl2]2– (Q) high spin complex(C) [Fe(NH3)6]2+ (R) optical isomerism(D) Na[PtBrCl(NO2)(NH3)] (S) geometrical isomerism

[Ans. (A) R, S; (B) Q; (C) Q; (D) P, S]

SubjectiveQ.12 Explain the following with appropriate reasons : [10](a) All octahedral complexes of Ni2+ ion must be outer orbital complexes.(b) [Co(NH3)6]2+ and [Co(NO2)6]4– ions are easily oxidisable(c) 4-coordinated complexes of Pd(II) and Pt(II) are diamagnetic square planar(d) Tetrahedral complexes do not show geometrical isomerism while square planar complexes do show this

kind of isomerism.(e) [Cu(NH3)4]2+ is coloured while [Cu(CN)4]3– is colour less.

ASSIGNMENT-23

One option is correctQ.1 Complex compound [Cr(NCS)(NH3)5] [ZnCl4] will be [3]

(A) Colourless & diamagnetic(B) Green coloured & diamagnetic(C*) Green coloured & shows coordination isomerism(D) Diamagnetic & shows linkage isomerism

Q.2 The correct order of energies of d-orbitals of metal ion in a square planar complex is [3]

(A) dxy = dyz = dzx > 22 yxd

= 2z

d (B) 22 yxd

= 2z

d > dxy = dyz = dzx

(C) 22 yxd

> 2z

d > dxy > dyz = dzx (D*) 22 yxd

> dxy > 2z

d > dzx = dyz

Q.3 4MnO is of intense pink colour, though Mn is in (7+) oxidation state, it is due to [3]

(A) Oxygen gives colour to it(B) Charge transfer when Mn (7+) gives its electron to oxygen & oxidise to Mn (8+) temporaily(C*) Charge transfer when oxygen gives its electron to Mn (7+) changing in Mn (6+) temporaily(D) none is correct explanation

Q.4 [CoCl2(NH3)4] + Cl– [CoCl3(NH3)3] + NH3 [3]If in the above reaction only one isomer of the product is obtained, which is true for the initial (reactant)complex(A) compound is in cis form (B*) compound is in trans form(C) compound is in both (cis and trans) forms (D) can't be predicted

Q.5 Aqueous solution of Ni2+ contains [Ni(H2O)6]2+ and its magnetic moment is 2.83 B.M.When ammoniais added in it, comment on the magnetic moment of solution [3](A*) It will remain same (B) It increases from 2.83 B.M.(C) It decreases from 2.83 B.M. (D) It can not be predicted theoretically

Q.6 The complex [Fe(H2O)5NO]2+ is formed in the brown ring test for nitrates when freshly prepared

FeSO4 solution is added to aq solution of 3NO followed by addition of conc.H2SO4. Select correct

statement about this complex: [3](A) colour change is due to charge transfer(B) it has iron in +1 oxidate state and nitrosyl as NO+

(C) it has magnetic moment of 3.87 B.M. confirming three unpaired electrons in Fe(D*) all are correct statements

Assertion ReasonQ.7 Statement-1 : Hund’s rule violates in [Co(CN)6]3– complex ion [3]

Statement-2 : Degeneracy of d orbitals is lost under any field ligand.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.8 Statement-1 : Complex ion [Co(NH3)6]2+ is readily oxidized to [Co(NH3)6]3+. [3]Statement-2 : Unpaired electron in complex ion [Co(NH3)6]2+ is present in 4p orbital.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

More than one correctQ.9 Select the correct statement. [4]

(A*) Chelation effect is maximum for five & six membered rings(B*) Greater the charge on the central metal cation, greater the value of (CFSE)(C) In complex ion [CoF6]3–, F– is a weak field ligand, so that oct < P (Pairing energy) and it is low spincomplex.(D*) [CoCl2(NH3)2 (en)]complex ion will have four different isomers.

Q.10 Which of the following statements is/are true [4](A*) In Ferrocyanide ion, the effective atomic number is 36(B*) Chelating ligands are atleast bidentate ligands

(C*) ])NH()CN(CrCl[ 2322 and [CrCl3(NH3)3] both have d2sp3 hybridisation(D*) As the number of rings in complex increases, stability of complex (chelate) also increases

Match the column:Q.11 Column-I Column-II [12]

(Pair of complex compounds) (Property which is different in given pair)(A) [Ni(CO)4] and K2[Ni(CN)4] (P) Magnetic moment(B) [Cu(NH3)4]SO4 and K3[Cu(CN)4] (Q) Oxidation no. of central metal(C) K2[NiCl4] and K4[Ni(CN)4] (R) Geometry(D) K2[NiCl4] and K2[PtCl4] (S) EAN of central metal

[Ans. (A) Q, R, S; (B) P, Q, R, S; (C) P, Q, S; (D) P, R, S]

Subjective:Q.12 Explain the following with appropriate reasons : [10](a) NH2–NH2 has two lone pairs. However, it does not act as chelating agent.(b) Cu(OH)2 is soluble in NH4OH where as it is insoluble in NaOH.(c) A blue coloured solution of [CoCl4]–2 ion changes to pink on reaction with HgCl2.(d) What happens when [Ni(H2O)6]2+ is mixed with ethylenediamine in excess?(e) [Fe(CN)6]3– ion has magnetic moment 1.73 BM, while [Fe(H2O)6]3+ has a magnetic moment 5.92 BM.

ASSIGNMENT-24

OXIDE ORE:

* ZnO Zincite

* Fe2O3 Haematite

* Fe3O4 Magnetite

* Al2O3·2H2O Bauxite

* Fe2O3·3H2O Limonite

* Cu2O Cuprite or Ruby copper

MnO2 Pyrolusite

SnO2 Tinstone or Casseterite

TiO2 Rutile

Fe·Cr2O4 (FeO + Cr2O3) Chromite ore

Na2B4O7·10H2O Borax or Tincal

Ca2B6O11·5H2O Colemanite

U3O8 Pitch Blende

FeO.TiO2 Ilmenite

SULPHURISED ORE:

* PbS Galena

HgS Cinnabar

* ZnS Zinc blende/sphalerite

* Cu2S Copper glance/Chalcocite

CuFeS2 Copper Pyrite (Chalcopyrite)

* FeS2 Iron pyrite or Fool's gold

Ag2S Silver glance or Argentite

HALIDE ORE:

NaCl Rock Salt

KCl Sylvine

CaF2 Fluorspar

Na3AlF6 Cryolite

AgCl Horn Silver

KCl·MgCl2·6H2O Carnalite

OXY SALT ORE:(1) CARBONATE ORE:

CaCO3 Lime stone

MgCO3 Magnesite

CaCO3·MgCO3 Dolomite

* FeCO3 Siderite

* ZnCO3 Calamine

* Cu(OH)2·CuCO3 Malachite or Basic CopperCarbonate

Cu(OH)2·2CuCO3 Azurite

PbCO3 Cerrusite

(2) SULPHATE ORE:

CaSO4·2H2O Gypsum

MgSO4·7H2O Epsom Salt

PbSO4 Anglesite

BaSO4 Baryte

Na2SO4·10H2O Glauber Salt

(3) NITRATE ORE:

KNO3 Indian Salt peter

NaNO3 Chile Salt peter

METALS IN LIVING ENTITIES :

(a) Magnesium is found in chlorophyll.

(b) Potassium is presnt in plant roots.

(c) Manganese, iron and copper are present inchloroplast.

(d) Zinc is present in eyes of cats and cows.

(e) Iron is present in haemoglobin.

(f) Calcium is present in bones.

(g) Vanadium is present in cucumbers.

(h) Chromium is present in pron.

(i) Cobalt is present in cynocobalamin (Vitamine)

ALLOYSNAME OF THE ALLOY COMPOSITION USES

1. Magnelium Al : 98%, Mg : 2% For making balance2. Duralumin Al: 95%, Cu : 4 % Air craft parts

Mg : 0.5 %, Mn : 0.5% boat machinary3. Aluminium bronze Al :10%, Cu : 90 % Making coins, photo frames

utensils, golden paints4. Almica Al : 20%, Ni : 20 % For making permanent magnet

Co: 10%, Steel : 50%5. -Alloy Al : 92%, Cu : 4% Pistons and machine parts

Mg : 1.5 %, Ni : 2.5 %6. Nickeloy Al : 95%, Cu : 4 %, Ni : 1% Air craft parts7. Pewter Pb : 20, Sn : 80 Utensils8. Solder Pb : 50, Sn : 50 Soldering9. Type metal Pb : 75, Sn : 5, Sb:20 Printing type10. Bell metal Cu : 80, Sn : 20 Bells making11. Babbit metal Sn : 90, Sb : 7, Cu : 3 Bearing of machinary12. Frary metal Pb : 97%, Ba: 2%, Ca: 1% Bearing of machine13. Lino type metal Pb : 83%, Sn : 3%, Sb:14% Printing type14. Brass Cu:70%, Zn:30% making utensils condenses tube

making15. Bronze Cu: 88-96%, Sn 4-12% utensils, coins, statues16. Monel metal Cu: 27%, Ni : 68%, Fe : 5% making pumps, turbines of

ships, boilers etc.17. German silver Cu: 50%, Zn: 30%, Ni: 20% Flower Vase & ornaments18. Electron Mg:95%, Zn:4.5, Cu: 0.5% Parts of aeroplane and motor

cars19. Dutch metal Cu: 80%, Zn: 20% Golden yellow colour used for

decorative purpose20 Nichrome Ni, Cr, Fe21. Gun Metal Cu : 87%, Zn:3%, Sn :10%22. Alnico Al, Ni, Co23. Con Stantan Cu : 60% , Ni : 40%24. Artifical Gold Cu : 90%, Al : 10%25. 14 Carat Gold Au : 54%, Ag : 14% to 30%, Cu : 12-28%26. 24 Carat Gold 100% Au

ALLOY OF STEEL1. Vanadium V : 0.2-1%2. Chromium Cr : 2- 4%3. Nickel Ni : 3-5%4. Manganese steel Mn : 10-18%5. Stainless steel Cr : 12-14 % and Ni : 2-4%6. Tunguston W : 10-20%7. Invar Ni : 36%

ASSIGNMENT-25

Q.1 SnO2 is reduced to metallic Sn on smelting oxide with anthracite, limestone and sand. In this, function ofsand is : [3](A) It acts as a flux (B) It removes basic impurities as slag(C*) Both are correct (D) None is correct

Q.2 (Ag + Pb) alloy addedis

zincandmelt (Ag + Pb + Zn) melt cool YLayer

XLayer [3]

Select correct statement based on above scheme :(A) Layer X contains zinc and silver(B) Layer Y contains lead and silver but amount of silver in this layer is smaller than in the layer X.(C) X and Y are immiscible layers(D*) All are correct statements

Q.3 AgCl + Na2CO3 Ag2CO3 X, X is : [3](A) Ag2O and CO2 (B*) Ag, O2 and CO2(C) Ag2O2 and CO2 (D) no effect

Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]

Incorrect statements about the plot is / are:(A) T1 and T2 are melting point & boiling point of Mg respectively.(B*) T1 and T2 are melting point & boiling point of MgO respectively.(C) Reduction of MgO by coke is possible above T3(D) Mg can be extracted from gaseous products by rapid cooling.

Comprehension (Q.5 to Q.8)Questions given below are based on the given diagram for extractive metallurgy. [12]

0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2

Temperature, °C The points noted by arrows are the melting and boiling points of the metals zinc and magnesium. G° asa function of temperature for some reactions of extractive metallurgy.

Q.5 At what approximate temperature, zinc and carbon have equal affinity for oxygen?(A*) 1000°C (B) 1500°C (C) 500°C (D) 1200°C

Q.6 At this temperature G° of the reaction is :ZnO + C Zn + CO

(A) – ve (B) +ve (C*) zero (D) nothing can be saidQ.7 To make the following reduction process spontaneous, temperature should be :

ZnO + C Zn + CO(A) < 1000°C (B*) > 1100°C (C) < 500°C (D) > 500°C but < 1000°C

Q.8 At 1100°C, which reaction is spontaneous to a maximum extent?(A) MgO + C Mg + CO (B) ZnO + C Zn + CO(C) MgO + Zn Mg + ZnO (D*) ZnO + Mg MgO + Zn

Q.9 Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false.(i) Cu metal is extracted from its sulphide ore by reduction of Cu2O with FeS. [3](ii) An ore of Tin containing FeCrO4 is concentrated by magnetic separation method.(iii) Auto reduction process is used in the extraction of Cu & Hg.(iv) Cassiterite and Rutile are oxide ores of the metals.(A) TFTT (B) TTFT (C*) FTTT (D) FFFT

Q.10 Electrolyte reduction of alumina to aluminium by Hall-Heroult process is carried out: [3](A) In the presence of NaCl(B) In the presence of BaF2(C*) In the presence of cryolite which forms a melt with lower melting temperature(D) In the presence of cryolite which forms a melt with higher melting temperature

Q.11 During the process of electrorefining of copper some metals present as impurity settle as anode mud.These are: [3](A) Sn and Ag (B) Pb and Zn (C*) Ag and Au (D) Fe and Ni

Q.12 The substance not likely to contain CaCO3 is: [3](A) Sea shells (B) Dolomite (C) Marble statue (D*) Calcined gypsum

Q.13 The most common elements present in the crust of the Earth are: [3](A*) oxygen, silicon, aluminium (B) oxygen, iron, magnesium(C) silicon, iron, potassium (D) oxygen, iron, silicon

Q.14 Extraction of silver from its ore involving NaCN, air and an active metal is known as: [3](A) Pattinson’s method (B) Amalgamation method(C*) Mc Arthur-forest method (D) Parke’s method

ASSIGNMENT-26

Q.1 Reduction of a metal oxide by excess carbon at high temperature is a method for the commercial preparationof some metals. This method can be successfully applied in the case of [3](A) BeO and Al2O3 (B*) ZnO and Fe2O3(C) CaO and Cr2O3 (D) BaO and U3O8

Q.2 In Goldschmidt aluminothermic process, thermite mixture contains: [3](A) 3 parts Fe2O3 and 2 parts Al (B) 3 parts Al2O3 and 4 parts Al(C) 1 part Fe2O3 and 12 part Al (D*) 3 parts Fe2O3 and 1 part Al

Q.3 Electric furances are lined with magnesia because: [3](A) It is not affected by acids (B) It liberates oxygen on heating(C*) It melts at very high temperature (D) It has no effect of electricity\

Q.4 Which pair of elements can form alloy: [3](A) Zn and Pb (B) Fe and Hg (C*) Fe and Cr (D) C and Pt

Question No. 5 & 6 (2 questions) [6]

Q.5 Step C (refining) involved in purification of Pb metal(A) Distillation (B) Bessemerization (C) Cupelation (D*) Electrolytic refining

Q.6 Which of the following metals are obtained by auto reduction method:Pb, Mn, Cu, Cr, Fe, Al.

(A) Cu, Fe (B) Cu, Pb, Mn (C) Mn, Cr, Pb (D*) Pb, Cu

Q.7 Match List-I with List-II and select the correct answer using the codes given below the lists. [3]List–I (Metals) List–II (Ores)

(a) Tin 1. Calamine(b) Zinc 2. Cassiterite(c) Titanium 3. Cerrusite(d) Lead 4. Rutile

(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4

Q.8 Match Column-I with Column-II and select the correct answer using the codes given below .Column-I (Metals) Column-II (Method used for refining)

(i) Iron & copper (P) Poling [3](ii) Zirconium & Titanium (Q) Bessemerisation(iii) Lead & Tin (R) Van-Arkel(iv) Copper & Tin (S) Liquation

(i) (ii) (iii) (iv)(A) P S R Q(B) Q S R P(C) P R S Q(D*) Q R S P

Q.9 On heating a mixture of Cu2O and Cu2S, we get: [3](A*) Cu + SO2 (B) Cu + SO3(C) CuO + CuS (D) Cu2SO3

Q.10 Select incorrect statement regarding silver extraction process. [3](A) When the lead-silver alloy is rich in silver, lead is removed by the cupellation process.(B) When the lead-silver alloy is rich in lead, silver is removed by parke's or pattinson's process.(C*) Zinc forms an alloy with lead, from which lead is separated by distillation.(D) Zinc forms an alloy with silver, from which zinc is separated by distillation.

Q.11 Froth floatation process for the concentration of sulphide ores is an illustration of the practical applicationof: [3](A*) Adsorption (B) Absorption (C) Sedimentation (D) Coagulation

Q.12 When ZnS and PbS minerals are present together, then NaCN is added to separate them in the frothfloatation process as a depressant, because [3](A) Pb(CN)2 is precipitated while no effect on ZnS(B*) ZnS forms soluble complex Na2[Zn(CN)4](C) PbS forms soluble complex Na2[Pb(CN)4](D) They cannot be separated by adding NaCN.

Q.13 When the sample of Cu with Zn impurity is to be purified by electrolysis, the appropriate electrodes are:Cathode Anode [3]

(A) Pure Zn Pure Cu(B) Impure sample Pure Cu(C) Impure Zn Impure sample(D*) Pure copper Impure sample

Q.14 Match List-I (Metal) with List-II (Process of Extraction) and select the correct answer using the codesgiven below the lists: [3]

List-I (Metal) List-II (Process of Extraction)(a) Aluminium (1) Blast furance(b) Iron (2) Mond process(c) Nickel (3) Bayer process(d) Copper (4) Cyanide process

(5) Froth floatation(a) (b) (c) (d)

(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1

Q.1 AgCl on fusion with Na2CO3 forms [3](A) Ag2CO3 (B) Ag2O (C*) Ag (D) Ag2C2

Q.2 Select the correct option for the given processes. [3](i) Process of heating steel to redness and then cooling it very slowly.(ii) Process of heating steel in presence of NH3 and producing hard coating of Iron Nitride on the surface of steel.(iii) Process of heating steel to redness and then cooling it suddenly by plunging it into water or oil.(iv) Process of heating quenched steel to a temperature well below redness and then cooling it slowly.

