Time Allowed: 3 Hours Maximum Marks: 70€¦ · · 2013-08-15Distinguish between ‘rate...

CBSE

XII

Board Paper – Set 1 (Delhi)

Chemistry Theory

[Time Allowed: 3 Hours] [Maximum Marks: 70]

General Instruction:

(i) All the questions are compulsory. (ii) Marks for each question are indicated against it. (iii) Question numbers 1 to 8 are very short answer questions and carry

one mark each. (iv) Question numbers 9 to 18 are short questions and carry two marks

each. (v) Question numbers 19 to 27 are also short answer question and

carry three marks each.

(vi) Question numbers 28 to 30 are long answer questions and carry five marks each.

(vii) Use Log Tables, if necessary. Use of calculation is not allowed.

1. ‘Crystalline solids are anisotropic in nature’. What does this statement mean?

2. Express the relation between conductivity and molar conductivity of a solution held in a cell.

3. Define ‘electrophoresis’.

4. Draw the structure of XeF2 molecule. 5. Write the IUPAC name of the following compound:

(CH3)3CCH2Br 6. Draw the structure of 3-methylbutanal. 7. Arrange the following compounds in an increasing order of their

solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2

8. What are biodegradable polymers? 9. The chemistry of corrosion of iron is essentially an electrochemical

phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.

10. Determine the values of equilibrium constant (Kc) and oG∆ for the

following reactions:

( ) ( ) ( ) ( )

( )

2 o

1

Ni s 2Ag aq Ni aq 2Ag s , E 1.05V

1F 96500 C mol

+ +

−

+ → + =

=

11. Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction. 12. State reasons for each of the following:

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(i) The N – O bond is -2NO is shorter than the N – O bond in

-3NO

(ii) SF6 is kinetically an inert substance. OR

State reason for the each of the following:

(i) All the P-Cl5 molecule are not equivalent. (ii) Sulphur has greater tendency for catenation than oxygen.

13. Assign reasons for the following: (i) Copper (I) ion is not known in aqueous solution. (ii) Actinoids exhibits greater range of oxidation states than

Lanthanoids. 14. Explain the following giving one example for each:

(i) Reimer-Tiemann reaction. (ii) Friedel Craft’s acetylation of anisole.

15. How would you obtain: (i) Picric acid (2, 4, 6-trinitrophenol) from phenol, (ii) 2-Methylpropene from 2-methylpaopanol?

16. What is essentially the difference between α -form of glucose and β -

form of glucose? Expain. 17. Describe what do you understand by primary structure and secondary

structure of proteins.

18. Mention two important uses of each of the following: (i) Bakelite

(ii) Nylon 6

19. Silver crystallizes in face-centred cubic unit cell. Each side of this

unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal

across the face of the unit cell. That is each face atom is touching the four corner atoms.)

20. Nitrogen pentaoxide decomposes according to equation:

( ) ( ) ( )2 5 2 22N O g 4NO g O g→ +

This first order reaction was allowed to proceed at 40 oC and the

data below were collected:

[N2O5] (M) Time (min)

0.400 0.00

0.289 20.0

0.209 40.0

0.151 60.0

0.109 80.0

(a) Calculate the rate constant. Include units with your answer. (b) What will be the concentration of N2O5 after 100 minutes? (c) Calculate the initial rate of reaction.



21. Explain how the phenomenon of adsorption finds applications in each of the following processes:

(i) Production of high vacuum (ii) Heterogeneous cataklysis (iii). Froth floation process

OR

(i) Micelles

(ii) Peptization (iii) Desorption

22. Describe the principle behind each of the following processes:

(i) Vapour phase refining of a metal. (ii) Electrolytic refining of a metal.

