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S.No Questions Answers Q.1 Write a point of distinction between a metallic solid and an ionic solid other

than metallic luster.

Ans.1 Ionic solids: They are poor conductor of electricity and therefore are insulators in solid state. Metallic solids: They are good conductors of electricity and heat in solid state.

Q.2 Which one of PCl4+ and PCl4-is not likely to exist and why?

Ans.2 PCl4- is not likely to exist because in this the oxidation state of P is +3 which is less stable.

Q.3 What is the role of graphite in the electrometallurgy of aluminum?

Ans.3 In the electrometallurgy of aluminium, a fused mixture of purified alumina (Al 2 O 3 ), cryolite (Na 3 AlF 6 ) and fluorspar (CaF 2 ) is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, Al is liberated at the cathode, while CO and CO 2 are liberated at the anode, according to the following equation.

Q.4 Arrange the following compounds in an increasing order of their reactivity in

nucleophilic addition reaction. Ethanol, propanal, propanone, butanone. Ans.4 Ethanol>Propanal>Propanone>Butanone.

Q.5 Draw the structural formula of 2methylpropan2ol molecule.

Ans.5

Q.6

Ans.6 3-Bromo-2 –methyl -1-Propene

Q.7 Define the term homo polymerisation giving an example. Ans.7 Polyethylene is an example of a homopolymer that is formed by polymerizing a single monomer. Copolymers are formed by polymerizing more than one monomer. Ethylene (CH2=CH2) and propylene (CH2=CHCH3) can be copolymerized, for example, to produce a polymer that has two kinds of repeating units.

Q.8 Arrange the following in the decreasing order of their basic strength aqueous solution.

Ans.8 (CH3)2NH> CH3NH2> (CH3)3N> NH3 Due to +I effect of alkyl group basic character 2o is high but in tertiary aminehave steric hinderence hence sequence of order is like this.

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Q.9

Ans.9 Ans. Elevation temperature (T)=100.18-100 =0.18K Now, i=mole fraction=T/Kbm i= 0.18/0.512x1=0.35

OR (i) Mole fraction. : the ratio of the number of molesof one component of a

solution or other mixture to the total number ofmoles representing all of the components.

(ii) An isotonic solution refers to two solutions having the same osmotic pressure across a semipermeable membrane. This state allows for the free movement of water across the membrane without changing the concentration of solutes on either side of the membrane.

(iii) The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of a substance as calculated from its mass.

(iv) ideal solute is a solution in which the interaction between molecules of the components does not differ from the interactions between the molecules of each component; usually : asolution that conforms exactly to Raoult's law.

Q.10 What do you understand by the ‘order of a reaction’? identify the reaction order from each of the following units of reaction rate constant:

(i) L-1molS-1 (ii) Lmol-1S-1

Ans.10 The order of reaction with respect to a given substance (such as reactant catalyst or product) is defined as the index, or exponent, to which its concentration term in the rate equation is raised.

(i) zero order reaction (ii) second order reaction

Q.11 Describe a conspicuous change observed when: (i) A solution of NaCl is added to a sol of hydrated ferric oxide. (ii) A beam of light is passed through a solution of NaCl and then

through a sol.

Ans.11 (i) When NaCl is added to ferric oxide sol, it dissociates to give Na+ and Cl−

ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl− ions.

(ii) When a beam of light is passed through a solution of NaCl, then scattering of light is be served. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution.

Q.12 What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation carried out.

Ans.12 The process of setting of colloidal particles is called coagulation of the sol. It is also known as precipitation. Following coagulation of lyophobic sols can be carried out.

(i) Electrophoresis: In this process, the colloidal particles move towards oppositely charged electrodes and get discharged

(ii) Mixing of two oppositely charged sols: When oppositely charged sols are mixed in almost equal proportions, their charges are neutralized. Both sols may be partially or completely precipitated as the mixing of ferric hydroxide (+ve sol) and arsenious sulphide (–ve sol) bring them in precipitated form. This type of coagulation is called mutual

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coagulation or meteral coagulation. (iii) Dialysis: Dialysis is the artificial process of eliminating waste

(diffusion) and unwanted water (ultrafiltration) from the blood. Our kidneys do this naturally.

Q.13 . Describe the following: (i) The role of cryolite in electro metallurgy of aluminium. (ii) The role of carbon monoxide in the refining of crude nickel.

Ans.13 (i) Cryolite is used in the electrolytic reduction of alumina so as to reduce its

melting point and make it a good conductor of electricity. (ii) Carbon monoxide is used in the purification of nickel because it reacts

with nickel to give a volatile complex called nickel tetracarbonyl, which on heating, decomposes to gives pure nickel metal.