(A) Tempering, Nitriding, Annealing & Quenching respectively(B) Quenching, Nitriding, Annealing & Case Hardening respectively(C) Tempering, Case harding, Quenching & Annealing respectively(D*) Annealing, Nitriding, Quenching & Tempering respectively

Q.3 A solution of Na2SO4 in water is electrolysed using inert electrodes. The products at cathode and anodeare respectively [3](A) O2 ; H2 (B) O2 ; Na (C*) H2 ; O2 (D) O2 ; SO2

Q.4 If impurity in a metal has a greater affinity for oxygen, then the purification of metal may be carried out by(A) Liquation (B) Distillation (C) Zone Refining (D*) Cupellation [3]

Q.5 Give the correct order of initials T or F for following statements. Use T if statement is true and F if it isfalse. [3](i) In Gold Schmidt thermite process aluminium acts as a reducing agent.(ii) Mg is extracted by electrolysis of aq. solution MgCl2.(iii) Extraction of Pb is possible by smelting.(iv) Red Bauxite is purified by Serpeck's process.(A) TTTF (B) TFFT (C) FTTT (D*) TFTF

Q.6 Among the following groups of oxides, the group containing oxides that cannot be reduced by C to givethe respective metal is [3](A*) CaO and K2O (B) Fe2O3 and ZnO (C) Cu2O and SnO2 (D) PbO and Pb3O4

Q.7 In the cyanide process for extraction of gold and silver from ores, the cyanide solution acts as a [3](A) reducing agent to reduce the gold and silver compounds present in the ores into the metallic states(B*) leaching agent to bring the gold and silver into solution as cyanide complexes and thus separatethese metals from the ores(C) leaching agent to dissolve all the other constituents of the ores leaving the gold and silver as metals(D) leaching agent to bring the ores into solution.

Q.8 By which process Pb and Sn are extracted respectively are: [3](A) Carbon reduction and self reduction(B*) Self reduction and carbon reduction(C) Electrolytic reduction and cyanide process(D) Cyanide process and electrolytic reduction

Q.9 Match the following: [3](I) Bauxite (a) Copper(II) Monazite (b) Uranium(III) Malachite (c) Thorium(IV) Pitch blends (d) Aluminium(A) I-d, II-b, III-c, IV-d (B) I-b, II-c, III-d, IV-a(C) I-c, II-a, III-b, IV-d (D*) I-d, II-c, III-a, IV-b

ASSIGNMENT-27

Q.10 Match List-I with List-II and select the correct answer using the codes given below the lists:List-I (Ore) List-II (Metal) [3]

(1) Carnallite (P) Zinc(2) Calamine (Q) Titanium(3) IImenite (R) Magnesium(4) Chalcopyrite (S) Copper

(1) (2) (3) (4)(A) P R Q S(B) P R S Q(C) R P S Q(D*) R P Q S

Q.11 During the production of iron and steel. [4](A) The oxide ore is primarily reduced to iron by solid coke according to the reaction.

2Fe2O3 + 3C 4Fe + 3CO2(B*) The oxide ore is reduced by the carbon monoxide according to the reaction

Fe2O3 + 3CO 2Fe + 3CO2(C*) Major silica impurities are removed as calcium silicate slag by addition of a fluxing agent lime stone.(D*) The converter slag containing phosphorus is used as a fertilizer.

Q.12 Consider the following metallurgical processes: [4](I) Heating impure metal with CO and distilling the resulting volatile carbonyl (b.p. 43°C) and finally

decomposing at 150°—200°C to get the pure metal(II) Heating the sulphide ore in air until a part is converted to oxide and then further heating in the

absence of air to let the oxide react with unchanged metal sulphide.(III) Electrolysis of the molten electrolyte containing approximately equal amounts of the metal chloride

and NaCl to obtain the metalThe processes used for obtaining magnesium, nickel and copper are respectively:(A) (I), (II) and (III) (B) (II), (III) and (I) (C*) (III), (I) and (II) (D) (II), (I) and (III)

Q.13 What products are formed during, the electrolysis of a concentrated aqueous solution of NaCl? [4](A*) Cl2 (g) (B*) NaOH (aq) (C*) H2 (g) (D) None

Q.14 Column-I (Ore) Column-II (Metal in Ore) [12](A) Ilmenite (P) Iron(B) Dolomite (Q) Magnesium(C) Carnalite (R) Potassium(D) Chromite (S) Titanium

[Ans. (A) P, S (B) Q (C) Q, R (D) P]

ASSIGNMENT-25




XLayer [3]



Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]



0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2













ASSIGNMENT-26











(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4













(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1











ASSIGNMENT-27













[Ans. (A) P, S (B) Q (C) Q, R (D) P]

ASSIGNMENT-25




XLayer [3]



Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]



0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2













ASSIGNMENT-26











(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4













(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1











ASSIGNMENT-27













[Ans. (A) P, S (B) Q (C) Q, R (D) P]

ASSIGNMENT-25




XLayer [3]



Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]



0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2













ASSIGNMENT-26











(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4













(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1











ASSIGNMENT-27













[Ans. (A) P, S (B) Q (C) Q, R (D) P]

ASSIGNMENT-25




XLayer [3]



Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]



0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2













ASSIGNMENT-26











(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4













(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1











ASSIGNMENT-27













[Ans. (A) P, S (B) Q (C) Q, R (D) P]

ASSIGNMENT-25




XLayer [3]



Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]



0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2













ASSIGNMENT-26











(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4













(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1











ASSIGNMENT-27













[Ans. (A) P, S (B) Q (C) Q, R (D) P]

ASSIGNMENT-25




XLayer [3]



Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]



0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2













ASSIGNMENT-26











(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4













(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1











ASSIGNMENT-27













[Ans. (A) P, S (B) Q (C) Q, R (D) P]

ASSIGNMENT-25




XLayer [3]



Q.4 2Mg +O 2MgO

2

2C +O 2CO2

T1 T2 T3

–1200–1100–1000–900–800–700–600–500–400–300–200–100

0

G

(kJ/

mol

)

Temperature

[3]



0

–1000–800–600–400

–200

0

500 1000 1500 2000G

°,kJ

m.p. b.p.

b.p.m.p.

2Mg+O 2MgO2

2C+O 2CO2 2Zn+O2ZnO

2













ASSIGNMENT-26











(a) (b) (c) (d) (a) (b) (c) (d)(A) 1 2 3 4 (B*) 2 1 4 3(C) 4 3 2 1 (D) 2 1 3 4













(A) 2 5 4 1(B*) 3 1 2 5(C) 2 1 4 5(D) 3 5 2 1











ASSIGNMENT-27













[Ans. (A) P, S (B) Q (C) Q, R (D) P]

Solubility chart of ionic compounds in aqueous solution.

1. All the cuprous salts are insoluble.2. Sulphides are usually insoluble, leaving those of alkali metals and ammonium sulphide, which are soluble.

Sulphides of alkaline earth metals and of ammonium are decomposed in aqueous solution.3. Compounds of alkali metals and ammonium salts are generally soluble.4. 2

3CO , OH–, 24CrO , 3

4PO & 23SO are found generally insoluble, leaving those of alkali metals and

ammonium cation, which are soluble.

Cation Anion Solublility Exception

Any

OCl–, 2ClO ,

3ClO , 4ClO ,

2NO , 3NO

CH3COO–, 3HSO ,

3HCO , HS–

All are soluble (i) 4ClO of K+, Rb+, Cs+

are insoluble. (ii) AgNO2 is insoluble (iii) CH3COOAg is partially soluble.

K+, Rb+, Cs+ &

4NH Any All are soluble [Pt Cl6]–2 &

[Co(NO2)6]3–

are insoluble

Na+ Any All are soluble ----- Any Cl–, Br–, I– All are soluble (i) Ag+, Pb2+, 2

2Hg , 2

2Cu are insoluble (ii) PbCl2, CuBr2, Cu2I2 are soluble on warming (iii) HgBr2 & HgI2 remain insoluble on warming.

Any 24SO All are soluble (i)Ba+2 ,Sr2+, Pb2+ are

insoluble

(ii) CaSO4, Ag2SO4 , SnSO4 & HgSO4 are partially soluble

Any 23SO , 2

3CO , 2

42OC , 34PO ,

O–2 ,OH– , F–

All are insoluble

(i) BeF2 & AgF are soluble (ii) K+ to Cs+ &

4NH are soluble (iii) O–2 & OH– of 2Sr & Ba+2 are soluble

Any CN–, OCN–,SCN–, S–2

All are insoluble

I group cation & II group cation are soluble

Any 24CrO is similar to 2

4SO Same as 2

4SO Same as 24SO but

SrCrO4 is soluble Any 1

4MnO is similar to

4ClO Same as

4ClO NaMnO4, KMnO4 are soluble

ASSIGNMENT-28

ASSIGNMENT-29

COLOURS OF DIFFERENT COMPOUNDS

Black Colour Compounds1. PbS2. Ag2S3. CuS4. Cu2S5. NiS6. CoS7. HgS8. FeS9. NiO10. MnO11. FeO12. CuO13. PbO214. MnO215. Mn3O416. Fe3O417. Co3O418. Ni(OH)319. Cu3P220. BiI321. Hg + Hg(NH2)Cl

Blue Colour Compounds

(a) Light Blue Compunds1. Cu(OH)22. Cu(NO3)23. CuCl24. CuSO4·5H2O (Blue Vitriol)5. Co(OH)2

(b) Deep Blue Compounds1. [Cu(NH3)4]SO4(Switzer’s reagent)2. [Cu(NH3)4](NO3)23. Fe4[Fe(CN)6]3 (Prussian’s blue)4. Fe3[Fe(CN)6]2 (Turnbull’s blue)5. Na4[Fe(CN)5(NOS)](Violet)

Green Colour Compounds1. Ni(OH)2 (green ppt.)2. Hg2I2 (green ppt.)3. Cr2O3 (green solid)4. Cr(OH)3 (green ppt.)5. Cr2(SO4)36. CrCl37. FeSO4 · 7H2O8. FeCl29. FeSO4· (NH4)2SO4·6H2O(Mohr’s salt)10. Na2MnO411. K2MnO412. B(OC2H5)3 (Burns with green edge flame)13. CoO · ZnO (Riemann’s green)White Colour Compounds1. AgCl2. Ag2CO3 (white)3. AgNO24. Ag2SO45. PbCl26. PbBr2 (White crystalline ppt.)7. Cu2I2 (White ppt.)8. PbSO49. SrSO410. BaSO411. Hg2SO412. BaCO313. SrCO314. CaCO315. MgCO316. Be(OH)217. Mg(OH)218. Ca(OH)219. Al(OH)3 (White gelatinous ppt.)20. Cd(OH)2 (Dirty White)21. Bi(OH)3 (White ppt.)22. Zn(OH)223. ZnCO324. ZnS25. ZnO26. CaC2O427. Ag2C2O428. Ag2S2O329. PbS2O330. Zn2[Fe(CN)6] (white ppt.)

Yellow Coloured Compounds1. As2S32. As2S53. CdS4. SnS2(Artificial gold)5. FeS2(Fool’s gold)6. (NH4)2Sx (where x = 2 to 5)(YAS)7. PbCrO48. BaCrO49. SrCrO410. AgBr (light yellow)11. AgI (Dark yellow)12. PbI213. PbO (in Cold)14. ZnO (in Hot)15. HgO (Yellow ppt.)16. Na2O2 (Pale yellow)17. Ag3PO418. Ag3AsO319. Cu(CN)220. K3[Co(NO2)6] (Fischer’s salt)21. (NH4)3PO4· 12MoO322. (NH4)3 AsO4 · 12 MoO323. Na2CrO424. CrO4

2– (Yellow in solution)

Red Colour Compounds1. Ag2CrO4 (Brick red)2. Hg2CrO4 (Brick red)3. HgI2 (Scarlet red)4. Pb3O4 (2PbO + PbO2)5. CrO2Cl2 (Reddish brown)6. Fe(CH3COO)3 (Blood red)7. Fe(SCN)3 (Blood red)8. AsI39. SbI310. SnI211. CuBr212. [Ni(DMG)2] (Rosy red)

Brown Coloured Compounds1. SnS2. Bi2S33. CdO4. PbO25. Fe(OH)3 (Reddish Brown)6. Fe2O3 (Reddish Brown solid)7. Fe2(CO3)38. Cu2O (Reddish Brown)9. Ag3AsO4 (Reddish Brown)

10. O NH2I

11. Cu2I2 +3I (Brown ppt.)

12. Cu2[Fe(CN)6] (Chocolate brown)13. NO2 (Brown gas)14. [Fe(H2O)5 (NO)]SO4 (Brown ring)

Orange Coloured Compounds1. Sb2S32. Sb2S53. KO34. CsO2

5. 272OCr (Orange in aq. solution)

Pink Coloured Compounds1. Mn(OH)22. MnS

3. 4MnO (Pink or purple in aq. solution)

4. CO(CN)25. (NH4)2 SnCl66.COCl2·6H2O (Pink on melting black also known as

sympathetic ink)

Gas1. Colourless odourless gas – CO2, N2, O22. Colourless gas having pungent smell

NH3, H2S, SO23. Coloured gas —— Cl2 (Yellowish green)

Br2 (Brown)NO2 (Brown)I2 (Violet fumes)

Choose the correct option. (Only one option is correct):Q.1 Which of the following pair of compounds cannot co-exist in aqueous solution.

(A) Na2CO3, K2CO3 (B*) NaHCO3, NaOH (C) Rb2CO3, KHCO3 (D) NaOH and KCl

Q.2 23CO and 2

3SO cannot be distinguished by(A) H2O2 solution (B) lime water (C) Cl2 water (D*) all of these

Q.3 Ag2CO3

A & B are(A) dil. HNO3 and boiling in water respectively.(B*) Direct heating and boiling in water respectively.(C) dil. HNO3 and NH3 respectively.(D) hot water and NH3 respectively.

Q.4 Calcium hydride, also known as hydrolith, is hydrolysed by water to produce(A*) Ca(OH)2 + H2 (B) CaO + O2 (C) CaO + H2 (D) does not hydrolyse

Q.5 Which of the following options is correct

I. HgCl2 can be used for the distinction of 3HCO and 2

3COII. Bisulphite solution is neutral towards litmus paper and pH is 7III. Aq. suspension of Ag2CO3 and Ag2SO3 both produces Ag2O on heatingIV. SO2 acts as reducing agent when passed through FeCl3 solution(A) FFTT (B) TFTT (C*) TTFT (D) FTFF

Q.6 Which of the following produces colourless gas with dil. H2SO4(A) Na2S2O3 (B) NaHCO3 (C) CH3CO2NH4 (D*) All

Q.7 Which of the following pair of salt produces odourless gas with dil. H2SO4

(A) 3HCO and

3HSO (B*) 3HCO and 2

3CO

(C) 232OS and

23COCH (D) 23CO and

23COCH

Q.8 Na2SO3 and Na2S2O3 cannot be distinguished by(A) addition of Pb(OAc)2 followed by heating(B) addition CaCl2 solution(C*) addition of AgNO3 followed by heating(D) All of these

Q.9 The colour of KMnO4/Hwill be decolourised by

(A) S2– solution (B) 23SO solution (C*) both (A) and (B) (D) None

ASSIGNMENT-30

Q.10 Comment True (T) or False (F) for the following statements.(I) Same gas comes out when Na2S is treated with dil. HCl and Na2SO3 is treated with

(Zn + dil. H2SO4)(II) Same observation when acidified nitrite solution is treated with urea or thiourea followed by

addition of FeCl3 solution.(III) Same observation when nitrite solution or acetate solution is treated with dil. H2SO4 solution(IV) With CaCl2 and BaCl2 both acetate and formate do not react.(A) FTTF (B) TFTT (C*) TFFT (D) FTTT

Question No. 11 to 15 (5 questions)Questions given below consist of two statements each printed as Assertion (A) and Reason (R); whileanswering these questions you are required to choose any one of the following four responses:(A) if both (A) and (R) are true and (R) is the correct explanation of (A)(B) if both (A) and (R) are true but (R) is not correct explanation of (A)(C) if (A) is true but (R) is false(D) if (A) is false and (R) is true

Q.11 Assertion : Initially there is no ppt. when AgNO3 is added to Na2SO3 solutionReason: Localised formation of Ag2SO3 is destroyed by soluble complex formation like [AgSO3]–

[Ans. A]Q.12 Assertion : NaHCO3 is the least soluble alkali bicarbonate.

Reason : massive H-bonding present in solid NaHCO3. [Ans. A]Q.13 Assertion : LiHCO3 cannot exist in solid form.

Reason : Li2O has the highest thermal stability among the alkali metals oxide. [Ans. B]

Q.14 Assertion : CO2 + K2Cr2O7 no reactionReason : C is already in maximum oxidation state [Ans. A]

Q.15 Assertion : When 3HCO and 2

3CO ions are present together, CaCl2 is to be used in excess to distinguished both

Reason : Excess amount of CaCl2 forms soluble complex with 3HCO of the formula of

463 ])HCO(Ca[ [Ans. C]

Choose the correct option. (Only one option is correct):Q.1 Soda extract is useful when given mixture has any insoluble salt, it is prepared by

(A) Fusing soda and mixture and then extracting with water(B) Dissolving NaHCO3 and mixture in dil HCl(C) Boiling Na2CO3 and mixture in dil HCl(D*) Boiling Na2CO3 and mixture in distilled water.