(iii) Recovery of silver after ore was leached with NaCN. 23. Complete the following chemical equations:

(i) 2

4 2 4MnO C O H− − ++ + →

(ii) heated

4KMnO →

(iii)2

2 7 2Cr O H S H− ++ + →

24. Write the name, stereochemistry and magnetic behaviour of the

following:

(At. Nos. Mn = 25, Co = 27, Ni = 28) (i) K4[Mn(CN)6]

(ii) [Co(NH3)5Cl] Cl2 (iii) K2[Ni(CN)4]

25. Answer the following:

(i) Haloalkanes easily dissolve in organic solvents, why? (ii) What is known as a racemic mixture? Give an example.

(iii) Of the two bromoderivatives, C6H5CH(CH3)Br and C6H5CH(C6H5)Br, which one is more reactive in SN1

substitution reaction and why? 26. (a) Explain why an alkylamine is more basic than ammonia. (b) How would you convert:

(i) Aniline to nitrobenzene (ii) Aniline to iodobenzene?

27 Describe the following giving one example for each: (i)Detergents (ii)Food preservatives

(iii)Antacids 28. (a) Difference between molarity and molality for a solution. How

does a change in temperature influence their values? (b) Calculate the freezing point of an aqueous solution containing

10.50 g of MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g)

(Kf for water = 1.86 K kg mol-1)



OR

(a) Define the terms osmosis and pressure. Is the osmotic pressure of a solution a colligative property? Explain.

(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water.(Kb for water = 0.512 K kg mol-1), Molar mass of NaCl = 58.44 g.

29. (a) Give the chemical test to distinguish between (i) Propanal and propanone,

(ii) Benzaldehyde and acetophenone. (b) How would you obtain

(i) But-2-enal from ethanal,

(ii) Butanoic acid from butanol, (iii) Benzoic acid from ethylbenzene?

OR

(a) Describe the following giving linked chemical equations: (i) Cannizzaro reaction

(ii) Decarboxylation (b) Complete the following chemical equations:

30. (a) Explain the following:

(i) NF3 is an exothermic compound whereas NCl3 is not. (ii) F2 is most reactive of all the four common halogens.

(b) Compete the following chemical equations:

( )

( )

2 4

4 2

2 2

(i) C H SO conc

(ii) P NaOH H O

(iii) C F

access

+ →

+ + →

+ →�

OR

(a) Account for the following: (i) The acidic strength decreases in the order HCl>H2S >PH3



(ii) Tendency to form pentahalides decreases down the group

in group 15 of the periodic table. (b) Complete the following chemical equations:

( )

4 2 2

2 2

2 3

(i) P SO Cl

(ii) XeF H O

(iii) I HNO

conc

+ →

+ →

+ →



CBSE

XII Solution to Board Paper – Set 1 (Delhi)

Chemistry Theory

Solution 1: This statement means that some of the physical properties of

crystalline solids such as electrical resistance or refractive index show different values when measured along different directions in the same

crystals. Solution 2. The Molar conductivity of a solution at a given concentration is

related to conductivity of that solution, by the following relation.

m kA

∧ =�

Where, m∧ is molar conductivity and k is the conductivity of the solution.

Solution 3. Electrophoresis is the phenomenon of movement of colloidal particles under the applied electric potential.

Solution 4. XeF2 is a linear molecule and adopts the following structure:

Solution 5.

The IUPAC name of the given structure is 2, 2-Dimethylbromopropane. Solution 6.



Solution 7. The solubility order of the given amines is as follows

C6H5NH2 < (C2H5)2NH < C2H5NH2 Reason:

The more extensive the H-bonding, the higher is the solubility. C2H5NH2

contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus C2H5NH2 undergoes more extensive H-bonding than (C2H5)2NH. Hence, the

solubility in water of C2H5NH2 is more than that of (C2H5)2NH. Further, the solubility of amines decreases with increase in the molecular

mass. This is because the molecular mass of amines increase with an increase in the size of the hydrophobic part. The molecular mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH. Thus, the solubility of

C6H5NH2 is less than that of C2H5NH2 and (C2H5)2NH.