Q.14 What is meant by (i) Pepetide linkage (ii) Bio-catalysts

Ans.14 (i) The chemical bond formed between amino acids, constituting the

primary linkage in all protein structures. In a peptide bond, the carboxyl group (COOH) of one amino acid bonds with the amino group (NH2) of another, forming the sequence CONH and releasing water (H2O).

(ii) Biocatalysts is the natural catalysts, such as protein enzymes, to perform chemical transformations on organic compounds.

Q.15 Explain the following giving an appropriate reason in each case: (i)O2 and F2 both stabilize higher oxidation state of metals but O2 exceeds F2 in doing so. (ii) Structures of Xenon Flourides cannot be explained by valence bond approach.

Ans.15 (i) Both O2 and F2 stabilize high oxidation states..but higher oxidation state of O2 exceeds that of F2 because oxygen has -2 charge whereas fluorine has -1 charge..therefore more force between O^-2 and Metal and hence higher oxidation state. (ii) Xenon has completely fill electronic filled configuration in the valence cell, whereas accordingvalence bond theory valence electron take part in bonding so, structure of XeF2 cannot be explained by valence bond approach.

Q.16

Ans.16 (i) Cr2O7

2- +14 H+ +6 I- 2Cr+3 + 7H2O + 3I2 (ii) 2MnO4

- + 5NO2 - + 6H+ 2Mn+2 + 5NO3

- + 3H2O

Q.17 Draw the strucuture of the monomer for each of the following polymers: (i)Nylon 6 (ii) polypropene

Ans.17

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Q.18 Write the main structural difference between DNA and RNA. Of the two bases, thymine and uracil, which one is present in DNA?

Ans.18 The structural differences between DNA and RNA are - (i) The sugar in DNA is deoxyribose lacking oxygen atom at 2’C, while that in RNA is ribose having oxygen atom at 2’C. (ii) DNA has a doublestranded helical structure, while RNA has a single stranded helical structure. (iii)DNA contains the base thymine, while RNA contains uracil.

Q.19 Tungsten crystallizes in BCC unit cell if the edge of unit cell is 315.5pm ,what is radius of tungsten?

OR Fe has a BCC unit cell with a cell dimension of 286.65 pm ,the dnsity of Fe is 7.874 gm/cm3. Use the information to calculate Avogadro number(atomic mass of Fe=55.845gm/mole)

Ans.19 For BCC unit cell r= √ a , (a=316pm) = 1.723x316.5/4 = 137.044pm.

OR Given cell dimension a= 286.65pm, number of atom in the BCC unit cell (Z) =2, atomic mass of Fe=55.84gm /mole, density of Fe = 7.874gm/cm3 Density = Z x atomic mass/a3xNo N0=Z x M/a3xdensity =2x55.84/(286.65x10-10)3x7.784 =6.09x1023mole-1

Q.20 A solution of glycerol (C3H8O3) in water was prepared by dissolving some

glycerol in 500 g of water. This solution has a boiling point of 100.42°C while pure water boils at 100°C. What mass of glycerol was dissolved to make the solution? (Kb = for water = 0.512 K kg mol−1)

Ans.20 WB=? WA = 500g Kb = 0.512 Kkg mol−1 ΔTb = 100.42°C − 100°C = 0.42°C MB = 3 × 12 + 8 × 1 + 3 × 16 = 36 + 8 + 48 = 92

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Q.21

Ans.21(a) Rate law is given by R=K[X]a[Y]b =K[NO]a[Cl2]b Now for experiment -1 and -2 R1=K[NO]a[Cl2]b =K[0.15]a[0.15]b =0.60 R2=K[NO]a[Cl2]b R2=K[0.15]a[0.30]b =1.20 R1/R2= K[0.15]a[0.15]b/ K[0.15]a[0.30]b = 0.60/1.20 Or (0.5)b =(0.5)1 So b=1 Thus order with respect to Cl =1 Now by comparing eq. 1 and 3 R1=K[NO]a[Cl2]b R1=K[0.15]a[0.55]b=0.60 R3=K[NO]a[Cl2]b R3= K[0.30]a[0.15]b=2.40 R1/R3= K[0.15]a[0.55]b/ K[0.30]a[0.15]b=0.60/2.40 ( ½)a=1/4 (1/2)a=(1/2)2 So a=2 So order with respect to NO is 2 Thus the rate law is R=K [NO]2[Cl2] (b)To calculate rate constant K=rate/[NO]2[Cl2] =0.60/(0.15)2(0.15)=177.77 mol-2 L2S-1 (c) Rate of reaction is given by Ex.4 R=K [NO]2[Cl2] =177.77 x (0.25)2 x(0.25) =2.77 mol L-1 min-1.