Q.2 Which of the following salts will not produce any observable changes when H2S is passed through itsaqueous solution(A) (CH3CO2)2Pb (B*) Na2[Fe(CN)5NO] (C) AgNO3 (D) none of these

Q.3 Choose the correct option for the following statementI. SnCl4 gives the chromyl chloride testII. Layer test is not applicable for chloride ionIII. In the layer test, if red colour comes first, it is confirmed that iodide is absent

IV. For the removal of I– from a mixture of I– + 3NO , AgNO3 can be used and then

3NO can betested

(A*) FTTF (B) FTFT (C) FFTF (D) TTTF

Q.4 In the K2Cr2O7 solution when alkali solution of BaCl2 is added, the yellow ppt. obtained is of(A) BaCr2O7 (B*) BaCrO4 (C) BaCrO4·2H2O (D) none

Q.5 .pptBrown)Bsolution(colourless(A) substance yellow a ofsolution .Aqaddedsolution

ClNH

.)alk(madesolution.conc

KI 4

Compound (A) is(A) PbI2 (B) AgI (C*) HgI2 (D) Hg2I2

Q.6 (X) KOH (Y) (gas turns red litmus blue) +(Z) KOHZn (Y) (gas)

(X) gas (supports in combustion)Identify (X) to (Z).(A) X = NH4NO2 Y = NH3 Z = KNO2

(B) X = (NH4)2Cr2O7 Y = NH3 Z = Cr2O3

(C) X = (NH4)2SO4 Y = NH3 Z = K2SO4

(D*) X = NH4NO3 Y = NH3 Z = KNO3Q.7 The co-ordination number of central ion of the complex obtained in the Na-nitroprusite test of sulphide

ion is(A) 5 (B*) 6 (C) 7 (D) 4

Q.8 BeC2O4 and BaC2O4 are heated separately with bunsen burner. The solid residue obtained arerespectively(A) BeO and BaO (B*) BeO and BaCO3 (C) BeCO3 and BaO (D) BeCO3 and BaCO3

ASSIGNMENT-31

Q.9 Unknown salt 'A' + solid K2Cr2O7 + conc. H2SO4 Reddish brown fumes.Which is the correct statement regarding the above observation(A) It confirms the presence of Cl– ion(B) It confirms the presence of Br– ion(C) It confirms the presence of both(D*) It neither confirms Cl– nor Br– unless it is passed through NaOH solution

Question No. 10 to 12 (3 questions)Questions given below consist of two statements each printed as Assertion (A) and Reason (R); whileanswering these questions you are required to choose any one of the following four responses:(A) if both (A) and (R) are true and (R) is the correct explanation of (A)(B) if both (A) and (R) are true but (R) is not correct explanation of (A)(C) if (A) is true but (R) is false(D) if (A) is false and (R) is true

Q.10 (A) 3NO and

2NO both do not give brown fumes with dil. H2SO4

(R) Protonation of 2NO is more easier compared to

3NO since 3NO is more stable by resonance

[Ans. D]Q.11 (A) Oxalate evolves gas with conc. H2SO4 in presence of MnO2

(R) MnO2 acts as catalyst over here [Ans. C]

Q.12 (A) In the brown ring compound, Fe is in the +1 oxidation state(R) Spin only magnetic moment found for this compound is 3.87 BM [Ans. A]

Question No. 13 to 15( 3 questions)Acetic acid is added to the solution of sodium carbonate the gas evolved does not turn purple colour ofKMnO4 but turns lime water milky forming a compound (M) which becomes soluble by passing thesame gas in excess forming another compound (N). But same observation is not obtained with boricacid

Q.13 Purple colour of KMnO4 has not changed because(A) the gas has no oxidising property since central atom is with minimum oxidation state(B*) the gas has no reducing property since the central atom is with maximum oxidation state(C) the gas has no precipitation characteristics(D) the gas precipitates CaCO3 from lime water

Q.14 The compound formed in the above sequence (M) and (N) are respectively(A) water soluble CaCO3 and water soluble Ca(HCO3)2(B) water insoluble CaCO3 and water insoluble Ca(HCO3)2(C*) water insoluble CaCO3 and water soluble Ca(HCO3)2(D) water soluble CaCO3 and water insoluble Ca(HCO3)2

Q.15 The correct order of increasing acidity is(A) Boric acid < Acetic acid < Carbonic acid (B) Acetic acid < Boric acid < Carbonic acid(C) Carbonic acid < Acetic acid < Boric acid (D*) Boric acid < Carbonic acid < Acetic acid

Choose the correct option. (Only one option is correct):Q.1 In the reaction sequence:

CrCl3 OHNH

ClNH

4

4 (A) lnsoONa 22 (B) acetate

lead (C). In this reaction sequence compound C is

(A) Na2CrO4 (B) Na2Cr2O7 (C) Cr(OH)3 (D*) PbCrO4

Q.2 Which is not easily precipitated from aqueous solution

(A) Cl– (B) 24SO (C*)

3NO (D) 23CO

Q.3 A sodium salt on treatment with MgCl2 gives white precipitate only on heating. The anion of the sodiumsalt is:

(A*) 3HCO (B) 2

3CO (C) 3NO (D) 2

4SO

Q.4 Which reaction is possible(A) KMnO4 + Na2SO4 (B) BaSO4 + KCl (C) SrSO4 + Ni(NO3)2 (D*) ZnSO4 + BaS

Q.5 Give the correct order of initials T or F for following statements. Use T if statement is true and F if it isfalse.

(i) Cu+ undergoes disproportionation to Cu and Cu2+ in aq. solution(ii) Hg2Cl2 does not impart chromyl chloride test(iii) [Fe(H2O)5NO]2+ complex is highly unstable due to the presence of weak field ligand(iv) Bond length of CO+ is greater than CO

(A) TFFT (B) TTFT (C) FTTF (D*) TTFF

Q.6 Fe2+ and Fe3+ can be distinguished by(A) K3[Fe(CN)6] (B) K4[Fe(CN)6] (C) KSCN (D*) All

Q.7 Mg3N2 OH2 A(gas) CuO

B + C(g) + H2O

C(g) can be obtained by heating(A) (NH4)2SO4 (B*) (NH4)2Cr2O7 (C) NH4NO3 (D) NH4Cl

Q.8 When K2Cr2O7 is treated with H2O2 in acidic medium, in presence and in absence of org. solvent. Theoxidation state of chromium finally:(A) increases and decreases respectively (B) decreases and increases respectively(C*) retained same and decreases respectively (D) retained same in both cases.

Q.9 Which of the following acid radical does not produce white ppt with Pb(OAc)2 solution.

(A) Br– (B) 232OS (C) 2

3CO (D*) None of these

Q.10 23CO and

3HCO can be distinguished by(A) Phenolphthalein (B) BaCl2 solution (C) HgCl2 solution (D*) All of these

ASSIGNMENT-32

Q.11 H2S is passed into BaCl2 solution. The ppt formed is(A) BaS (B) Ba(SH)2 (C) Ba(OH)2 (D*) no ppt.

Q.12 Colour obtained in test of 232OS + [Ni(en)3] NO3 is

(A) Red (B*) Violet (C) Indigo (D) Yellow

Q.13 Which of following is / are reduced by thiosulphate solution.(I) Fe+3 solution (II) I2 solution (III) Cu+2 solution (IV) Hg2+

(A*) I, II, III only (B) I, III only (C) IV only (D) I, III, IV only


Aq. solution of ‘A’

Solid ‘A’ heating

on anh. ‘A’’

Q.14 ‘A’ is(A) K2S2O3 (B) CaS2O3 (C) PbS2O3 (D*) Na2S2O3

Q.15 ‘A’ on strong heating produces compound(s) has/have(A) chain structure (B) Tetrahedral structure(C*) both (D) none

Choose the correct option. (Only one option is correct):

Q.1 Na2S2O3. 5H2O is heated strongly to produce M and N and water. Both are consisting of S. Mentionthe average oxidation state of 'S' in M and N respectively.

(A) –2, +5 (B) –2 ,+ 4 (C*) 52

, +6 (D) –2, +6

Q.2 A compound (X) on decomposition gives a colourless gas. The residue is dissolved in water to obtain(Y). Excess CO2 is bubbled through aqueous solution of (Y) and (Z) is formed. (Z) are gentle heatinggives back (X). The (X) is(A*) CaCO3 (B) Ca(HCO3)2 (C) NaHCO3 (D) Na2CO3

Q.3 BaCO3(s) + AcOH

SolutionOCNa 422 ?

Comment on the product of this reaction.(A) BaCO3 remains unaffected.(B) BaC2O4 will be precipitated as white precipitate(C) Ba(OAc)2 will be precipitated as white precipitate(D*) Clear solution

Q.4 Statement-1: Brown ring test can be done for 3NO in presence of

2NOStatement-2: Oxidation state of iron is changing from +2 to +1 in the brown ring complex.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D*) Statement-1 is false, statement-2 is true.

Q.5 Statement-1: 4NH and K+ cations can be distinguished by using Nesseler's reagent followed by

adding HClO4.Statement-2:

4NH gives brown ppt. with Nesseler's reagent where as K+ forms white ppt. of KClO4

with perchloric acid.(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

ASSIGNMENT-33

Question No. 6 & 7 (2 questions)

Q.6 Water of crystallisation in compound P(A*) 4 moles (B) 7 moles (C) 10 moles (D) 5 moles

Q.7 Blue compound Q is(A) Co(BO2)2 (B*) NaCoPO4 (C) CoO·Al2O3 (D) CoO·MgO


Q.8 Compound (D) + I– + H+ — GasEvolved gas is similar to(A) Gas-B (B) Gas-G (C*) Gas-H (D) None

Q.9 Yellow ppt. of compound (I) is insoluble in(A) NaOH (B*) CH3COOH (C) dil. HNO3 (D) None

Q.10 Type of hybridization of complex (E)(A*) sp3d2 (B) d2sp3 (C) sp3 (D) dsp2

Q.11 Type of hybridization of central atom of gas (B)(A) sp (B*) sp2 (C) sp3 (D) No hybridization

Q.12 Match the column-I with column-II.Note that column-I may have more than one matching options in column-II.

Column - I Column -II(A) Sodium nitroprusside (P) = 0 B.M.(B) Brown ring complex (Q) octahedral

(C) Complex of Ag formed during its extraction (R) = 15 B.M.(D) Potassium ferrocyanide (S) NO+ ligand

[Ans. (A) P,Q,S, (B) Q,R,S, (C) P, (D) P,Q]

Choose the correct option. (Only one option is correct):Q.1 Match the following list and choose the correct option.

List ‘A’ List ‘B’(I) HgI2 (a) Yellow ppt obtained when H2S is passed in its dil. HCl solution(II) Na+ (b) Yellow solution(III) FeCl3 (c) Yellow flame(IV) As2O3 (d) Yellow sublimate(A) I-(a), II-(c), III-(d), IV-(b) (B*) I-(d), II-(c), III-(b), IV-(a)(C) II-(d), III-(c), I-(b), IV-(a) (D) II-(d), I-(c), III-(b), IV-(a)

Q.2 ‘A’ (white substance) swells up first heating

strong contracted into amorphous powder..

A is(A) Na2B4O7.10H2O (B) Na2B4O7(C*) K2SO4.Al2(SO4)3.24H2O (D) Na(NH4)HPO4.4H2O

Q.3 Aqueous solution of A + AcOH + K2CrO4 yellow ppt. The above information is not correct forwhich of the following cations:(I) Pb+2 (II) Ba2+ (III) Ca2+ (IV) Sr2+

(A) I & III (B) I, II & IV (C) I, II (D*) III, IV

Q.4 Calcium imide on hydrolysis will give gas (B) which on oxidation by bleaching powder gives gas (C) gas(C) on reaction with magnesium give compound (D). (D) on hydrolysis gives again gas (B). (B), (C) and(D) are(A*) NH3, N2, Mg3N2 (B) N2, NH3, MgNH(C) N2 , N2O5 , Mg(NO3)2 (D) NH3, NO2 , Mg(NO2)2

Q.5 The compound present in brorax bead is(A) B2O3 (B) NaBO2 (C) NaBO3 (D*) NaBO2 + B2O3

Q.6 When NH4OH is added in Hg2(NO3)2 solution, the ppt formed is(A) Hg2O (B) Hg + HgO(C*) 2Hg + HgO·Hg(NH2)NO3 (D) HgO·Hg(NH2)NO3

Q.7 A reddish pink substance on heating gives off a vapour which condenses on the sides of the test tube andthe substance turns blue. It on cooling water is added to the residue it turns to its original colour. Thesubstance is(A) Iodine crystals (B) Copper sulphate crystals(C*) Cobalt chloride crystals (D) Zinc oxide

ASSIGNMENT-34

Q.8 Brown ppt. (A) dissolve in HNO3 gives (B) which gives white ppt (C) with NH4OH. (C) on reactionwith HCl gives solution (D) which gives white turbidity on addition of water. What is (D).(A*) BiCl3 (B) Bi(OH)3 (C) BiOCl (D) Bi(NO3)3

Q.9 When conc. H2SO4 was treated with K4[Fe(CN)6], CO gas was evolved. By mistake, somebody useddilute H2SO4 instead of conc. H2SO4, then the gas evolved was(A) CO (B*) HCN (C) N2 (D) CO2

Q.10 An inorganic Red colour compound (A) on heating, gives a compound (B) and a gas (C). (A) ontreatment with conc. HNO3, gives compound (D), brown colour substance (E) and a neutral oxide (F).Compound (D) on warming gives off again gas (C). Then, (E) will be:(A) Mn3O4 (B*) PbO2 (C) Pb3O4 (D) Fe2O3

Q.11 In this sequence X, Y, Z are respectively(A) Acidified H2O2 ; Alkaline H2O2 ; Acidified H2O2(B) Alkaline H2O2 ; Acidified H2O2 : Zn / HCl(C) Acidified H2O2 ; Heat ; Alkaline H2O2(D*) Alkaline H2O2

; Acidified H2O2 ; On standing

PHYSICAL CHEMISTRY ASSIGNMENTS

ASSIGNMENT-1

Q.1 50 gm ethylamine reacts with 60 gm 2ONNa

& 30 gm HCl. Where N is N15 (isotope). After the

completion of reaction, gas is evolved. Then select correct option. [3](A) Pure 10 gm N2 evolved out(B) 10 gm of hybrid N2 evolved out(C) 5 gm C2H5OH is formed(D*) None of these

Q.2 Which of the following mixtures can be regarded as buffer ? [3](A) 500 ml of 0.2 M AcOH + 1000 ml 0.2 M NaOH(B*) 500 ml of 0.2 M AcOH + 300 mL of 0.2 M NaOH(C*) 0.2 M Na2B4O7 · 10H2O(D) 500 ml of 0.2M AcOH + 500 ml 0.2 M NaOH

Q.3 If the value of Kp for the reaction [5]2H2S(g) 2H2(g) + S2(g) is 50% of total pressure at equilibrium than show that degree of

dissociation of H2S will be less than 66.6% but more than 50%.Q.4 100 gm Ca2B6O11 is mixed with 50gm Na2CO3. The resultant is filtered and washed. The mother liquor

collected and CO2 gas is passed through it. Find out the total weight of sodium compound formed in thesolution. [7]

[Ans. 71 gm ]Q.5 Separate the extensive and intensive variables enthalpy, refractive index, density, visocity, volume, pressure,

dipolemoment, heat capacity, elevation in BP, chemical potential universal gas constant R, molar volume,vapor pressure. [5][Ans. Intensive : ref index, density, visocity, pressure, dipole, elevation in boiling point chemical point, R,molar vol, vapor pressure ]

Q.6 FeS2 on oxidation by air yields Fe2O3 and SO2. If the equivalent of oxygen used in the reaction is X thanhow many equivalent of Fe2O3 and SO2 are formed after the reaction [5]

[Ans. 10x/11 ]

Q.7 Certain amount of ion X reacts with cyclopentadiene in basic solution and forms complex Y. X is obtainedby the reaction of FeCl3 with Na2S2O3. If 100 gm of ion X reacts with 200 gm of cyclopentadiene inbasic solution. How much Y is obtained if reaction efficiency is 63 %. Also comment in the symmetry ofY & its stability. [7]

[Ans. 0.9689 mol. ]

Q.8 A current of 1.7 A is passed through 300 mL of 0.160 M solution of ZnSO4 for 230 sec. with currentefficiency of 90%. Find the molarity of Zn2+ after the deposition of Zn. Assume volume remains constant.

[5][Ans. 0.154 M]

Q.9 How much amount of Fe0.93 O1.00 is completely oxidized by 50 ml 0.02 M Ba(MnO4)2 in presence ofnon oxidising acid. [5]

[Ans. 0.86 gms ]Q.10 Consider the following reaction,

As2S3 + NO3¯ AsO43– + SO4

2– + NOIf number of equivalent of NO3¯ used is X. Calculate the number of equivalent of AsO4

3– & SO42–

formed. [5][Ans. 24x/28 ]

ASSIGNMENT-2

Q.1 Which of the following statement are true. [3](A) CV is independent of temperature for a perfect gas.(B) It neither heat nor matter can enter or leave a system, that system must be isolated.

(C*) dF = 0 where F = E, H, G, A, S, C(D) A process in which final temperature equals to initial temperature must be an isothermal process.

Q.2 Predict the sign of q, w, U & H for given process. [2 × 7 = 14]

q w DU DHMelting of solid benzene at 1 atm and normal melting point. > 0 < 0 > 0 > 0

Melting of ice at 1 atm and 0°C. > 0 > 0 > 0 > 0

Adiabatic expansion of one mole of ideal gas. 0 < 0 < 0 < 0

Adiabatic expansion of ideal gas into vacuum. 0 0 0 0

Isothermal expansion of a perfect gas. > 0 < 0 0 0

Heating of perfect gas at constant P. > 0 < 0 > 0 > 0

Cooling of perfect gas at constant volume. < 0 0 < 0 < 0

Sign ofProcess Remark

Q.3 Which of the statement are true. [5]

(A*) CP – CV = PT T

VPVE

(B*) CP – CV = VT T

PPHV

(C*) CP – CV = nR for gases only (D*) TV

H

= 0 for ideal gas

Q.4 Select the False statement [5](A*) E = q + w for every thermodynamic system at rest in the absence of external field.(B*) q = 0 for every cyclic process(C) In isothermal free expansion q = 0, w = 0, E = 0, H = 0

(D) TP

H

= Vm where Vm is molar volume of solid, liquid or gas.

Q.5 Which of following statement are true. [5](A) (–w)rev – (–w)irrev > 0 for isothermal expansion.(B) Heat absorbed in isothermal reversible process is greater than isothermal irreversible process.(C) (w)irev > (w) rev for isothermal compression.(D) Heat rejected in isothermal process is greater in (n–1)th stage as compare to nth stage work.