Solution 8. A polymer that can be decomposed by bacteria is called a biodegradable polymer. For example: poly- β -hydroxybutyrate-CO-β -hydroxyvalerate (PHBV) is

biodegradable aliphatic polyester.

2 2

3 2 3 n

O CH CH C O CH CH C

| || | ||

CH O CH CH O

PHBV

− − − − − − − −

Solution 9. The process of corrosion is a redox reaction that involves

simultaneous oxidation and reduction reactions. It can therefore be referred to as an electrochemical reaction.

In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is can be written as follows.

Anodic reaction: 2Fe(s) Fe (aq) 2e+ −→ +

Electrons released at the anodic spot move through the metallic object and

go to another spot of object. There, in the presence of H+ ions, the electrons reduce molecular oxygen. This spot behaves as the cathode. There H+ ions



come either from H2CO3 , which are formed due to the dissolution of carbon

dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.

The reaction corresponding at the cathode is written as follows.

Cathodic reaction: 2 2O (g) 4H (aq) 4e 2H O(l)+ −+ + →

Thus, the overall reaction is: 2

2 22Fe(s) O (g) 4H (aq) 2Fe (s) 2H O(l)+ ++ + → +

Also, ferrous ions are further oxidized by atmosphere oxygen to ferric ions.

These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide (Fe2O3, xH2O) i.e., rust.

Solution 10.

( ) ( ) ( ) ( )2 o

s aq aq s

2

ocell

1

o or cell

o 1r

Ni 2Ag Ni 2Ag , E 1.05V

The galvanic cell of the given cell reaction is depicted as :

Ni(s)|Ni || Ag | Ag(s)

Now, the s tandard cell potential is

E 1.05V

n 2

F 96500 C mol

G nFE

G 2 96500 Cmol 1.0

+ +

+ +

−

−

+ → + =

=

=

=

∆ = −

∆ = − × ×

1

1

5 V

202650 Jmol

202.65 kJmol

−

−

= −

= −

Solution 11. The rate expression can be defined as an expression in which

the rate of reaction is given as the product of the molar concentration of the reactants, with each term raised to some power, which may or may not be the stoichiometric coefficients of the reacting species in a balanced chemical

equation. The rate constant can be defined as the rate of reaction when the

concentration of each of the reactant is taken as unity.

Example: 2 22NO(g) O (g) 2NO (g)+ →

The rate expression for the above reaction can be written as follows: 2

2Rate k[NO] [O ]= (Experimentally determined)

Now, if the concentration of NO and O2 is taken to be unity, then the rate constant is found to be equal to the rate of the reaction.



Solution 12.

(i) The Shorter N – O bond in 2NO− is due to the existence of resonance

in 2NO− . The resonating structure can be drawn as follows.

Due to resonance in 2NO−, the two bonds are equivalent. This leads to a

decrease in bond length. Thus, the N – O bond length in 2NO− resembles a

double bond.

Now, the resonating structure for 3NO− can be drawn as:

As seen from the above resonating structure of 3NO− , the three oxygen

atoms are sharing two single bonds and one double bond. So the real N-O bond length resembles a single bond closely.

This explains the existence of shorter bond length of the N-O bond in 2NO−

than in 3NO− .

(ii) The kinetic inertness of SF6 can be explained on the basis of its structure.



As seen from the above structure, the six fluorine (F) atoms protect the

sulphur atom from attack by the regents to such an extent that even thermodynamically most favourable reactions like hydrolysis do not occur.

OR

(i) In gaseous and liquid state, PCl5 has a trigonal bipyramidal structure. In this structure, the two axial P – Cl bonds are longer

and less stable than the three equatorial P – Cl bonds. This is because of the greater bond pair – bond pair repulsion in then axial bonds. Hence, all the bonds in PCl5 are not equivalent.

(ii) Because of stronger S-S bonds as compared to O-O bonds, sulphur has a greater tendency for catenation than oxygen.