Q.22 State a reason for each of the following situations: (i) Co2+ is easily oxidized to Co3+ in presence of a strong ligand. (ii) Co is a stronger complexing reagent than NH3. (iii) The molecular shape of Ni(CO)4 is not the same as that of

[Ni(CN)4]2−.

Ans.22 (i) In Co+2 crystal field stabilization energy(CFSE) is less due to presence unpaired e- in higher eg0 level whereas in Co+3 has higher CFSE due to absence of unpaired e- in this higher eg0 level in presence of strong ligand. (ii) Co is a stronger complexing reagent than NH3because back bonding in case of Co which is a good sigma donor and 휋 acceptor. In Co donation of e- from filled d-orbital of metal to π* MO of Co takes place. (iii)In Ni (CO)4 , Ni is in the zero oxidation state i.e., it has a configuration of 3d84s2.

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Q.23 How would you account for the following? (i) With the same dorbital configuration (d4) Cr2+ is a reducing

agent while Mn3+ is an oxidizing agent. (ii) The actinoids exhibit a larger number of oxidation states than the

corresponding members in the lanthanoid series. (iii) Most of the transition metal ions exhibit characteristic in colours

in aqueous solutions.

Ans.23 (i) The configuration of Cr+2 is d4 so it is strong oxidizing agent. While acting

as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be written as configuration, which is a more stable configuration. In the case of Mn3+d4), it acts as an oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half filled dorbital and is highly stable.

(ii) Because actinoids have 5f, 6d and 7s energy levels which are comparable energies. Therefore all these three sub shells can participate. But the most common oxidation state of actinoids is +3.

(iii) It is due to: (a)charge transfer transitions. An electron may jump from a predominantly ligand orbital to a predominantly metal orbital, giving rise to a ligand-to-metal charge-transfer (LMCT) transition. These can most easily occur when the metal is in a high oxidation state.

(b) d-d transitions. An electron jumps from one d-orbital to another. In

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complexes of the transition metals the d orbitals do not all have the same energy. The pattern of splitting of the d orbitals can be calculated using crystal field theory.

Q.24 Write chemical equation for the following conversions: (i) Nitrobenzene to benzoic acid. (ii) Benzyl chloride to 2­phenylethanamine. (iii) Aniline to benzyl alcohol.

Ans.24

(ii) Benzyl chloride to 2­phenylethanamine.

(iii) Aniline to benzyl alcohol.

Q.25 What are the following substances? Give one example of each one of them.

(i) Tranquilizers (ii) Food preservatives (iii) Synthetic detergents

Ans.25 (i) Tranquilizers- a medicinal drug taken to reduce tension or anxiety. Eg-

chlordiazepoxidehydrochloride (ii) Food preservatives-A preservative is a substance that is added to products

such as foods, pharmaceuticals, paints, biological samples, wood, etc. to prevent decomposition by microbial growth or by undesirable chemical changes. In general preservation is implemented in two modes, chemical and physical.Eg-Sodium benzoate,Vineger.

(iii) Synthetic detergents- synthetic detergent. Noun. Any synthetic substance, other than soap, that is an effective cleanser and functions equally well as a surface-active agent in hard or soft water. Eg- : Sodium p­dodecylbenzenesulphonate.

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Q.26 Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidizing agent is used.

(i) CH3CH2CH2CH2OH (ii) 2­butenol (iii) 2­methyl­1­propona

Ans.26 (O) (i) CH3CH2CH2CH2OH CH3CH3CH2CO2H. (O) (ii) ) 2­butenol CH3-CO-CH=CH2 (3-Oxobutene) (O) (iii) 2­methyl­1­propona CH3-CH-COOH CH3 (2- methyl propanoic acid)

Q.27 Although chlorine is an electron withdrawing group, yet it is ortho­, para­directing in electrophilic aromatic substitution reactions. Explain why it is so?