Q.6 Two moles of an ideal monoatomic gas undergoes cyclic process ABCA as shown in figure. Find theratio of temperature at B & A. [5]

[Ans. 27/4]

60° 30°

B

A C

V0 6V0

3P0

P0

Q.7 The average degree of freedom per molecule for a gas is 6. The gas perform 25 J of work when itexpands at constant pressure. Find the heat absorbed by the gas. [5]

[Ans. 100 J]

Q.8 A transition metal X forms an oxide of formula X2O3. It is found that only 50% of X atoms in thiscompound are in +3 oxidation state. The only other stable oxidation state of X are +2 and +5. What %of X atoms are in the +2 oxidation state in this compound. [5]

ASSIGNMENT-3Q.1 The minimum work which must be done to compress 16 gm of oxygen at 300 K from a pressure of

1.01325 × 103 N/m2 to 1.01325 × 105 N/m2 is : [3](A*) 5744 J (B) 8622 J (C) 3872 J (D) 7963 J

Q.2 An ideal gaseous sample at initial state i (P0, V0, T0 ) is allowed to expand to volume 2V0 using twodifferent process, in the first process the equation of process is 2PV2 = K1 and in second process theequation of the process is PV = K2. Then [3](A) Work done in first process will be greater than work in second process (magnitude wise)(B) The order of values of work done cannot be compared unless we know the value of K1 and K2(C*) Value of work done (magnitude) in second process is greater in above expansion irrespective ofthe value of K1 and K2.(D) 1st process is not possible.

Q.3 Select the correct statement. [3](A*) When equilibrium is established macroscopic properties do not change with time.(B*) At equilibrium the system should be at rest without the assistance of any external source.(C) Steady state and equilibrium state is same(D*) Steady state is dynamic on macroscopic as well as microscopic level but equilibrium is dynamic onmicroscopic level only.

Q.4 Internal energy includes. [3](A*) Rotational energy (B*) Bond energy (C*) Relativistic energy (D*) Vibrational energy

Q.5 In the PV-diagram shown in figure, ABC is a semicircle. Find the workdone in the process ABC. Where P is in atm and V in litres. [4] P

3

1

1 2 V

A

B

C

[Ans. /2 ]Q.6 P–V graph for an ideal gas undergoing polytropic process PVm = constant is shown is here. Find the

value of m. [4]

37°

2×10 (Pa)5

4×10 (m )5 3

P

V

[Ans. 3/2 ]

Q.7 A portion of helium gas in a vertical cylindrical container is in thermodynamic equilibrium with thesurroundings. The gas is confined by a movable heavy piston. The piston is slowly elevated by a distanceH from its equilibrium position and then kept in the elevated position long enough for the thermodynamicequilibrium to be reestablished. After that, the container is insulated and then the piston is released. Afterthe piston comes to rest, what is the new equilibrium position of the piston with respect to initial position?

[Ans. The piston ends up 0.4 H above its initial position.][4]Q.8 An ideal gas has a molar heat capacity CV at constant volume. What is molar capacity of this gas in a

function of volume if gas undergoes process T = T0eV. [4]Q.9 A 10 g mixture of isobutene and isobutane requires 20 g of Br2 (in CCl4) for complete addition. If 10 g

of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the presence oflight at 127°C, which exclusive product and how much of it would be formed? (Atomic weight ofbromine= 80) [Ans. 24.51 g C4H9Br] [4]

Q.10 How much (in gms) of AlCl3 (solid) must be added to 250 ml 0.01 M HCl solution so that Cl¯ ionconcentration is increased to 220 %. [4]

[Ans. 0.1335 gms.]

ASSIGNMENT-4

ExplainQ.1 How was the absolute zero of temperature determined ?Q.2 Can a tyre be inflated without a rise in temperature, If yes, how ?

Q.1 2 mol of 4 atoms combine to form 1 mol of H2 molecule. If the internal energy change for the process is–432.6 kJ. What will be the theoretical decrease in mass for the process? [3](A) – 3 × 10–2 kg. (B*) – 4.8 × 10–12 kg (C) – 1.4 × 10–12 kg (D) – 1.4 × 10–6 kg

ComprehensionConsider the system

CH (g) +4C H (g)4 10

4CO (g)+5H O(g)2 2

Rigid impermeable cylinder

Massless Frictonless Adiabatic piston

O (g)2+2O (g)2

CO (g)+2H O(g)2 2

213

The temperature is maintained at 400 K and the complete combustion of methane and butane withrequisite amount of O2 is done., as shown. If initially the piston was present at the centre of the cylindershown above. Given : Total volume of cylinder = 10 litre R = 8.3 mol/k.

Q.2 Calculate the final position of the piston after the combustion has taken place, if c/s area ofcylinder = 1 dm2. [3](A) 25 cm from Right (B) 75 cm from Left (C*) 75 cm from Right (D) 125 cm from Left

Q.3 Calculate the net work done during the process ? [3](A) + 6.9 kJ (B) –12.11 kJ (C*) – 5.21 kJ (D) + 0.42 kJ

Q.4 A vessel containing 1 mol of an ideal gas with Pi = 1 bar and Cpm = 25

R is in thermal contact with a

water bath. Treat the vessel, gas and water bath as being in thermal equilibrium initially at 298 k, and asseparated by adiabatic walls from the rest of the universe. The vessel gas and water bath have anaverage heat capacity of Cp = 7500 J/K. The gas is compressed reversibly to Pf = 10.5 bar. What is thetemperature of the system after thermal equilibrium has been established? [3](A*) 299.2 K. (B) 763.2 K. (C) 479.2 K (D) 182.9 K

Q.5 Which of the following can be considered true in regard to quasi-static process? [5](A*) An example of quasi-static process can be the movement of heat from hot body to cold body

separated through an almost vacuum.(B*) The system undergoes a major change in terms of a directed path along a sequence of states in

which the system and surrounding are in internal equilibrium.(C*) Reversible and irreversible processes are two classes of quasi-static process(D) A quasi-static process can only be plotted on a P-V-T curve.

Q.6 A bottle at 21°C contains an ideal gas at a pressure of 126.4 × 103 Pa. The rubber stopper closing thebottle is removed. The gas expands adiabatically against Pexternal = 101.9 × 103 Pa and some gas isexpelled from the bottle in the process when P = Pexternal , the stopper is quickly replaced. The gasremaining in the bottle slowly warms up to 21°C what is the final pressure[in Pa] in the bottle for amonoatomic gas. [3]

[Ans. 110.5 × 103 Pa.]

Q.7 If a monoatomic ideal gas undergoes the cyclic process as shown below. [5]

12

P

V

43

(a) Calculate the efficiency of this cycle if 1

3

VV

= 2.

(b) Plot the same cycle on P–T and V–T graph. [Ans. (a) 68.35 %]

Q.8 One mole of N2 in a stale defined by Ti = 300 k and Vi = 2.5 litre undergoes an isothermal reversibleexpansion until Vf = 23 litre, What is the percentage error, in calculation of work, using the ideal gasequation instead of the Vander Waals equation. Given a = 1.370 dm6 bar mol–2, b = 0.0387 dm3mol–1.

[5][Ans 0.25%]

Q.9 Regard H as a function of T and P use the Cyclic rule to obtain the expression. [3]

HTp P

TPHC

Q.10 U is a state function and obeys Euler’s reciprocity theorem. Using this statement show that 0VC

T

v

for an ideal gas. [3]

ASSIGNMENT-5Q.1 For a perfectly crystalline solid Cp,m = aT3, where a is constant. If Cp,m is 0.42 J/K-mol at 10 K, molar

entropy at 10 K is [3](A) 0.42 J/K-mol (B*) 0.14 J/K-mol (C) 4.2 J/K-mol (D) zero

Q.2 Entropy change for process at 1 atm and – 10°C, )s(OH)(OH 22 l is [3][Given: Hfusion(0°C) = – 6000 J/mol, Hfusion(–10°C) = – 5523 J/mol

Cp,m(H2O, s) = 36 J/K-mol, Cp,m(H2O, l) = 75 J/K-mol, ln

263273

= 0.04]

(A) – 23.47 J/K-mol (B*) – 20.41 J/K-mol(C) 21 J/K-mol (D) 20.47 J/K-mol

Q.3 Select the correct statement. [3](A) Entropy of XeF5

+ is maximum when structure is square pyramidal (perfect)(B) Entropy of XeF5

+ is maximum when structure is distorted square pyramidal.(C) Entropy of XeF5

+ is maximum when structure is distorted octahedral.(D) Entropy of XeF5

+ is greater in distorted octahedral as compared to distorted square pyramidal.Q.4 Select the correct statement. [3]

(A) Every system which is left to itself, will on average change towards condition of maximum probability.(B) Gain in information is loss in entropy.(C) On increasing temperature entropy will always increases.(D) At 0° K only one micro state is possible.

Q.5 A diatomic gas is used in a carnot engine as a working substance. In an adiabatic expansion, volume ofgas increases two times then what is the percentage efficiency of carnot engine? [4]

[Ans. 0024]Q.6 A diatomic gas (ideal) undergoes change according to law T3V2 = constant from initial volume 2.5 litres

to 4.5 litres at initial temperature T1 = 300K. Find out entropy change in the system. [4]Q.7 Two identical bodies of constant heat capacity one at same initial temperature. A refrigerator operates

between these two bodies until one is cooled to temperature T2. if the bodies remains at constantpressure and undergoes no change in phase. Then calculate the minimum work needed. [4]

W = CP

12

2

21 T2T

TT

Q.8 An ideal gas undergoes change such that pressure drop to 4 times and volume increases to 2 times. Thencalculate the entropy change in system during the process. [4]

Q.9 The expansion process of n = 2 moles of argon proceeds so that gas pressure increases in direct proportionto it volume. Find the entropy increment of the gas in this process provided its volume increases twotimes. [4]

Q.10 The enthalpy of vaporization of liquid A at 42°C is 32 kJ/mole. If the normal boiling point of liquid is47°C, using the given data, find Svaporisation at 42°C (in J/K mole) [4]

CP [A (l)] = 62 J/K moleCP [A (g)] = 30 J/K moleA(l) A(g)

take ln

320315

=

3205

[Ans. 100 J/K mole]

ASSIGNMENT-6

Q.1 Consider the system shown below : [3]

CO (g)2RigidAdiabaticBoundary

A

Massless-frictionless-Non-RigidAdiabatic piston

O (g)2 He(g)

1mtr.

B C

The three chambers shown in the figure contain one mole of each gas and piston A and B are adjustedmechanically (using pin/holders) in such a way that each gas occupies same volume. A student by mistakepushes piston C towards left such that the pin/holder breaks and all the three pistons are free to move.

The ratio 4.1

)Final(CO

66.1)Final(He

2V

V = x 3

1. Calculate x .

Consider the c/s area of piston (A) and (B) to be negligible.(A) 2.5 (B) 3 (C) 4.5 (D) 1.5

Q.2 Which of the following reactions is an example of complete combustion ? [3]

(A) N2(g) + 2O2 (g) N2O4 (g) (B*) 81

S8(g) + 23

O2(g) SO3(g)

(C) Cl2(g) + 23

O2(g) Cl2O3(g) (D) All of these

Q.3 Calculate the Enthalpy of hydrogenation of ,if the f H of and are –37 and –156 kJ/

mol respectively. [3](A) –199 kJ/mol (B*) –238 kJ/mol (C) –59.5 kJ/mol (D) –476 kJ/mol

Q.4 Which of the following statements is/are incorrect ? [3](A) The enthalpy of formation of elements in standard state is taken as zero, because the total energy ofthe universe is constant.(B) The absolute entropy of elements in standard state can not be taken as zero, because the entropy ofthe universe is always increasing.(C*) Refrigerator can be considered equivalent to Carnot Engine.(D*) Cr+ in its ground state has 6 possible microstates, considering quantum mechanical model of atom.

Q.5 Which of the following can be a criteria of spontaneity ? [3](A*) S(U,V) > 0 (B*) U(S,U) < 0 (C*) A(T,V) < 0 (D*) G(T,P) < 0

Q.6 It is estimated that an average human brain consumes the equivalents of 10 gm glucose per hour estimatethe power output of Brain in Watt. If 1W =1 J/sec. (combH(glucose) = – 2818.8 kJ/mol).

[Ans. 43.5 W] [4]

Q.7 The free energy change for the conversion of malate into fumarate is 3 kJ. In metabolism this reaction iscoupled with fumarate asparate. Which is exergonic by 15.5 kJ. Calculate G° for the overallreaction and interpret the result. [4]

Q.8 The vapor pressure of decane is 10 torr at 55.7 °C and 400 torr at 150.6 °C. Calculate Hvap., Svap.at 100 °C. Also find out normal boiling point. [4]

Q.9 Compound (A) is mixed with (B) in presence of acid catalyst and converted into cyclic compound (C).Calculate the enthalpy change for the reaction. [4]

A + B CCompound (B) is prepared by NaNO2 + NaHSO3 + SO2 + H2O NaHSO4 + (B). (A) is used inthe preparation of Caprolactum.If EC—H = 414 kJ/mol EC—N = 230 kJ/mol

EC=O = 312 kJ/mol EC—C = 340 kJ/molEN—H = 198 kJ/mol EN—O = 210 kJ/molEO—H = 150 kJ/mol

Q.10 n-Butane is heated with AlCl3. Calculate the enthalpy change for the reaction if EC—H = 414 kJ/mol andEC—C = 340 kJ/mol. [4]

ASSIGNMENT-7

Q.1 In the following graph : [3]

A B

S

T

Select the correct options(s)(A) PA > PB (B) VB > VA (C) PB > PA (D) VA > VB

Q.2 The free energy of gas vs temperature is represented as follows [3]

DC

AG ° at P°g

Gg at PB

TO

Gg

Which of the following is/are true?(A) P° > P (B) P > P° (C) SC > SD (D) SD > SC

Q.3 Two identical bodies A and B of constant heat capacity at temperature TA and TB[TA > TB], are connectedusing a heat engine. Which of the following statements is/are correct in this regard ? [3](A) At maximum attainable temperature, the work done by the engine is minimum.(B) At minimum attainable temperature, the Suniv.= 0.

(C) The final temperature ranges between 2TT 21 and 21 T.T .

(D) The total work delivered by the engine is the difference between heat absorbed from hot body andheat rejected to cold body.

Q.4 Statement - 1 : On increasing the temperature, enthalpy of products and reactants changes by differentextents. [3]Statement - 2 : CP values of products and reactants are different.(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Q.5 500 ml decimolar ACOH is mixed with 1000 ml M/20 NH3. What is heat of neutralization of themixture if ionization constant of acid and base are same. [4]

Q.6 When glucose is burnt in a bomb calorimeter, the reaction is found to be exothermic andE = – 2880 kJ/mol the entropy change is found to be 182.4 J/K-mol. Calculate free energy change at298 K for the process. How much of the energy change can be extracted as Heat and How much can beextracted as work. [4]

Q.7 CaC2 an hydrolysis gives a gas (A) which is also hydrolysed in presence of promoters and catalyst andresults a liquid which is purified and added in base catalyst and product (B) is obtained after slightheating. Find out enthalpy change in the reaction (A) to (B) assuming reaction take place in gas phase. IfECC= 900 kJ/mol, ECC= 600 kJ/mol, EC= 700 kJ/mol , EO–H= 243 kJ/mol, EC–C= 300 kJ/mol. [4]

Q.8 An ideal gas cycle is represented by a rectangle on PV diagram. If P1 and P2 are the lower and higherpressure and V1, V2 are smaller and larger volumes respectively, then show that [4]

)VV(V

)PP(P

1

12

1

12

2

Q.9 R–Br is mixed with NaSCN, using polar protic solvent (water) resulting in the formation of 79% A and16% B. Calculate the f H of A and B, Given [4]Assume , R = CH3 EC—S = 380 kJ/mol EC—H = 414 kJ/mol

EC=S = 520 kJ/mol ECN = 680 kJ/molEC=N = 620 kJ/mol

What are compounds A and B and justify its percentage ?

Q.10 What is the probability that 2 moles of water originally at 50 °C will spontaneously separate into 1 molewater at 49°C and 1 mol water at 51°C. CP,m = 75 J/K mol [4]

Q.1 Find the n factor in following non-redox interaction. [10](a) Of base

(i) Al(OH)3 + HCl Al(OH)Cl2 + H2O(ii) Al(OH)3 + H2SO4 Al(OH)(HSO4)2 + H2O(iii) Ba(OH)2 + HCl BaCl2 + H2O

(iv)22

22

NHCH|

NHCH

+ HCl

ClNHCH|

ClNHCH

32

32

(v) + HCl

[Ans. (a) (i) n = 2, (ii) n = 2, (iii) 2, (iv) 2, (v) 1 ]

(b) Of acid

(i) H3SbO4 KOH KH2SbO4 + H2O(ii) H3SbO4 + KOH K2HSbO4 + H2O(iii) H2SO4 + KOH KHSO4 + H2O

(iv) KOH

(v) KOH

[Ans.(b) (i) 1, (ii) 2, (iii) 1, (iv) 1, (v) 2]

Q.2 Find the n factor of underlined compound in following redox interaction [6](i) Pb(NO3)2 + Cr2(SO4)3 PbSO4 + Cr(NO3)3

(ii) KMnO4 + MnSO4MnO2

(iii) P4 H2PO2¯ + PH3

(iv) N2H4 N2 + NH3

(v) KMnO4 K2MnO4 + MnO2 + O2

(vi) CH4 + O2 CO2 + H2O

[Ans. (i) 2,6,2,3 (ii) 3,2,6/5 (iii) nf = 3, nf = 1 , nf = 3, (iv) n = 4/3, (v) n = 2, (vi) n = 8, ]

ASSIGNMENT-12

Q.1 Find the n factor in following non-redox interaction. [7](i) MgAl(OH)5 + HCl MgCl + AlCl3 + H2O(ii) Ba(OH)2 + HCl Ba(OH)Cl + H2O(iii) R–NH2 + HCl R–NH3Cl–(iv) H2SO4 + KOH K2SO4 + H2O(v) H3PO4 + NaOH Na3PO4 + H2O

(vi) + NaOH + H2O

(vii) NaOH + H2O

[Ans. (i) 5, (ii) 1,(iii) 1, (iv) 2, (v) 3, (vi) 1, (vii) 2]Q.2 Find the n factor of underlined compound in following redox interaction [2]

(a) Cl2 ClO4– + Cl–

(b) S + NaOH Na2S + Na2S2O3 + H2O [Ans. (a) n = 7/4 (b) n = 1]Q.3 Given the reaction: [4]

S2O82– + 2e¯ 2SO4

2–

Mn2+ + 4H2O MnO4¯ + 8H+ + 5e¯How many moles of S2O8

2– ions are required to oxidise 1 mole of Mn2+ ? [Ans. 2.5 mol]Q.4 NH2OH reacts with ferric sulphate as follows: [4]

2NH2OH + 4Fe3+ N2O + H2O + 4Fe2+ + 4H+

What is the equivalent weight of NH2OH in this reaction? [Ans. M/2]Q.5 20 ml of 0.1 M solution of metal ion reacted with 20 ml of 0.1 M SO2 solution. SO2 reacted according

to the equation: [4]SO2 + 2H2O SO4

2– + 4H+ + 2e¯If the oxidation number of metal ion was +3, then what is the new oxidation number of the metal?