Solution 13.

(i) In aqueous solution, Cu+ ion undergoes oxidation to Cu2+ ion. The

relative stability of different oxidation states can be seen from their electrode potentials.

( )+ o

redCu (aq) e Cu (s), E 0.52V−+ → =

( )2+ o

redCu (aq) 2e Cu(s), E 0.34V−+ → =

Due to more reduction electrode potential value of Cu+, it undergoes oxidation reaction quite feasibly. Hence, Copper (I) ion

is not known in aqueous solution. (ii) The actinoids show a larger number of oxidation states because of

very small energy gap between the 5f, 6d and 7s sub-shells. Hence all their electrons can take part in bond formation.

Solution 14. Riemer-Tiemann reaction: Riemer-Tiemann reaction involves the treatment of phenol with chloroform in the presence of aqueous

sodium hydroxide at 340 K followed by hydrolysis of the resulting product to give 2-hydroxybenzaldehyde (salicylaldehyde). The chemical reaction can be represented as follows.

(i)



(ii)Friedel-Crafts acetylation of anisole: Friedel-Crafts acetylation of

anisole involves the treatment of anisole with either acetyl chloride or acetic anhydride to give 2-methoxyacetophenone (as a mirror product) and 4-

methoxyacetophenone (as a major product), the chemical reaction can be represented as follows.

Solution 15.

(i) Phenol on reaction with concentrated HNO3 results in the formation

of picric acid.

(ii) 2-Methyl propene can be obtained from 2-methyl propanol by the

reaction of the later with alc. KOH



Solution 16. The α-form of glucose and β -form of glucose can be

distinguished by the position of the hydroxyl group on the first carbon atom. In open chain α -glucose, the hydroxyl group on the first carbon atom is towards the right whereas, in the closed ring α - glucose, the hydroxyl group on the first carbon atom is below the plane of the ring. On the other hand, in open chain β -glucose, α -glucose, the hydroxyl group

on the first carbon atom is towards the left whereas, in the closed ring α -glucose, the hydroxyl group on the first time atom is above the plane of the

ring. The structures of open and closed α -form and β - form of glucose can be

drawn as follows.

Solution 17.

I. Primary structure of proteins: Each polypeptide chain in

a protein has amino acids linked with each other in a specific



sequence. This sequence of amino acids is said to be the

primary structure of proteins. II. Secondary structure of proteins: The secondary structure

of proteins refers to the shape in which a long polypeptide can exist. The two different secondary structure possible are α -Helix structure and β – pleated sheet structure.

• α -Helix structure: In α -Helix structure, a polypeptide chain forms all possible hydrogen bonds by twisting into a helix with – NH group of each amino acid residue and hydrogen bonded to >C=O of an adjacent turn of helix.

• β -Helix structure: In a β -pleated structure, all peptide

chains are stretched out of nearly maximum extensions and then laid side by side which are held together by intermolecular hydrogen bonds.

Solution 18. (i) Uses of Bakelite:

(a) It is used for making combs. (b) It is used for manufacturing electrical switches.

(ii) Uses of Nylon 6:

(a) It is used for making tyre cords. (b) It is used for making fabrics and mountaineering

ropes. Solution 19. Given, silver crystallizes in fcc unit cell

aSo, r

2 2=

Where e is the radius of the silver atom and a is the edge length

Now, edge length = 400 pm = 10400 10−× cm

Thus, 10

10

400 10 cmr

2 1.414

141.44 10 cm

141.4 pm

−×=

×

= ×

=

Thus, the radius of the silver atom was found to be 141.4 pm

Solution 20. (a)The plot of [N2O5] v/s t is as follows



[N2O5] (M) Time (min) log[N2O5]

0.400 0.00 -.03979

0.289 20.0 -0.5391

0.209 40.0 -0.6798

0.151 60.0 -0.8210

0.109 80.0 -0.9625

From the plot, log [N2O5] v/s t, we obtain

( )0.70 0.60Slope

40 20

0.70 0.60 0.10

20 20

− − −=

−

− + −= =

Also, slope of the line of the plot = k

2.303

−



3 1

k 0.10

2.303 20

0.10or k 2.303 0.01151

20

or, k 1.15 10 min− −

−∴ =

= × =

= ×

(b) After 100 min

2 5 o

2 5 t

1

[N O ]2.303k log

t [N O ]

After 100 min

2.303 0.400k log

t 0.098

0.1406 min−

=

=

=

(c) The initial rate of reaction

2 5

3

4 1

r k[N O ]

1.15 10 0.400

4.6 10 s

−

− −

=

= × ×

= ×

Solution 21. (i) Production of high vacuum: Traces of air can be adsorbed by

charcoal from a vessel, evacuated by a vacuum pump to give a very high vacuum.

(ii) Heterogeneous catalysis: The gaseous reactants are adsorbed

on the surface of the solid catalysts. As a result, the concentration of the reactants increase on the surface and hence the rate of the

reaction increases. (iii) Froth floatation process: This process is used to remove gangue

from sulphide ores. The basic principle involved in this process is

adsorption. In this process, a mixture of water pine oil is taken in tank. The

impure powdered sulphide ore is dropped in through hopper and the compressed air is blown in through the agitator is rotator is

rotated several times. As a result, froth is formed and the sulphide ores get adsorbed in the froth. The impurities settled down and are let out through an outlet at the bottom. The froth formed is

collected in froth collector tank. After sometime, the ore particles in the froth collecting tank start settling gradually, which are then

used for further metallurgical operations. OR



(i) A micelle is an aggregate of surfactant molecules dispersed in a a

liquid. A micelle in aqueous solution forms as aggregate such that the hydrophilic “head” regions are in the centre of micelle.

(ii) Peptization is the process of conversion of a precipitate into a colloidal sol by shaking it with the dispersion medium in the presence of an electrolyte. The electrolyte used in this reaction is

known as a peptizing agent. (iii) Desorption is the process of removing an adsorbed substance from

the surface through which it was adsorbed. Solution 22.

(i) Vapour phase refining : Vapour phase refining is the process of refining metal by converting it into its volatile compound and then,

decomposing it to obtain a pure metal. The basic principle involved in this process are: (a) The metal should form a volatile compound with an available

reagent, and (b) The volatile compound should be easily decomposable so that

the metal can be easily. Nickel, zirconium, and titanium are refining using this method.

(ii) Electrolytic refining of a metal is the process of refining impure

metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution

of a solution salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal

from the anode dissolves into the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the

anode. This is known as anode mud. n

n

Anode : M M ne

Cathode : M ne M

+ −

+ −

→ +

+ →



(iii) In the process of leaching, the finely divided silver is treated with

dilute solution of sodium cyanide while a current of air is

continuously passed. As a result, silver pass into the solution forming solution dicyanoargenate(I) while the impurities remain

unaffected which are filtered off. ( )2 2

2Ag S 4NaCN 2Na[Ag CN ] Na S

Sodium dicyanoarg enate (I)

+ → +

Solution 23.

(i) 2 2

2 4 4 2 25C O 2MnO 16H 2Mn 8H O 10CO− − + ++ + → + +

(ii) Heated

4 2 4 2 22KMnO K MnO MnO O→ + +

(iii) 2 3

2 7 2 2Cr O 8H 3H S 2Cr 7H O 3S− + ++ + → + +

Solution 24. (i) K4[Mn(CN)6]

Name: Potassium hexacyanomanganate(II) Stereochemistry – Does not show geometric or optical isomerism Magnetic behaviour – Paramagnetic

(ii) [Co(NH3)5Cl] Cl2 Name: Pentaamminedchloridocobalt (III) chloride

Stereochemistry – Does not geometric isomerism but is optically active Magnetic behaviour – Paramagnetic

(iii) K2[Ni(CN)4] Name: Potassium tetracynoinickelate (II)

Stereochemistry – Does not show geometric or optically isomerism Magnetic behaviour – Diamagnetic

Solution 25.