[Ans. +1]Q.6 How many ml of 0.15 M Na2CrO4 will be required to oxidise 40 ml of 0.5 M Na2S2O3. [4]

Given: CrO42– + S2O3

2– Cr(OH)4¯ + SO42– [Ans. 355 ml]

Q.7 It requires 40.05 ml of 1M Ce4+ to titrate 20ml of 1M Sn2+ to Sn4+. What is the oxidation state of thecerium in the product. [4]

[Ans. + 3]Q.8 A volume of 12.53 ml of 0.05093 M SeO2 reacted with exactly 25.52 ml of 0.1M CrSO4. In the

reaction, Cr2+ was oxidized to Cr3+. To what oxidation state was selenium converted by the reaction.[4]

[Ans. zero]Q.9 Potassium acid oxalate K2C2O4 · 3H2C2O4·4H2O can be oxidized by MnO4

– in acid medium. Calculatethe volume of 0.1M KMnO4 reacting in acid solution with one gram of the acid oxalate. [4]

[Ans. V = 31.68 ml]

Q.10 50gm of a sample of Ca(OH)2 is dissolved in 50ml of 0.5N HCl solution. The excess of HCl was titratedwith 0.3N – NaOH. The volume of NaOH used was 20cc. Calculate % purity of Ca(OH)2. [4]

[Ans. 1.406%]

ASSIGNMENT-13

Q.1 Calculate the number of moles of KMnO4 which will react with 180 gm H2C2O4 according to givenreaction

KMnO4 + H2C2O4 2CO2 + Mn2+

and also calculate the volume of CO2 produced at STP. [4]

Q.2 Calculate the mass of K2Cr2O7 required to produced 254 gm I2 from KI solution.Given : K2Cr2O7 + 2KI 2Cr3+ + I2 [4]

Q.3 Calculate the moles of KCl required to produced 10 mol Cl2 by the reaction with KClO3. [4]

Q.4 3 mol of a mixture of FeSO4 and Fe2(SO4)3 required 100 ml of 2M KMnO4 solution in acidic medium.What is the mole fraction of FeSO4 in the mixture? [4]

[Ans. 1/3]

Q.5 5 ml of N–HCl, 20 ml of 2N

H2SO4 and 30 ml of 3N

HNO3 are mixed together and the volume made

to 1 litre. What is the weight of pure NaOH required to neutralize the solution? [4][Ans. 1 gm]

Q.6 A 1 g sample of H2O2 solution containing x % H2O2 by mass requires x cm3 of a KMnO4 solution forcomplete oxidation under acidic conditions. Calculate the normality of KMnO4 solution. [4]

[Ans. 0.588 N]

Q.7 A mixture of FeO and Fe2O3 is reacted with acidified KMnO4 solution having a concentration of 0.2278M, 100 ml of which was used. The solution was then titrated with Zn dust which converted Fe3+ of thesolution to Fe2+. The Fe2+ required 1000 ml of 0.13 M K2Cr2O7 solution. Find the % of FeO & Fe2O3.

[4][Ans. FeO = 13.34%; Fe2O3 = 86.66%]

Q.8 25 mL of a solution containing HCl was treated with excess of M/5 KIO3 and KI solution of unknownconcentration where I2 liberated is titrated against a standard solution of 0.021M Na2S2O3 solutionwhose 24 mL were used up. Find the strength of HCl and volume of KIO3 solution consumed. [4]

[Ans. VKIO3 = 0.42 mL, [HCl] = 0.0168N]

Q.9 A 0.517g sample containing Ba(SCN)2 was dissolved in a bicarbonate solution. 50.0 mL of 0.107 Niodine was added, and the mixture was allowed to stand for five minutes. The solution was then acidified,and the excess I2 was titrated with 16.3 mL of 0.0965 M sodium thiosulphate. Write a balanced equationfor the oxidation of SCN into SO4

2 and HCN . Calculate the percent Ba(SCN)2 in the sample. [4][Ans. SCN+3I2+ 4H2O l SO4

2 + HCN + 7H+ + 6I, 15.68%]

Q.10 The neutralization of a solution of 1.2 g of a substance containing a mixture of H2C2O4. 2H2O, KHC2O4.H2O and different impurities of a neutral salt consumed 18.9 ml of 0.5 N NaOH solution. On titrationwith KMnO4 solution, 0.4 g of the same substance needed 21.55 ml of 0.25 N KMnO4. Calculate the% composition of the substance. [4]

[Ans. H2C2O4. 2H2O = 14.36%, KHC2O4. H2O = 81.71% ]

ASSIGNMENT-14

Q.1 520 gm mixture of Fe2O3 & FeO reacts completely with 158 gm KMnO4 in acidic medium. Calculatethe composition of the mixture by moles. [4](Given : Mn+7 Mn+2)

Q.2 Calculate the concentration of H2O2 solution if 20 ml of H2O2 solution reacts completely with 10ml of 2M KMnO4 in acidic medium. [4]

Q.3 How many moles of thiosulphate (S2O32–) will be required to react completely with one mole I2 in acidic

medium (where it gets oxidised to SO42–)? [4]

[Ans. 1/4]Q.4 Metallic tin in the presence of HCI is oxidized by K2Cr2O7 to stannic chloride, SnCl4. What volume of

deci-normal dichromate solution would be reduced by 1g of tin. [4][Ans. 337 mL]

Q.5 5g sample of brass was dissolved in one litre dil. H2SO4. 20 ml of this solution were mixed with KI,liberating I2 and Cu+ and the I2 required 20 ml of 0.0327 N hypo solution for complete titration. Calculatethe percentage of Cu in the alloy. [4]

[Ans. 41.53%]Q.6 A 1.0 g sample of Fe2O3 solid of 55.2% purity is dissolved in acid and reduced by heating the solution

with zinc dust. The resultant solution is cooled and made upto 100.0 mL. An aliquot of 25.0 mL of thissolution requires 17.0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of molesof electrons taken up by the oxidant in the reaction of the above titration. [4]

[Ans. 6.07 6]Q.7 5g of pyrolusite (impure MnO2) were heated with conc. HCl and Cl2 evolved was passed through

excess of KI solution. The iodine liberated required 40 mL of 10N

hypo solution. Find the % of MnO2 in

the pyrolusite. [4][Ans. 0.174g; 3.48%]

Question No. 8 to 10 (3 questions)A steel sample is to be analysed for Cr and Mn simultaneously. By suitable treatment Cr is oxidized asCr2O7

2 and the Mn to MnO4.

Cr Cr2O72

MnMnO4

A 10 gm sample of steel is used to produce 250.0 mL of a solution containing Cr2O72 and MnO4

.A 10 mL portion of this solution is added to a BaCl2 solution and by proper adjustment of the acidity, thechromium is completely precipitated as BaCrO4; 0.0549 g is obtained.

Cr2O72

H BaCrO4

A second 10 mL portion of this solution requires exactly 15.95 mL of 0.0750M standard Fe2+ solutionfor its titration (in acid solution). [9]

Q.8 % of chromium in the steel sample(A) 1.496 (B*) 2.82 (C) 1.96 (D) 5

Q.9 Equivalent of Fe2+ required for reduction of MnO4— is

(A*) 5.44 × 10–4 (B) 0.544 × 10–2 (C) 1.196 × 10–3 (D) 11.96 × 10–4

Q.10 Amount of BaCl2 required for conversion of Cr2O72 to BaCrO4 in steel sample

(A) 0.045 (B) 0.0549 (C*) 1.125 (D) 2.82

ASSIGNMENT-15

Single CorrectQ.1 How many times solubility of CaF2 is decreased in 4 × 10–3 M KF (aq.) solution as compare to pure

water at 25°C. Given Ksp (CaF2) = 3.2 × 10–11 [3](A) 50 (B*) 100 (C) 500 (D) 1000

Q.2 During the electrolysis of aqueous Zn (NO3)2 solution: [3](A) Zn plates out at the cathode (B) Zn plates out at the anode(C) H2 gas is evolved at the anode (D*) O2 gas is evolved at the anode

Q.3 The salt Al(OH)3 is involved in the following two equilibria, [3]Al(OH)3(s) Al3+ (aq.) + 3OH– (aq.), Ksp

Al (OH)3(s) + OH– (aq.) Al(OH)–4 (aq.), Kc

Which of the following relationship is correct at which solubility is minimum ?

(A) 3/1

c

sp

KK

OH

(B) 4/1

sp

cKKOH

(B) 4/1

c

sp

KK

OH

(D*) None of these

Q.4 The mass of P4O10 produced if 440 gm of P4S3 is mixed with 384 gm of O2 is [3]P4S3 + O2 P4O10 + SO2

(A) 568 gm (B*) 426 gm (C) 284 gm (D) 396 gm

Q.5 The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is :(Given At. mass of 'A' = 10 ; At. mass of 'B' = 120) [3]

5A2 + 2B4 2AB2 + 4A2B(A*) 2120 gm (B) 1060 gm (C) 560 gm (D) 1660 gm

More than one correct

Q.6 The reduction potential of hydrogen half-cell will be positive if [4](A) p(H2) =1 atm and [H+] = 1 M (B*) p(H2) = 1 atm and [H+] = 2M(C) p(H2) = 2 atm and [H+] =1 M (D*) p(H2) = 2 atm and [H+] = 2M

ASSIGNMENT-16

ComprehensionNaBr, used to produce AgBr for use in photography can be self prepared as follows : [9]

Fe + Br2 FeBr2 ....(i)FeBr2 + Br2 Fe3Br8 ....(ii) (not balanced)Fe3Br8 + Na2CO3 NaBr + CO2 + Fe3O4 ....(iii) (not balanced)

Q.7 Mass of iron required to produce 2.06 × 103 kg NaBr(A) 420 gm (B*) 420 kg (C) 4.2 × 105 kg (D) 4.2 × 108 gm

Q.8 If the yield of (ii) is 60% & (iii) reaction is 70% then mass of iron required to produce 2.06 × 103 kgNaBr(A) 105 kg (B) 105 gm (C*) 103 kg (D) None

Q.9 If yield of (iii) reaction is 90% then mole of CO2 formed when 2.06 × 103 gm NaBr is formed(A) 20 (B*) 10 (C) 40 (D) None

Subjective:Q.10 Find the EMF of cell formed by connecting two half cells: [5]

Pt(s) | 4MnO (0.1 M) ; MnCl2 (0.2 M) HCl (1M)

Pt(s) | 272OCr (0.1 M) ; CrCl3 (0.2 M) HCl (0.7 M)

Given 2

4 Mn/MnOE = 1.51 volt ;

3272 Cr/OCr

E = 1.33 volt [Ans. 0.2 Volt]

Single CorrectQ.1 Calculate the pH of 6.66 × 10–3 M solution of Al(OH)3, it's first dissociation is 100 % where as second

dissociation is 50% and third dissociation is negligible. [3](A) 2 (B*) 12 (C) 11 (D) 3

Q.2 10 ml of 0.1 M tribasic acid H3A is titrated with 0.1 M NaOH solution. What is the ratio of ]A[]AH[

33 at

2nd equivalence point. Given K1 = 7.5 × 10–4, K2 = 10–8, K3 = 10–12 [3](A) ~10–4 (B) ~ 10+4 (C*) ~ 10–7 (D) ~ 10+6

Q.3 Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and56 lit. O2 at 1 atm & 273 K according to given reaction, is [3]

C(s) + H2(g) + O2 (g) C12H22O11(s)(A) 138.5 (B*) 155.5 (C) 172.5 (D) 199.5

Q.4 The number of carbon atoms present in a signature, if a signature written by carbon pencil weights1.2 × 10–3 g is [3](A) 12.04 × 1020 (B*) 6.02 × 1019 (C) 3.01 × 1019 (D) 6.02 × 1020

Q.5 An aqueous solution of Na2SO4 was electrolysed for 10 min. 82 ml of a gas was produced at anodeover water at 27°C at a total pressure of 580 torr. Determine the current that was used. [3](Vapour pressure of H2O at 27°C = 10 torr) (R = 0.082 atm lit./ mol / K)(A) 0.1 Amp (B) 1.25 Amp (C) 2.23 Amp (D*) 1.61 Amp

More than one correctQ.6 A + 2B C [4]

2C 3DWhich of the following option(s) is / are incorrect for above reaction.(A) maximum no. of moles of ‘D’ which can be produced by 2 moles of ‘A’ is 3.(B*) maximum no. of moles of ‘D’ which can be produced by 6 moles of ‘B’ is 4.(C*) maximum no. of moles of ‘D’ which can be produced by 2 moles of A and 6 moles of B is 4.5(D*) All above options are correct.

ComprehensionFor questions No. 7 & 8 use following information: [6]

8 mole of a mixture of N2, NO2 and N2O4 has a mean molecular mass of 8378

. On heating to atemperature at which N2O4 dissociated completely (N2O4 2NO2), the mean molecular mass

becomes 9378

. [Given : Atomic mass N= 14, O = 16]Q.7 The ratio of number of moles of N2 : NO2 : N2O4 in original mixture is

(A) 3 : 4 : 1 (B*) 2 : 5 : 1 (C) 3 : 5 : 2 (D) 4 : 2 : 1

Q.8 The ratio of number of moles of N2 and NO2 after heating is(A) 1 : 1 (B) 3 : 9 (C*) 2 : 7 (D) 1 : 2

ASSIGNMENT-17

Match the ColumnQ.9 Column-I and column-II contains four entries each. Entry of column-I is to be matched with only one

entry of column-II. [Given : Atomic mass Ag = 108, S = 32] [10]Column I Column II

(A) 5 m NaOH solution (P) 16 M(dsolution = 0.6 gm /ml).Molarity of solution is

(B) 250 ml of H2O2 solution (Q) 1 Mprovides 64 gm O2 .Molarity of H2O2 solution is

(C) 100 ml , 1 M H2SO4 solution (R) 2.5 M(dsolution = 1.5 gm/ml) is mixedwith 400 ml of water, density of finalsolution 1.25 gm/mlFinal Molarity of H2SO4 solution is

(D) 100 ml, 6 M NaCl solution is mixed with (S) 0.227 M100 ml of 17% (w/w) AgNO3[dsolution= 8 gm/ml] solution.The molarity of Ag+ ions in final solution is [Ans. (A) R; (B) P; (C) S; (D) Q]

SubjectiveQ.10 1.0 L of solution which was in equilibrium with solid mixture of AgCl and Ag2CrO4 was found to contain

1 ×10–4 moles of Ag+ ions, 1.0 × 10–6 moles of Cl– ions and 8.0 × 10–4 moles of 24CrO ions. Ag+ ions

are added slowly to the above mixture (keeping volume constant) till 8.0 × 10–7 moles of AgCl gotprecipitated. How many moles of Ag2CrO4 were also precipitated? [5]

[Ans.7.68 × 10–4 mole]

Single CorrectQ.1 The half life of the unimolecular elementary reaction

A(g)B(g) + C(g) is 6.93 min. How long will it take for the concentration of A to be reduced to 10% of the initial value? [3]

(A) 10.053 min (B) 4.6 min (C) 46 min (D*) 23.03 min

Q.2 Volume of water which must be added to 8M HCl to make 2 lit. of 7.3 % w/v HCl solution is [3](A*) 1.5 lit. (B) 0.5 lit. (C) 1 lit. (D) 2 lit.

Q.3 For the reaction A (aq) B (g) + C (g), the rate law is –

dtAd

= k[A].