(i) Haloalkanes can easily dissolve in organic solvents of low polarity

because the new forces of attraction set up between haloalkanes and the solvent molecules are of same strength as the forces of

attraction being broken. (ii) A mixture of equal amounts of two enantiomers is known as

racemic mixture.

For example: When a 3o halide undergoes substitution with KOH, the reaction proceeds through SN1 mechanism forming the racemic

mixture in which one of the products has the same configuration as a reactant, while the product has an inverted configuration.

(iii) The SN1 substitution reaction involves the formation of carbocation,

which is not affected by the presence of bulky groups. Thus, C6H5CH(C6H5)Br will be more reactive towards SN1

substitution reaction forming racemic mixture.

Solution 26. (a) The basicity of amines depends on the +I effects of the alkyl group.

The presence of –CH3 group in alkylamine increases the electron

density on the nitrogen atom and thus increases the basicity. Hence, alkylamine is more basic than ammonia

CH3NH2 > NH3 (b) (i)



(ii)

Solution 27. (i) Detergents: A detergents is a surfactant or a mixture of

surfactants having cleaning properties in dilute solution. Commonly, detergent refers to alkylbenzenesulphonates. For example: Sodium dodecylbenzene sulphonate

(ii) Food preservatives: Food preservatives are chemicals that prevent food from spoilage due to microbial growth. Table salt, sugar,



vegetable oil, sodium benzoate (C6H3COONa), and salts of propanoic acid are

some examples of food preservatives. (iii) Antacids: Any drug that is used to counteract the effects of excess

acid in the stomach and raise the pH to an appropriate level is called an antacid. Example: Omeprazole

Solution 28. (a) Molartiy is defined as the number of moles of solute dissolved per litre of solution.

Mathematically M = 3

Number of moles of solute

Volume of solution in litres (dm )

Molality of a solution is defined as the number of moles of solute

dissolved in 1000 grams of solvent.

Mathematically, m = Number of moles of the solute

Mass of solvent in kg

While molarity decreases with an increase in temperature, molality

is independent of temperature. This happens because molality involves mass, which deos not change with a change in

temperature, while molarity involves volume, which is temperature dependent.

(b) Given w2 = 10.50 g

w1 = 200g

Molar mass of MgBr2 (M2) = 184 g

Using the formula,

fT∆ = f 2

1 2

1000 k w

w M

× ×

×

= 1000 1.86 10.50

200 184

× ×

×

= 19.530

200 184× = 0.53

Now, Tf = To - fT∆

= 273 – 0.53 = 272.47 K

OR



(a) Osmosis : The process of flow of solvent molecules from pure

solvent to solution or from solution of lower concentration of solution of higher concentration through a semi – permeable

membrane is called osmosis.

Osmotic pressure : The pressure required to just stop the flow of

solvent due to osmosis is called osmotic pressure (π) of the solution. Yes, the osmotic pressure of a solution is colligative property. The osmotic pressure is expressed as.

nRT

Vπ =

Where, π = osmotic pressure

n = number of moles of solute

V = volume of solution

T = temperature

From the equation, it is clear that osmotic pressure depends upon

the number of moles of solute ‘n’ irrespective of the nature of the solute. Hence, osmotic pressure is a colligative property.

(b) Given, Kb = 0.512 k kg mol-1

w2 = 15.00 g

w1 = 250.0 g

M2 = 58.44 g

Using the formula,

b 2b

1 2

1000 K wT

w M

× ×∆ =

×

= 1000 0.512 15.00

250.0 58.44

× ×

×

= 7.680

14.600 = 0.52

Now, Tb = To + bT∆

= 373 + 0.53 = 373.53 K

Ans29. (a)



(i) Propanal (CH3CH2CHO) can be distinguished from propanone

(CH3COCH3) by iodoform test.