Find the value 'X' in data table provided [3]time (min) 0 10 30conc. of A(M) 1 0.7 X

(A) 0.1 (B) 0.243 (C*) 0.343 (D) None of these

Q.4 The instantaneous rate of disappearance of the MnO4¯ ion in the following reaction is 4.56 ×10–3 Ms–1.2MnO4¯ + 10 I¯ + 16 H+ 2Mn2+ + 5I2 + 8H2O [3]The rate of appearance of I2 is

(A) 1.14 × 10–3 Ms–1 (B) 5.7 × 10–3 Ms–1

(C) 4.56 × 10–4 Ms–1 (D*) 1.14 × 10–2 Ms–1

Q.5 For which of the following solution, minimum amount of NaOH is required for complete neutralisation.[Assume complete dissociation of acid ] [3](A) 50 ml H2SO4 solution [basicity = 2] (dsolution = 2 gm/ml) , % w/w = 98(B) 50 ml H3PO4 solution [basicity = 3] (dsolution = 2 gm/ml) , % w/w = 98(C) 100 ml HCl solution [basicity = 1] (dsolution = 1.5 gm/ml) , % w/w = 36.5(D*) 100 ml HNO3 solution [basicity = 1] (dsolution = 1.5 gm/ml) , % w/w = 36.5

Q.6 For the reaction [3][Ag(CN)2]¯ l Ag+ + 2CN¯ ; Kc at 25°C is 4 × 10–19

The equilibrium concentration of Ag+ is, in a solution which is 0.1 M in KCN & 0.03 M in AgNO3originally.(A) 7.5 × 10–16 M (B*)7.5 × 10–18 M (C) 1.25 ×1019 M (D) 1.25 × 10–17 M

Q.7 2 mole of an ideal monoatomic gas at 27°C undergoes adiabatic free expansion until it volume becomes5 times of initial volume. The change in entropy of the gas is ( ln 5 = 1.60 ; R = 8.3 J/K-mole ) [3](A) Zero (B*) 26.56 J/K (C) 7968 J/K (D) 39.84 J/K

ASSIGNMENT-18

Q.8 For a gas reaction A P at T (K) the rate is given by [3]rate = k1

2Ap atm/hr

rate = 2A2Ck mol / litre / hr

the relation between k1 and k2 is

(A) k2 = k1 (B*) k2 = k1 RT (C) k2 = k1/RT (D) k2 = k1 (RT)2

Assertion & ReasonQ.9 Statement -1 : For a very dilute solution molality & molarity are always approximately equal

Statement -2 : Mass of solution is always approximately equal to mass of solvent for a very dilutesolution. [3]

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

More than one correct:Q.10 A 1 litre solution containing NH4Cl and NH4OH has hydroxide ion concentration of 10–6 mol/lit. Which

of the following hydroxides could not be precipitated when the solution is added to 1 litre solution of 0.1M metal ions [5](A*) Ag(OH) (KSP = 5 × 10–3) (B*) Ca(OH)2 (KSP = 8 × 10–6)(C*) Mg(OH)2 (KSP = 3 × 10–11) (D) Fe(OH)2 (KSP = 8 × 10–16)

ASSIGNMENT-19

Single correctQ.1 For acid catalysed hydrolysis of ester rate law obtained is rate = k [ester] [H+],

where k = 0.01 M–1 hr – 1.What is the half-life if the initial concentrations are 0.02 M for the ester and0.05 M for the catalyzing acid : [3](A) 1429 hours (B) 5000 hours (C*) 1386 hours (D) 2 hours

Q.2 For a reaction of order n, what is the relationship between t3/4 and t1/2 , where t3/4 is the time required forconcentration (C) to become 1/4 C0 where C0 and C are the values of the reactant concentration at thestart and after time t, respectively. [3](A*) t3/4 = t1/2 [2n–1 + 1] (B) t3/4 = t1/2 [2n–1 – 1](C) t3/4 = t1/2 [2n+1 – 1] (C) t3/4 = t1/2 [2n+1 + 1]

Q.3 100 ml of H2SO4 solution having molarity 1M is mixed with 400 ml of water. Resultant molarity ofH2SO4 solution is: [3](A) 4.4 M (B) 0.145 M (C) 0.227 M (D*) 0.2 M

Q.4 For the reactions (i) A IK P.. [3]

& (ii) B IIK Q, following observation is made.

0.5 M 0.5 M

0 030 6060

1 M 1 M

0.25 M

Con

cent

ratio

n

Con

cent

ratio

n

Time (min) Time (min)

[A] [B]

Calculate II

I

KK

, where KI and KII are rate constant for the respective reaction.

(A*) 2.772 (B) 1 (C) 0.36 (D) None of these

Q.5 There exist an equilibrium between solid BaSO4 and Ba2+ and 24SO ion in aqueous medium. The

possible equilibrium state are shown in figure as thick line. Now, if equilibrium is disturbed by addition ofBaNO3 and (b) K2SO4 and dotted line represent approach of system toward’s equilibrium. Match thecolumns given below: [3]

(a) addition of BaNO3 (c) (e)

(b) addition of K2SO4 (d) (f)

(A) ac, bd (B*) ad, be (C) af, be (D) ad, bc

More than one may be correctQ.6 In a chemical reaction A(g) is converted to B(g), following observation is made. Identify the correct

statement(s). [4]

10 20 30

2

4

6

8Pr

essu

re (a

tm)

time (min.)

PB

PA

(A*) Chemical reaction should be 21

A(g) B(g).

(B) t1/2 of the reaction is independent of initial partial pressure of A.(C*) Reaction must be complex reaction.(D*) Time of completion for the reaction is 40 minutes.

ComprehensionComprehension (Q.7 to Q.9)

Concentrations measured as a function of time when gaseous N2O5 at initial concentration of 0.0200 Mdecomposes to gaseous NO2 and O2 at 50°C. The change in concentration with time is given by thefollowing graph. [9]

0 100 200 300 400 500 600 700 800

0.01

0.00

0.02

0.03 NO2

O2

N O2 5

[NO ]=0.004M2

[NO ]=0.003 M

2

Time(s)

Con

cent

ratio

n(M

)

= tan (2.5 × 10 )–1 –5

Q.7 The instantaneous rate of disappearance of N2O5 at the begining of the reaction is(A) 2 × 10–5 M/s (B) 1 × 10–5 M/s (C*) 5 × 10–5 M/s (D) zero

Q.8 The rate of formation of O2 during the period 600 – 700 s is(A) 4 × 10–5 M/s (B) 3 × 10–5 M/s (C) 5 × 10–5 M/s (D*) 7.5 × 10–6 M/s

Q.9 The rate of decomposition of N2O5 during the period 300 – 400 s is(A) 4 × 10–4 M/s (B) 3 × 10–5 M/s (C*) 2 × 10–5 M/s (D) 4 × 10–8 M/s

SubjectiveQ.10 50 ml of a gaseous hydrocarbon is mixed with 200 ml oxygen. After sparking, evolved gaseous mixture

was passed through KOH solution which result in 100 ml contraction, when further passed throughalkaline pyrogallcol solution shows a contraction of 50 ml. Find molecular formula of hydrocarbon. [5]

[Ans. C2H4 ]

ASSIGNMENT-20

Single CorrectQ.1 The dissociation of HI molecules as shown below, occurs at a temperature of 629 K. The rate constant,

k = 3.02 × 10–5 M–1s–1

)g(I21)g(H

21)g(HI 22

what is the reaction order. [3](A) 0 (B) 1 (C*) 2 (D) Can not be predicted

Q.2 For the first order reaction [3]

N NCl

Cl + N (g)2

following observation is made :

10 20 30

2345678

40time (min.)

log (V – V) t

where Vt(in cc) is volume of N2 collected at time t & V (in cc) is volume of N2 collected after a longtime. What is the time taken in minute for 75% reaction ?(A) 2.5 (B) 5.0 (C*) 3 (D) 10

Q.3 At 1800 K ethane gas decomposes to ethene and hydrogen. Rate constant for the reaction is10–3 Pa–1 sec–1. If initial pressure of ethane is 3 × 105 Pa, how many sec. would it take for the pressureto reach 5 × 105 Pa? [3](A) 1800.2 sec. (B) 3.33 × 10–2 sec. (C*) 6.66 ×10–3 sec. (D) 1000.4 sec.

Q.4 Two first order reaction have half-lives in the ratio 8 : 1. Calculate the ratio of time intervals t1 : t2. The

time t1 and t2 are the time period for th

41

and th

43

completion. [3]

(A) 1 : 0.301 (B) 0.125 : 0.602 (C*) 1 : 0.602 (D) None of these

Q.5 10 ml of 0.1 M CH3COO¯Na+ is mixed with 11 ml of 0.1 M HCl. The concentration of CH3COO¯ ionat equilibrium is [Take Ka (CH3COOH) = 10–5] [3](A) 2.1 ×10–4 M (B) 21 ×10–4 M (C*) 10–4 M (D) 10–8 M

Q.6 For any sparingly soluble salt [M(NH3)4 Br2] H2PO2, What will be the solubility product constant ?

[Given 0Br)NH(M 243

= 100 S-m2- mol–1, 0POH –

22 = 50 S-m2- mol–1

Specific resistance of saturated [M(NH3)4 Br2] H2PO2 solution is 200 -cm ] [3](A*) 1.11 × 10–11 (B) 1.11 × 10–3 (C) 3.33 × 10–6 (D) None of these

Assertion & ReasonQ.7 Statement -1 : Molality of pure ethanol is lesser than pure water. [3]

Statement -2 : As density of ethanol is lesser than density of water.[Given : dethanol = 0.789 gm/ml; dwater = 1 gm/ml]

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement-1.(C) Statement-1 is false, statement-2 is true.(D) Statement-1 is true, statement-2 is false.

More than one may be correctQ.8 Select the option(s) which have equal number of 'O' atom. [4]

(A) 10 ml H2O(l) (B*) 0.15 mole of V2O5(C*) 12 gm O3(g) (D) 12.044 ×1022 molecules of CO2

SubjectiveQ.9 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant

volume. [5]SO2 Cl2 (g) SO2 (g) + Cl2 (g)

6.010025.001

atm/pressureTotalTime(s)Experiment

Calculate the rate of the reaction when total pressure is 0.65 atm. [Ans. 7.8 × 10–4 atm s–1]

Q.10 Would H2O2 behave as oxidant or reductant with respect to the following couples at standard condition.

(a) I2 / I– (b) 282OS / 2

4SO (c) Fe3+ / Fe2+ [7]

Given

222 OH/H,OE = + 0.69 volt ;

OH/H,OH 222E = + 1.77 volt

I/I2

E = 0.535 volt ;

24

282 SO/OS

E = 2.0 volt

23 Fe/Fe

E = 0.77 volt[Ans. (a) oxidising agent, (b) reducing agent, (c) both oxidising and reducing agent]

ASSIGNMENT-21

Single CorrectQ.1 The acid catalysed hydrolysis of an organic compound A at 30°C has a time for half change of 100 min.

when carried out in a buffer solution at pH = 5, and 10 min. when carried out at pH = 4. Both times ofhalf change are independent of the initial concentration of A. If the rate constant K is given by

dt]A[d

=K[A]a[H+]b , what are the values of a and b? [3]

(A*) a = 1, b = 1 (B) a = 2, b = 1 (C) a = 0, b = 1 (D) a = 1, c = 0

Q.2 What is the aq. ammonia concentration of a solution prepared by dissolving 0.15 mole of 4NH CH3COO–

in 1L H2O.[ )COOHCH(a 3K =1.8×10–5 ; )OHNH(b 4

K =1.8×10–5] [3]

(A*) 8.3 ×10–4 (B) 0.15 (C) 6.4 ×10–4 (D) 3.8 × 10–4

Q.3 AgCl can be obtained: [3]

(A) by extrapolation of the graph and C to zero concentration

(B*) by known values of of AgNO3, HCl and HNO3(C) both (A) and (B)(D) none of these

Q.4 Benzoic acid solution is titrated with NaOH conductometrically, graphical representation of the titration is:

(A) (B) (C) (D*) [3]

Q.5 C6H5OH (g) + O2 (g) CO2 (g) + H2O (l) [3]Magnitude of volume change if 30 ml of C6H5OH (g) is burnt with excess amount of oxygen, is(A) 30 ml (B*) 60 ml (C) 20 ml (D) 10 ml

Q.6 Two radioactive materials X1 and X2 have decay constants 10 and respectively. If initially they have thesame number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be1/e after a time [3](A) 1/(10) (B) 1/(11) (C) 11/(10) (D*) 1/(9)

Q.7 In a radioactive decay sequence [3]

A 1 B 2 CWhat is the ratio of numbers of atoms of A to the number of atoms of B after long time interval startingwith pure A. The mean lives of A and B are 20 hours and 30 minutes respectively.(A) 2/3 (B) 10 (C) 20 (D*) 40

Q.8 A certain weak acid has a dissociation constant of 1.0 × 10–4. The equilibrium constant for its reactionwith a strong base is : [3](A) 1.0 × 10–4 (B) 1.0 × 10–10 (C*) 1.0 × 1010 (D) 1.0 × 1014

More than One may be correctQ.9 Which of the following statements is correct. [4]

(A*) On bombarding 7N14 Nuclei with -particle, the nuclei of the product formed after release ofproton would be 8O17.

(B*) Nuclide and it's decay product after -emission are called isobars(C*) Nuclide and it's decay product after -emission are called isodiaphers.(D) Half life of radium is 1580 years. Its average life will be 1097.22 years.

SubjectiveQ.10 A 5.0 gm sample containing Pb3O4, PbO2 and some inert impurity is dissolved in 250 ml

dil. HNO3 solution and 2.68 g of Na2C2O4 was added so that all lead converted into Pb2+.A 10 ml portion of this solution required 8.0 ml, 0.02 M KMnO4 for titration of excess of oxalate. In ananother experiment, 25 ml of solution was taken and excess oxalate was removed by extraction, thisrequired 10 ml of a permanganate solution for oxidation of Pb2+ to Pb4+, 10 ml of this permanganatesolution is equivalent to 4.48 ml, 5V H2O2 solution.Calculate mass % of PbO2 & Pb3O4 in the original sample. [At. wt. of Pb = 207] [6]

[Ans. % Pb3O4 = 68.5, % PbO2= 23.9]

ASSIGNMENT-22

Single CorrectQ.1 A radioactive nuclide is produced at a constant rate of -per second. It decay constant is . If N0 be the

no. of nulcei at time t = 0, then maximum no. of nuclei possible are [3]

(A*)

(B) N0 –

(C) / (D)

+ N0

Q.2 Among the following ion which is best oxidising agent? [3]

Given 46

36 ])CN(Fe/[])CN(Fe[

0E = 0.36 V

23 Fe/Fe0E = 0.77 V

(A*) Fe3+ (B) Fe2+ (C) [Fe(CN)6]4– (D) [Fe(CN)6]3–

Q.3 The reaction: 2NO + 2H2 N2 + 2H2O has been assigned to follow following mechanism [3]I. NO + NO l N2O2 (fast)II. N2O2 + H2 N2O + H2O (slow)III. N2O + H2 N2 + H2O (fast)The rate constant of step II is 1.2 × 10–4 mole–1L min–1 while equilibrium constant of step I is1.4 × 10–2. What is the rate of reaction when concentration of NO and H2 each is 0.5 mole L–1.(A*) 2.1 × 10–7 mol L–1 min–1 (B) 3.2 × 10–6 mol L–1 min–1

(C) 3.5 × 10–4 mol L–1 min–1 (D) none of these

Q.4 When a car starts following chemical reaction takes place in the battery for 5 sec. to deliver 48.25 Acurrent:- [3]Pb (s) + PbO2 (s) + 2H2SO4 + 2e– 2PbSO4(s) + 2H2O + 2e–

Calculate mass of H2SO4 consumed during the process (in mg)(A) 0.245 (B*) 245 (C) 0.490 (D) 490

ComprehensionQuestion No. 5 to 7 (3 questions)Following titration method is taken to compute stepwise ionisation constant of a weak dibasic acid

A:

p-hydroxybenzoic acidA has two ionisation proton and there can be stepwise neutralisation by NaOH25 ml of a dilute aqueous solution of A is titrated with 0.02 M NaOH (aq) and pH measured. [9]

point)equivalentfirstat(02.7ml24.16II57.4ml12.8I

pHaddedNaOHofVolumeStep

Q.5 Which H+ is removed in step I?

(A) + H2O l + OH3 (B*) + H2O l + OH3

(C) both 50% in each part (D) H2O + H2O l H3O + OHr

(autoprotolysis of H2O)

Q.6 1apK (= – log 1aK ) of p-hydroxybenzoic acids is:

(A*) 4.57 (B) 9.47 (C) 4.90 (D) 7.00

Q.7 2apK (= – log 2aK ) of p-hydroxy benzoic acid is nearly(A) 4.5 (B) 7 (C*) 9.5 (D) 5

More than one may be correctQ.8 Choose the correct option(s): [4]

(A*) Melting point of ice decreased on increasing pressure(B*) 2O3(g) 3O2(g), S > 0(C) pH is an extensive property(D) Decrease in atomic no. by one unit due to emission of positron e0

1 is as a result of conversion of

one neutron into proton.

Q.9 Iodine liberated was titrated against 0.1 M hypo solution 40 ml of which was required. Concentration ofH2O2 solution may be expressed as. [5]

(A*) 0.68 gm / L (B*) 0.02 M (C*) 0.04 N (D) 6.8 % w/v

SubjectiveQ.10 A gaseous compound 'A' reacts by three independent first order processes

(as shown in figure) with rate constant 2 × 10–3, 3 × 10–3 and1.93 × 10–3 sec–1 respectively for products B, C and D respectively. Ifinitially pure 'A' was taken in a closed container with P = 2 atm, findthe partial pressure of 'B' (in atm) after 100 sec from start of experiment. [5]

[Ans. 2.9 × 10–1]

ASSIGNMENT-23

Single CorrectQ.1 During electrolysis at acidified water, O2 gas is formed at the anode. To produce O2 gas at the anode at

the rate of 0.224 cm3 per second at STP, current passed is : [3](A) 0.224 A (B) 2.24 A (C) 9.65 A (D*) 3.86 A

Q.2 Ionic conductances of H+ and –24SO at infinite dilution are x and y S cm2 equiv–1. Hence, equivalent

conductance of H2SO4 at infinite dilution is: [3](A*) x + y (B) 2(x + y) (C) 2x + y (D) x + 2y

Q.3 The radioisotope 15 P32 is used in biochemical studies. A sample of this isotope has an activity 1000times the detectable limit. How long could an experiment be run with the sample before the radioactivitycould no longer be detected ( 2/1t =14.2 days) [3](A) ~ 28 days (B) ~ 102 days (C*) ~ 142 days (D) ~ 50 days

Q.4 A radioactive mixture containing a short lived species A and short lived species B. Both emitting -particles at a given instant, emits at rate 10,000 -particles per minute. 10 minutes later, it emits at therate of 7000 particles per minute. If half lives of the species are 10 min and 100 hours respectively, thenthe ratio of activities of A : B in the initial mixture was [3](A) 3 : 7 (B) 4 : 6 (C*) 6 : 4 (D) 10 : 4

Q.5 1.5 gm mixture of SiO2 and Fe2O3 on very strong heating leave a residue weighing 1.46 gm. Thereaction responsible for loss of weight is

Fe2O3 (s) Fe3O4 (s) + O2(g)What is the percentage by mass of Fe2O3 in original sample. [3]

(A*) 80% (B) 20% (C) 40% (D) 60%

Q.6 For the reaction [3]

R – X + OH¯

R – OH + X¯

R – OH + X¯

k2

k1

II ordernd

I orderst

the rate is given byRate = k1 [RX] [OH¯] + k2 [RX]Calculate % of RX which reacts by 2nd order mechanism when [OH]¯ is 0.1 M.