Being a methyl ketone, propanone on treatment whith I2/NaOH

undergoes iodoform reaction to give a yellow ppt. of iodoform

CH3COCH3 + 3NaOI → CHI3 + CH3COONa + 2NaOH

Propanone Iodoform

Propanal on the other hand does not give this test.

CH3CH2CHO NaOI

→ No yellow ppt. of

Propanal Iodoform

(ii) Benzaldehyde (C6H5CHO) and acetophenone (C6H5COCH3) can

be distinguished by iodoform test.

Acetophenone, being a methyl ketone on treatment with I2/NaOH

undergoes iodoform reaction to give a yellow ppt. of iodoform. On the other hand, benzaldehyde does not give this test.

C6H5COCH3 + 3NaOI

Acetophenone

C6H5COONa + CHI3 + 2NaOH

Iodoform

C6H5CHO NaOI

→ No yellow ppt of iodoform

Benzaldehyde

(b)

(i)



(ii) CH3CH2CH2CH2OH 2 2 7 2 4

2 4

(i)K Cr O /H SO

(ii) Dil.H SO→CH3CH2CH2COOH

Butanol Butanoic acid

(iii)

OR

(i) Cannizaro reaction

In this reaction, the aldehydes which do not have an α -

hydrogen atom, undergo self oxidation and reduction



(disproportionation) reaction on treatment with a concentrated

alkali.

Example:

(ii) Decarboxylation

The decarboxylation reaction can be carried out either by using soda lime or by electrolysis

• Using soda lime

Sodium salts of carboxylic acids when heated with soda lime (NaOH + CaO) in the ratio 3:1 undergoes decarboxylation

reaction to yield alkanes.

R – COONa NaOH CaO

Heat

−→ R – H + Na2CO3

(Alkane)

• Electrolytic decarboxylation

Electrolysis of aqueous solutions of sodium of potassium salts

of carboxylic acids give alkanes having twice the number of carbon atoms present in the alkyl group of acid. This is

known as Kolbe’s decarboxylation.

2RCOONa → 2RCOO- + 2Na+

H2O → 2OH- + 2H+

At Anode:-

2H+ + 2e- → H2

(b)

(i)



(ii)

(iii)

C6H5CONH2 3

H O

heat

+

→ C6H5COOH

Benzoic acid

Ans30. (a)

(i) As we move down the group 17, the size of the atom increases

from fluorine to chlorine. The larger difference in the size of N and Cl results in the weakness of strength of N – Cl bond. On the other hand,

the difference in size of N and F is small; consequently the N – F bond is quite strong. As a result, NF3 is an exothermic compound.

(ii) Due to the small size of F atom, the three lone pair of electrons on each F atom F – F molecule repels the bond pair. As a result, F – F is

most reactive of all the four common halogens.

(b)

(i) C + 2H2SO4 → 2SO2 + CO2 + 2H2O

Sulphur dioxide

(ii) P4 + 3NaOH + 3H2O 2

,CO∆→ PH3 + 3NaH2PR2

Phosphine

(iii) Cl2 + 3F2 → 2ClF3



(excess) Chlorine trifluoride

OR

(a)

(i) In a period, the electro negativity decreases in the order Cl > S >

P. As a result, the loss of H+ ions decreases.

Thus, the acidic strength of the hydrides decreases in the following

order: HCl > H2S > PH3

(ii) The tendency to form pentahalides decreases down the group 15

due to inert pair effect i.e., in Bi the s-electrons remain inert and do not take

part in bonding.

(b)

(i) P4 + 10SO2Cl2 → 4PCl5 + 10SO2

(ii) 2XeF2 + 2H2O → 2Xe + 4HF + O2

(iii) I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O

(conc).



Date post:	27-May-2018
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