(A) 21

1

kk100k

(B) 21

2

kk100k

(C) 21

1

kkk10 (D*)

21

1

k10kk100

ComprehensionParagraph for question nos. 7 to 9

A factory, producing methanol, is based on the reaction :CO + 2H2 l CH3OH

Hydrogen & carbon monoxide are obtained by the reactionCH4 + H2O l CO + 3H2

Three units of factory namely, the "reformer" for the H2 and CO production, the "methanol reactor" forproduction of methanol and a "separator" to separate CH3OH from CO and H2 are schematicallyshown in figure.

Reformer Reactor SeparatorValve-1 Valve-2 Valve-3

CH OH3

The flow of methanol from Valve-3 is 103 mol/sec. The factory is so designed that 32

of the CO is

converted to CH3OH. Assume that the reformer reaction goes to completion.CO + 2H2 l CH3OH Hr = – 100 R

Q.7 What is the flow of CO and H2 at Valve-1? [3](A) CO : 1500 mol / sec.; H2 : 2000 mol/sec.(B) CO : 1500 mol / sec.; H2 : 3000 mol/sec.(C) CO : 1000 mol / sec.; H2 : 2000 mol/sec.(D*) CO : 1500 mol / sec.; H2 : 4500 mol/sec.

Q.8 What is the flow of CO and H2 at Valve-2? [3](A) CO : 500 mol / sec.; H2 : 1000 mol/sec. (B*) CO : 500 mol / sec.; H2 : 2500 mol/sec.(C) CO : 500 mol / sec.; H2 : 2000 mol/sec. (D) CO : 500 mol / sec.; H2 : 1500 mol/sec.

Q.9 Amount of energy released in methanol reactor in 1 minute? [3](A)1200 kcal (B*) 12000 kcal (C) 6000 kcal (D) None of these

More than one may be correctQ.10 The order of reaction A product can be given by the expression(s) [5]

[where r = rate of reaction ; [A]1 = concentration at time t1 ; [A]2 = concentration at time t2 ]

(A*) 12

12

]A[ln]Aln[rlnrln

(B) 12/122/1

1020

]t[ln]tln[]A[ln]A[ln

(C*) ln

dt.k]A[d

/ ln [A] (D*) ]Aln[)k/r(ln

ASSIGNMENT-24

Single CorrectQ.1 In the redox reaction [3]

a CrCl3 + b H2O2 + c NaOH 2 Na2CrO4 + x NaCl + y H2OFind the value of a, b, c in the given equation :

(A*) a = 2, b = 3, c = 10 (B) a = 1, b = 2, c = 8(C) a = 2, b = 2, c = 8 (D) a = 2, b = 3, c = 4

Q.2 An equimolar mixture of Na2C2O4 and H2C2O4 required V1 litre of 0.1 M KMnO4 in acidic mediumfor complete oxidation. The same amount of the mixture required V2 litre of 0.2 M NaOH forneutralization. The ratio of V1 to V2 (V1/V2) is : [3]

(A) 2 : 5 (B) 1 : 2 (C*) 4 : 5 (D) None of these

Q.3 Ionisation constant of a weak acid (HA) in terms of m and m is: [3]

(A) Ka = )(C

m

m

(B*) Ka = )(

C

mmm

2m

(C) Ka = )()(C

mmm

2m

(D) none of these

Q.4 1 mole of AgNO3 is added to 10 litre of 1 M NH3. What is the concentration of Ag(NH3)+ in solution?

[Given : For 23)NH(Ag , 3

f 100.2K1

; 4f 10K2 ] [3]

(A) 8 × 10–5 (B*) 1.25 × 10–5 (C) 4 × 10–6 (D) 1.25 × 10–4

Q.5 The enthalpy changes of the following reactions at 27°C are [3]

Na(s) + 21

Cl2 (g) NaCl (s) rH = –411 kJ/mol

H2(g) + S (s) + 2O2 (g) H2SO4 (l) rH = –811 kJ/mol2Na(s) + S(s) + 2O2 (g) Na2SO4 (s) rH = –1382 kJ/mol

21

H2(g) + 21

Cl2(g) HCl (g) rH = –92 kJ/mol; R = 8.3 J/K-mol

from these data, the heat change of reaction at constant volume ( in kJ/mol) at 27°C for the process2NaCl (s) + H2SO4 (l) Na2SO4 (s) + 2HCl (g) is(A) 67 (B*) 62.02 (C) 71.98 (D) None

Q.6 What volume of 0.2 M RNH3Cl solution should be added to 100 ml of 0.1 M RNH2 solution toproduce a buffer solution of pH = 8.7? [3]Given : pKb of RNH2 = 5 ; log 2 = 0.3(A) 50 ml (B*) 100 ml (C) 200 ml (D) None of these

Assertion & Reason

Q.7 Statement -1 : Partial volume of CO2 and CO in a mixture are 117

litre and 1 litree respectively. The

mass of CO is same at that of CO2 in the mixture. [3]Statement -2 : Ratio of total number of atoms of C to O is 7 : 11 in above mixture.

(Assuming gases to be ideal)(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C*) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

More than One may correct :Q.8 For the first order parallel reactions of A starting with 1M initial concentration of A : [4]

A 2C2k

B

3Dk

3k

Which of the following is / are true.(A*) [B]t : [C]t : [D]t is independent of time(B*) [C]t > [B]t = [D]t(C*) [C] = 0.67 M(D) If k = 0.5 min–1 then [B] = 1 M after 1.386 min.

Q.9 NO2 is involved in the formation of smog and acid rain. A reaction that is important in the formation ofNO2 is O3(g) + NO(g) l O2(g) + NO2(g) Kc= 6 × 1034 , if the air over Bansal Classes contained1 × 10–6M O3, 1×10–5 M NO, 2.5 ×10–4 M NO2 and 8.2 × 10–3 M O2, then which is incorrect ?(A*) There will be a tendency to form more NO and O3 [4](B) There will be a tendency to form more NO2 and O2(C*) There will be a tendency to form more NO2 and O3(D*) There will be no tendency for change because the reaction is at equilibrium.

SubjectiveQ.10 10 mole of liquid 'A' and 20 mole of liquid 'B' is mixed in a cylindrical vessel containing a piston arrange-

ment. Initially a pressure of 2 atm is maintained on the solution. Now, the piston is raised slowly andisothermally. Assume ideal behaviour of solution and A and B are completely miscible.

atm9.0Pandatm6.0P oB

oA

What is the pressure at which 31

th of the total amount (by mol) of liquid solution taken initially, will be

present in the vapour form. [6]

[Given : 57 = 7.55 ] [Ans. 0.791 atm]

ASSIGNMENT-25

Single CorrectQ.1 15 g of a solute in 100 g water makes a solution freeze at –1°C. 30 g of a solute in 100 g of water will

give a depression in freezing point equal to [3](A) –2°C (B) 0.5°C (C*) 2°C (D) 1°C

Q.2 In cold countries, ethylene glycol is added to water in the radiators of cars during winters. It results in(A) lowering of boiling point (B) reduced viscosity [3](C) reduced specific heat (D*) lowering of freezing point

Q.3 Which represents correct difference for an ideal solution containing non-volatile solute. [3]

(I) (II)

(III)

(A*) I, II, III (B) I, III (C) II, III (D) I, II

Q.4 The reaction N2 + O2 l 2NO, follows the kinetics Rate [N2][O2]. The order of reactions when bothN2 and O2 are in small amount, N2 is in large excess as compared to O2 and both N2 and O2 are in largeexcess are [3](A*) 2, 1 and 2 (B) 2, 1 and 1 (C) 2, 1 and 0 (D) 1, 1 and 0

Q.5 During an experiment an ideal gas is found to obey an additional law PV3 = constant. The initial temperatureof gas is 600 K what will be final temperature if gas expands to it’s double volume. [3](A) 1200 K (B) 2400 K (C) 300 K (D*) 150 K

Q.6 For two parallel first order reactions, what is the overall activation energy of reaction? The yields of Band C in products are 40% and 60%, respectively. [3]

BA mol/kcal20Ea CA mol/kcal40Ea

(A) 60 kcal/mol (B*) 32 kcal/mol (C) 28 kcal/mol (D) 20 kcal / mol

More than one may be correctQ.7 The cell potential for the unbalanced chemical reaction. [4]

Hg22+ + NO3¯ + H+ Hg2+ + HNO2 + H2O ; V02.0Eo

cell

is measured under standard state conditions in the electrochemical cell.

06.0

FRT303.2

Then which is the correct statement:(A*) Solution in cathodic compartment is acidic.(B) Solution in Anodic compartment is acidic

(C*) Cell potential will be zero at pH = 92

, if activity of other components are equal to one.

(D*) 0.6 moles of electron pass through the circuit when 0.6 moles of Hg2+ are produced in the cell.

Q.8 E = 0 for which process, must be zero [5](A*) Cyclic process (B*) Isothermal ideal gas expansion(C) Isochoric process (D) Adiabatic process

Q.9 The incorrect statement(s) regarding 2M MgCl2 aqueous solution is/are (dsolution = 1.09 gm/ml) [4](A) Molality of Cl¯ is 4.44 m(B*) Mole fraction of MgCl2 is exactly 0.035(C) The conc. of MgCl2 is 19% w/v(D*) The conc. of MgCl2 is 19 × 104 ppm

SubjectiveQ.10 Kinetic data for hydrolysis of a 0.3 M solution of ethyl acetate in presence of HCl as catalyst is given in

the following table. A constant volume of reaction mixture is taken at different time and titrated againststandard alkali solution at different time.

Time / min 0 10 t = Vol. of Alkali / ml 25.00 28.00 40.00

If the rate law is given by rate = k[Ester]1[H+]1, where H+ is a catalyst ,find the rate (in M/min) of esterhydrolysis in a solution which is 0.1 M each in ester and acid HCl (M/min)

Use: ln

45

= 0.225 [5]

[Ans. 4.5 × 10–04]

ASSIGNMENT-26

Single CorrectQ.1 A graph was plotted between the molar conductance of various electrolytes (HCl, KCl and CH3COOH)

and root of their concentrations in mole per litre. [3]

Which of the following is correct match?(A*) I (CH3COOH) ; II (KCl) ; III (HCl) (B) I (HCl) ; II (KCl) ; III (CH3COOH)(C) I (CH3COOH) ; II (HCl) ; III (KCl) (D) I (KCl) ; II (CH3COOH) ; III (HCl)

Q.2 For a zero order reaction and a Ist order reaction half life are in ratio of 4 : 1. Calculate ratio of time takento complete 87.5 % reaction for zero order : first order reaction respectively. [3](A*) 7 : 3 (B) 3 : 7 (C) 4 : 1 (D) 5 : 3

Q.3 One litre water in a bucket is placed in a closed dry air room having dimension 4 × 2 × 1.5 m3 at 300 K.If vapour pressure of water at 300 K is 38 mm & density of water at 300 K is 0.9 gm/ml. Then volumeof water left in liquid state will be (R = 0.08 L atm mol–1) [3](A) 450 ml (B*) 500 ml (C) 875 ml (D) 900 ml

More than one may be correctQ.4 Which of the following statement true for one mole of an ideal gas [3]

(A*) RTE

TH

VP

(B*) VP T

ETH

(C*) TV

E

= 0 (D*) 0VH

T

ComprehensionQuestion No. 5 to 8 (4 questions)Properties whose values depend only on the concentration of solute particles in solution and not on theidentify of the solute are called colligative properties.

Q.5 Acetic acid in benzene solution forms dimer due to intermolecular H-bonding. For this case van’t Hofffactor is: [3](A) i =1 (B) i > 1 (C*) i < 1 (D) None of these

Q.6 An aqueous solution of 0.01 M CH3COOH has van’t Hoff factor of 1.01. If pH = – log [H+], pH of0.01 M CH3COOH solution would be: [3](A) 2 (B) 3 (C*) 4 (D) 5

Q.7 In which case van’t Hoff factor is maximum? [3](A) KCl, 50% ionised (B) K2SO4 40% ionised(C) SnCl4, 20% ionised (D*) FeCl3, 30% ionised

Q.8 A complex containing K+, Pt (IV) and Cl– is 100% ionised giving i = 3. Thus, complex is: [3](A) K2 [PtCl4] (B*) K2[PtCl6] (C) K3[PtCl5] (D) K[PtCl3]

Q.9 Match the column : [8]Column I (Order) Column II

(A) Zero (P)

1(a–x)

2

t

(B) First (Q)log(t )1/2

log a

45°

(C) Second (R)(a–x)

t

(D) Third (S)log(t )1/2

log a

(T)(a–x)

t [Ans. (A) Q,R (B) S,T, (C) T, (D) P,T]

SubjectiveQ.10 A solution of Na2S2O3 is to be standardised by titration against iodine liberated from standard KIO3

solution. The later is prepared by dissolving 4.28 gm of KIO3 in water and making upto 500 ml. 20 mlof this solution are then mixed with excess KI solution and the following reaction occurs. [6]

3IO + 5I ¯ + 6H+ 3I2 + 3H2O

The resulting iodine is titrated with the Na2S2O3 solution according toI2 + 2

32OS2 2 I ¯ + 264OS

and it is found that 48 ml are needed.Calculate the concentration (in millimoles / L) of Na2S2O3 solution. [Ans. 0100.00 Ans. ]

ASSIGNMENT-27

Single CorrectQ.1 An element has FCC structure with edge length 200 pm. Calculate density if 200 g of this element

contains 24 × 1023 atoms. [3](A) 4.16 g cm–3 (B*) 41.6 g cm–3 (C) 4.16 kg m–3 (D) 41.6 kg m–3

Q.2 The boiling point of 1.0 m K4[Fe(CN)6] is (Kb for water = 0.52 K kg mol–1) [3](A) 100.52°C (B) 100.10°C (C) 100.26°C (D*) 102.6°C

Q.3 For CrCl3. xNH3, elevation in boiling point of one molal solution is double than that of one molal ureasolution, hence x would be (complex is 100% ionized) [3](A*) 4 (B) 5 (C) 6 (D) none of these

Q.4 The solubility of gases in liquids: [3](A) increases with increase in pressure and temperature(B) decreases with increase in pressure and temperature(C*) increases with increase in pressure and decrease in temperature(D) decreases with increase in pressure and increase in temperature

Q.5 A reaction at 300 K with G° = – 1743 J /mol consists of 3 mole of A (g), 6 mole of B (g) and 3 moleof C (g). If A, B and C are in equilbrium in 1 litre container then the reaction may be [3][Given : 2 = e0.7, R = 8.3 J/K - mol](A) A + B l C (B) A l B + 2C (C*) 2A l B + C (D) A + B l 2C

Q.6 pH of 0.1 M BOH (a weak base) is found to be 12. The solution at temperature TK will display anosmatic pressure equal to: [3](A) 0.01 RT (B) 0.01 RT (C*) 0.11 RT (D) 1.1 RT

Q.7 Equal volumes of 1.0 M KCl and 1.0 M AgNO3 are mixed. The depression of freezing point of theresulting solution will be (Kf(H2O) = 1.86 K kg mol–1, 1 M = 1m) [3](A) 3.72 K (B*) 1.86 K (C) 0.93 K (D) none of these

More than One may be correct :

Q.9 The activation energies of two reactions I and II are Ea and 2Ea respectively. If the temperature of thereacting systems is increased from T to T ', predict which of the following alternative is/are correct?(k represent rate constant) [5]

(A*) 1k/k I'I (B*) 1k/k II

'II (C*) II

'III

'I k/kk/k (D) II

'III

'I k/k2k/k

Subjective :Q.10 A gas has been subjected to an isochoric and isobaric cycle. Plot the graph

of this cycle in the pressure-density, V-T and P-T co-ordinates.

[6]

[Ans. (i) (ii) (iii) 2

34

1

P

T

]

ASSIGNMENT-28

Single correctQ.1 Select incorrect statements: [3]

(A) NH3 is soluble in water due to hydrogen bonding as well as due to formation of ions(B) Gases which can be liquefied easily are more soluble in water than the gases which cannot beliquefied(C*) The solute follows Henry law at all pressure of gas(D) The solute follows Henry law at low pressure of gas

Q.2 For an ionic crystal of the general formula AX and the coordination number 6, the values of radius ratiowill be [3](A) Greater than 0.73 (B*) In between 0.73 and 0.41(C) In between 0.41 and 0.22 (D) Less than 0.22

Q.3 The standard EMF of the cell in which the reaction, 4MnO + 5Fe2+ + 8H+ Mn2++ 5Fe3+ + 4H2O

occurs is 0.59 V at 25°. The equilibrium constant for the given reaction is approximately. [3](A) 50 (B) 10 (C*) 1050 (D) 105

Q.4 Based upon the technique of reverse osmosis the approximate pressure required to desalinate sea watercontaining 2.5% (mass/volume) NaCl at 27°C will be: [3](A) 10.5 atm (B*) 21 atm (C) 2.1 atm (D) 1.05 atm

ComprehensionQuestion No. 5 to 7 (3 questions)Freezing point of a liquid is defined as that temperature at which it is in equilibrium with its solid state.

Phase diagram for a pure solvent and solution for depression in freezing point.Q.5 Freezing point of the following system is: [3]

liquid solvent l solid solvent

(A) S.TGH

(B*) SH

(C) SG

(D) HS

Q.6 Freezing point of an ideal solution containing a non-volatile solute is smaller than that point of a solvent.It is due to: [3](A) H of solution and solvent is almost identical since intermolecular forces between solvent moleculesare involved(B*) S of solution is larger than that for the solvent(C) S of the solution is smaller than that of the solvent(D) H of the solution is much higher than of solvent but S of solution is smaller than that of the solvent

Q.7 60 g of urea is dissolved in 1100 g solution. To keep T/ Kf as 1 mol/kg, water separated in the form of ice is:(A*) 40 g (B) 60 g (C) 100 g (D) 200 g [3]

More than one may be correctQ.8 In a AB unit crystal of NaCl types assuming Na+ forming FCC [4]

(A*) The nearest neighbour of A+ is 6 B– ion (B*) The nearest neighbour of B– is 6 A+ ion

(C*) The second neighbour of A+ is 12 A+ (D) The packing fraction of AB crystal is 83

Q.9 Which of the following statement(s) is/ are correct? [4](A*) The coordination number of each type of ion in CsCl crystal is 8.(B) A metal that crystallizes in bcc structure has a coordination no. of 12.(C*) A unit cell of an ionic crystal shares some of its ions with other unit cells.

(D*) The length of the unit cell in NaCl is 552 pm ( pm181Clr,pm95Nar )SubjectiveQ.10 A solution of a nonvolatile solute in water freezes at 0.30°C. The vapor pressure of pure water at

298K is 23.51mmHg and Kf for water is 1.86 degree/molal. Calculate the vapor pressure of this solutionat 298K. [4]

[Ans. 23.44 mm Hg]

ASSIGNMENT-29

Single correctQ.1 How many nearest neighbours Cs+ are present in CsCl structure [3]

(A) 6 (B*) 8 (C) 12 (D) 4

Q.2 If the positions of Na+ and Cl– are interchanged in NaCl, the crystal lattice with respect to Na+ and Cl– is :[3]

(A*) unchanged (B) changes to 8:8 coordination from 6:6(C) additivity of ionic radii for “a” is lost (D) none

Q.3 In the closest packing of atoms, [3](A) the size of tetrahedral void is greater than that of octahedral void(B*) the size of tetrahedral void is smaller than that of octahedral void(C) the size of tetrahedral void is equal to that of octahedral void(D) the size of tetrahedral void may be greater or smaller or equal to that of octahedral void dependingupon the size of atoms.

Q.4 A mineral having the formula AB2 crystallises in the cubic close-packed lattice, with the A atoms occupyingthe lattice points. The co-ordination number of the A atoms, that of B atoms and the fraction of thetetrahedral sites occupied by B atoms are [3](A*) 8, 4, 100% (B) 2, 6, 75% (C) 3, 1, 25% (D) 6, 6, 50%

Question Number 5 to 9 (5 questions)In haxagonal close packing second row of spheres are placed in depressions of first row whereas thirdrow is vertically aligned with first, 2nd with fourth forming a pattern AB.AB...... The number of nearestatoms is called the coordination number.

Q.5 What is the coordination number of a central sphere in hcp [3](A) 8 (B*) 12 (C) 6 (D) 4

Q.6 If –r

r lies between 0.225 to 0.414, cation occupies [3]

(A*) tetrahedral void (B) octahedral void (C) trigonal void (D) cubic void

Q.7 The coordination number of octahedral void is [3](A) 8 (B) 4 (C*) 6 (D) 3

Q.8 Which of the following packing are most efficient [3](A) HCP (B) FCC (C*) A & B both (D) BCC

Q.9 ABC.ABC....... is called [3](A) HCP (B*) FCC (C) BCC (D) simple cubic

SubjectiveQ.10 10 gm of NH4Cl (mol. weight = 53.5) when dissolved in 1000 gm of water lowered the freezing point by

0.637°C. Calculate the degree of cationic hydrolysis of the salt if degree of dissociation of salt is 0.75.The molal depression constant of water is 1.86 kg mol–1 K. [5]

[Ans. h = 0.109]

ASSIGNMENT-30

Single correct

Q.1 Graph depicting correct behaviour of ideal gas & H2 gas will be (neglect a) [3]

(A*) (B) (C) (D)

Q.2 For a real gas, behaving ideally the pressure may be : [3]

(A) abVmolar (B) ab

Vmolar (C*) bVa

molar(D)

molaraVb

More than one may be correctQ.3 Following represents the Maxwell distribution curve for an ideal gas at two temperatures T1 & T2.

Which of the following option(s) are true? [4]

(A*) Total area under the two curves is independent of moles of gas(B*) If dU1= f Umps1 & dU2 = f Umps2 then A1 = A2(C) T1 > T2 and hence higher the temperature, sharper the curve.(D*) The fraction of molecules having speed = Umps decreases as temperature increases.

Q.4 Select the correct option(s). [4](A*) Fraction of molecule in the range Uavg ± f Uavg is same for SO2 and O2 at same Temperature(0 < f < 1).(B) Fraction of molecule in the range Umps ± 100 (m/sec) is same for SO2 and O2 at same Temperature.(C*) Fraction of molecule in the range Uavg ± f Uavg is same for O2 at 300 K and at 200 K (0 < f < 1).(D) None of these

SubjectiveQ.5 The gas having higher value of Vander Waal's constant "a" will be __________ compressible than the

one having lower value of "a", provided Vander Waal's constant "b" is same for both the gases. [2][Ans. more]

Q.6 At what temperature in °C, the Urms of SO2 is equal to the average velocity of O2 at 27°C. [4][Ans. 236.3°C]

Q.7 Calculate the fraction of N2 molecules at 101.325 kPa and 300 K whose speeds are in the range ofump – 0.005 ump to ump + 0.005 ump. [4] [Ans. 8.303 × 10–3]

Q.8 The density of mercury is 13.6 g/cm3. Estimate the b value. [4] [Ans. 58.997 cm3]

Q.9 The molar volume of He at 10.1325 MPa and 273 K is 0.011075 of its molar volume at 101.325 KPaat 273 K.Calculate the radius of helium atom. The gas is assumed to show real gas nature. Neglect thevalue of a for He. [4] [Ans. r = 1.33 × 10–8]

Q.10 N2 molecule is spherical of radius 100 pm. [4](a) What is the volume of molecules is one mole of a gas?(b) What is the value of vander waal's constant b?

[Ans. (a) 2.52 × 10–3 l mol–1, (b) 10.08 × 10–3 dm3 mol–1]

ASSIGNMENT-31

Single correct

Q.1 Correct option regarding a container containing 1 mol of a gas in 22.4 litre container at 273 K is [3](A) If compressibility factor (z) > 1 then 'P' will be less than 1 atm.(B*) If compressibility factor (z) > 1 then 'P' will be greater than 1 atm.(C) If 'b' dominates, pressure will be less than 1 atm.(D) If 'a' dominates, pressure will be greater than 1 atm.

Q.2 The compressibility of a gas is less than unity at STP. Therefore [3] [JEE 2000](A) Vm > 22.4 L (B*) Vm < 22.4 L (C) Vm = 22.4 L (D) Vm = 44.8 L

More than one may be correctQ.3 Which of the following statement(s) are true about Z vs P graph for a real gas at a given temperature.

[4]

(A) dPdZ 0 as P 0 for most real gases

(B*) dPdZ = –ive as P 0 for most real gases

(C) dPdZ 0 at a pressure where repulsive and attractive forces are comparable.

(D*) dPdZ = +ive for real gases at extremely high pressure.

Q.4 Select the incorrect statement(s): [4](A*) At Boyle's temperature a real gas behaves like an ideal gas irrespective of pressure.(B*) At critical condition, a real gas behaves like an ideal gas.(C) On increasing the temperature four times,collision frequency (Z11) becomes double at constant volume.(D) At high pressure Vander Waal's constant 'b' dominates over 'a'.

Q.5 Match gases under specific conditions listed in Column I with their properties / laws in Column II.Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [8]

Column I Column II(A) Hydrogen gas (P = 200 atm, T = 273 K) (P) Compressibility factor 1(B) Hydrogen gas (P ~ 0, T = 273 K) (Q) Attractive forces are dominant(C) CO2 (P = 1 atm, T = 273 K) (R) PV = nRT(D) Real gas with very large molar volume (S) P (V– nb) = nRT [JEE 2007]

[Ans. (A) P, S; (B) R; (C) P, Q; (D) R ]

SubjectiveQ.6 Sign of initial slope of compressibility factor (z) versus P curves is ________ if a gas is below its Boyle's

temperature and ________ if it is above its Boyle's temperature. [4] [Ans. –ive, +ive]

Q.7 At 273.15 K and under a pressure of 10.1325 MPa, the compressibility factor of O2 is 0.927. Calculatethe mass of O2 necessary to fill a gas cylinder of 100 dm3 capacity under the given conditions. [4]

[Ans. 15.40 kg]

Q.8 Using Vander Waals equation, calculate the constant "a" when 2 moles of a gas confined in a 4 litre flaskexerts a pressure of 11.0 atmp at a temperature of 300 k. The value of "b" is 0.05 litre mol 1. [4]

[JEE 1998][Ans. 6.46 atmp L2 mol–2]

Q.9 The compression factor (compressibility factor) for one mole of a vander Waals gas at 0° C and 100atmosphere pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculatethe vander waals constant 'a'. [4] [JEE 2001]

[Ans. 1.2544 atmp L2 mol–2]

Q.10 1 mole of CCl4 vapours at 27°C occupies a volume of 40 lit. If Vander Waals constant are 24.6 L2 atmmol–1 and 0.125 Lmol–1. Calculate compressibility factor under [6](a) Low pressure region (b) High Pressure region[Take R = 0.082 lit-atm/mol/K] [Ans. (a) 0.975; (b) 1.003]

ASSIGNMENT-32

Single correctQ.1 The term that corrects for the attractive forces present in a real gas in the Vander Waals equation is [3]

[JEE 2009]

(A) nb (B*) 2

2

Van

(C) 2

2

Van

(D) –nb

Q.2 Positive deviation from ideal behaviour takes place because of [3] [JEE 2003]

(A) molecular attraction between atoms and nRTPV

> 1

(B) molecular attraction between atoms and nRTPV

< 1

(C*) finite size of atoms and nRTPV

> 1

(D) finite size of atoms and nRTPV

< 1

More than one may be correct

Q.3 A gas described by van der Waals equation [4] [JEE 2008](A*) behaves similar to an ideal gas in the limit of large molar volumes(B) behaves similar to an ideal gas in the limit of large pressures(C*) is characterised by van der Waals coefficients that are dependent on the identity of the gas but areindependent of the temperature(D*) has the pressure that is lower than the pressure exerted by the same gas behaving ideally

SubjectiveQ.4 The density of water vapour at 327.6 atm and 776.4 K is 133.2 gm/dm3.

Determine the molar volume, Vm of water and the compression factor. [4][Ans. Molar vol = 0.1353 L/mol; Z = 0.6957 ]

Q.5 The vander waals constant for O2 are a = 1.36 atm L2 mol–2 and b = 0.0318 L mol–1. Calculate thetemperature at which O2 gas behaves, ideally for longer range of pressure. [4] [Ans. 521 K]

Q.6 The vander Waals constants for gases A, B and C are as follows [4]Gas a/dm6 kPa mol–2 b/dm3 mol–1

A 405.3 0.027B 1215.9 0.030C 607.95 0.032

Which gas has (i) the highest critical temperature, (ii) the largest molecular volume.[Ans. (i) B, (ii) C]

Q.7 For a real gas obeying van der Waal's equation a graph is plotted between PVm (y-axis) and P(x-axis)where Vm is molar volume. Find y-intercept of the graph. [4] [JEE 2004]

[Ans. RT]

Q.8 For a real gas (mol. mass = 30) if density at critical point is 0.40 g/cm3 and its Tc = 821

102 5K, then

calculate Vander Waal's constant a (in atm L2mol–2). [4] [Ans. 1.6875 ]

Q.9 A commercial cylinder contains 6.91 m3 of O2 at 15.18 M Pa and 21°C. the critical constants for O2 areTC = –118.4°C , PC = 50.1 atmp. Determine the reduced pressure and reduced temperature for O2under these conditions. [4] [Ans. PR = 2.99 , TR = 1.90]

Q.10 The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 Kg m–3. The vapoureffuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. [6]Determine(i) mol. wt.; (ii) molar volume; (iii) compression factor (z) of the vapour and(iv) which forces among the gas molecules are dominating, the attractive or the repulsive

[JEE 2002][Ans. (i) 18.1 g/mol , (ii) 50.25 L mol–1 , (iii) 1.224 , (iv) repulsive, ]

ASSIGNMENT-33

Single correctQ.1 A particle A moving with a certain velocity has a de-Broglie wavelength of 1 Å. If particle B has mass

25% of that A and velocity 75% of that of A, the of B will be approximately [3](A) 1 Å (B*) 5.3 Å (C) 0.2 Å (D) 3 Å

Q.2 If the radius of first Bohr orbit is x, then de-Broglie wavelength of electron in 3rd orbit is nearly [3](A) 2 x (B*) 6 x (C) 9 x (D) x / 3

Q.3 If E1, E2 and E3 represent respectively the kinetic energies of an electron, an alpha particle and a protonrespectively each having same deBroblie wavelength then [3](A*) E1 > E3 > E2 (B) E2 > E3 > E1 (C) E1 > E2 > E3 (D) E1 = E2 = E3

Q.4 The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H–like atom to difference inwavelength for 2nd and 3rd lines of same series is: [3](A) 2.5 : 1 (B*) 3.5 : 1 (C) 4.5 : 1 (D) 5.5 : 1

ComprehensionQuestion No. 5 to 7 (3 questions)The French physicist Louis de Broglie in 1924 postulated that matter, like radiation, should exhibit a dualbehaviour. He proposed the following relationship between the wavelength of a material particle, itslinear momentum p and planck constant h.

= ph

= mvh

The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases.It also implies that for a given velocity heavier particles should have shorter wavelength than lighterparticles. The waves associated with particles in motion are called matter waves or de Broglie waves.

Q.5 The correct order of wavelength of Hydrogen (1H1), Deuterium (1H2) and Tritium (1H3) moving withsame kinetic energy is [3](A*) H > D > T (B) H = D = T (C) H < D < T (D) H < D > T

Q.6 The transition, so that the de-Broglie wavelength of electron becomes 3 times of its initial value in He+ ionwill be [3](A) 2 5 (B) 3 2 (C*) 2 6 (D) 1 2

Q.7 If the uncertainty in velocity & position is same, then the uncertainty in momentum will be [3]

(A*) 4

hm(B)

4hm (C)

m4h

(D) 4

hm1

Match the columnQ.8 Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate

your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [12][JEE 2008]

Column I Column II(A) Orbital angular momentum of the (P) Principal quantum number

electron in a hydrogen-like atomic orbital

(B) A hydrogen-like one-electron wave (Q) Azimuthal quantum numberfunction obeying Pauli principle

(C) Shape, size and orientation of hydrogen (R) Magnetic quantum numberlike atomic orbitals

(D) Probability density of electron at the nucleus (S) Electron spin quantum numberin hydrogen-like atom

[Ans. (A) Q,R (B) P, Q, R,S (C) P, Q, R (D) P, Q]

SubjectiveQ.9 A proton is accelerated to one- tenth of the velocity of light. If its velocity can be measured with a

precision + 1%. What must be its uncertainity in position. [5][Ans. 1.05×10–13 m]

Q.10 What is de Broglie wavelength associated with an e– accelerated through potential difference = 100 KV.[5]

[Ans. 3.88 pm]

Q.11 Calculate the de-broglie wavelength associated with motion of earth (mass 6 × 1024 Kg) orbiting aroundthe sun at a speed of 3 × 106 m/s. [5]

[Ans. 3.68 × 10–65 m]

ASSIGNMENT-34

Single correctQ.1 Three energy levels P, Q, R of a certain atom are such that EP < EQ < ER. If 1, 2 and 3 are the wave

length of radiation corresponding to transition R Q ; Q P and R P respectively. The correctrelationship between 1, 2 and 3 is [3]

(A) 1 + 2 = 3 (B*) 213

111

(C) 3 = 21 (D)

213

112

Q.2 Number of possible spectral lines which may be emitted in bracket series in H atom, if electrons presentin 9th excited level returns to ground level, are [3](A) 21 (B*) 6 (C) 45 (D) 5

Q.3 The wavelength associated with a golf weighing 200g and moving at a speed of 5m/h is of the order [3](A) 10–10m (B) 10–20m (C*) 10–30m (D) 10–40m

Q.4 What is uncertainity in location of a photon of wavelength 5000Å if wavelength is known to an accuracyof 1 pm? [3](A) 7.96 × 10–14 m (B*) 0.02 m (C) 3.9 ×10–8 m (D) none

Q.5 Assuming Heisenberg Uncertainity Principle to be true what could be the minimum uncertainty in de-brogliewavelength of a moving electron accelerated by Potential Difference of 6 V whose uncertainty in position

is 227

n.m. [3]

(A) 6.25 Å (B) 6 Å (C*) 0.625 Å (D) 0.3125 Å

Assertion and Reason :Q.6 It is a data sufficiency problem in which it is to be decided on the basis of given statements

whether the given question can be answered or not. No matter whether the answer is yes or no.Question : Is the orbital of hydrogen atom 3px?

Statement-1: The radial function of the orbital is R(r) = 2/2/3

0

e)4(a69

1 , = 2r

Statement-2: The orbital has 1 radial node & 0 angular node. [3](A) Statement (1) alone is sufficient. (B*) Statement (2) alone is sufficient(C) Both together is sufficient. (D) Neither is sufficient

More than one may be correct :Q.7 Correct statement(s) regarding y3P orbital is/are [4]

(A) Angular part of wave function is independent of angles ( and )(B*) No. of maxima when a curve is plotted between 4r2R2(r) vs r are '2'(C*) 'xz' plane acts as nodal plane(D) Magnetic quantum number must be '–1'

Match the columnQ.8 Column I & column II contain data on Schrondinger Wave–Mechanical model, where symbols have

their usual meanings.Match the columns. [12]Column I Column II (Type of orbital)

(A) (P) 4s

(B) (Q) 5px

(C) (,) = K (independent of &) (R) 3s(D) atleast one angular node is present (S) 6dxy

[Ans. (A) P, (B) P,Q,S, (C) P, R (D) Q, S]

SubjectiveQ.9 A base ball of mass 200 g is moving with velocity 30 × 102 cm/s. If we can locate the base ball with an

error equal in magnitude to the of the light used (5000 Å), how will the uncertainty in momentum becompared with the total momentum of base ball. [5]

[Ans. 1.75 × 10–29]

Q.10 An electron has a speed of 40 m/s, accurate up to 99.99%. What is the uncertainity in locating itsposition. [5]

[Ans. 0.0144 m]

Q.11 A cylindrical source of light which emits radiation radially (from curved surface) only, placed at thecentre of a hollow, metallic cylindrical surface, as shown in diagram.The power of source is 90 watt and it emits light of wavelength 4000 Å only. The emitted photons strikethe metallic cylindrical surface which results in ejection of photoelectrons. All ejected photoelectronsreaches to anode (light source). The magnitude of photocurrent is[Given : h = 6.4 × 10–34 J/sec.] [5]

[Ans.10 amp]

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