Ans.27 This is due to availability of e- on Cl- in the chlorobenzene ,Cl- act as O- and P- directing that means it will send the incoming electrofile (E+) on the O- and P- position of benzene . The O- and P- position carry high e- charge density due to resonance which can be shown as (X=Cl)

Q.28 (a) Complete the following chemical reactions equations: (i)P4 + SO2Cl2 (ii)XeF6 + H2O (b) Predict the shape and the asked angle (90° or more or less) in each of the following cases: (i) SO3

2-and the angle O − S − O (ii) ClF3 and the angle F − Cl – F

OR (a) Complete the following chemical equations: (i)NaOH(hot &conc.) + Cl2 (ii)XeF 4 + O2F2 (b) Draw the structures of the following molecules: (i) H3PO2 (ii) H2S2O7 (iii)XeOF4

Ans.28(a) (i) P4 + 10SO2Cl2 4PCl5 + 10SO2 (ii) XeF6 + H2O XeOF4 + 2HF

XeF6 +2H2O XeO2F2 +4HF (b)(i)

(ii)

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OR

(a) (i) 6NaOH(hot &conc.) + 3Cl2 5NaCl + NaClO3 +3H2O (ii) XeF 4 + O2F2 XeF6 + O2 (b)

(i) (ii) (iii)

Q.29 (a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it. (b) In the button cell, widely used in watches, the following reaction takes place Determine E° and ΔG° for the reaction

OR

(a) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte. (b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10−3 S cm−1?

Ans.29 (a) A lead storage battery has a secondary cell. Thus, it can be recharged by passing direct current through it. Therefore, it can be reused. It is used in automobiles. In a lead storage cell, the anode is made of spongy lead and the cathode is a grid of lead packed with lead dioxide. The electrolyte used is H2SO4.

Pb Pb2+ + 2 e–

reaction at the anode is therefore:

Pb + SO42– PbSO4 (electrode) + 2 e–

reaction at the cathode is:

2 e– + PbO2 + 4 H+ + SO42– PbSO4 (electrode) + 2 H2O

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Overall reaction is: Pb + PbO2 +4H+ + SO4

2- 2 PbSO4 + 2H2O b) From the given reaction, it is known that zinc is oxidised and silver is reduced in the button cell. E0 = Ered - Eox E0=0.80-(0.76) =0.80 +0.76 =1.56V

G=-n F E0 = -2 × 96500 × 1.56 =-301.080 KJ OR

(a)Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution. Its units are siemens per meter per molarity, or siemens meter-squared per mole. The usual symbol is a capital lambda, Λ, or Λm.

the molar conductivity is given by

where:

κ is the measured conductivity c is the electrolyte concentration.

For strong electrolytes, such as salts, strong acids and strong bases, the molar

conductivity depends only weakly on concentration.

where

is the molar conductivity at infinite dilution (or limiting molar

conductivity)

K is the Kohlrausch coefficient

For weak electrolytes (i.e. incompletely dissociated electrolytes), the molar

conductivity strongly depends on concentration: The more dilute a solution, the

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greater its molar conductivity, due to increased ionic dissociation.

where:

is the molar ionic conductivity of ion i.

is the number of ions i in the formula unit of the electrolyte .

(b)Conductivity, K=0.146 × 10-3 Scm-1 Resistance ,R =1500 Ohm Cell Constant =k × R = 0.146 × 10−3 × 1500 == 0.219 cm−1

Q.30 (a) Give a plausible explanation for each one of the following: (i) There are − NH2 groups in semicarbazide. However, only one such group is involved in the formation of semicarbazones. (ii) Cyclohexanone forms cyanohydrins in good yield but 2, 4, 6trimethylcyclohexanone does not. (b) An organic compound with molecular formula C9H10O forms 2, 4, DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1, 2benzenedicarboxylic acid. Identify the compound. OR (a) Give chemical tests to distinguish between (i) Phenol and Benzoic acid (ii) Benzophenone and Acetophenone (b) Write the structures of the main products of following reactions: (i)

Ans.30 (i) Semicarbazide has two –NH2 groups. The attachement of the two –NH2

groups on the C=O is different. One –NH2 is directly attached to C=O while the other –

NH2 is not attached directly but attached through –NH.Therefore due to resonance the

electron density on this –NH2 group decreases and therefore cannot act as a

nucleophile.The lone pair of electrons in –NH2 which is attached through –NH is not

involved in resonance and therefore is available for nucleophillic attack on the C=O .The

resonance structures of semicarbazide is shown below.

(ii) 2,2,6-trimethylcyclohexanone has three methyl group at the alpha position .There is a lot of steric hinderance in this molecule.Due to steric hinderance the CN- ion do not attack the molecule.There is no steric hinderance in Cyclohexanone.Therefore it forms cyanohydrin in good percentage.The structure of both the compounds are given below.

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(b) The compound is aldehyde since it give Cannizaro reaction therefore –CHO group is directly attach to benzene ring hence compound is-

The given reactions are as follows:

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OR

(i) ) Phenol and Benzoic acid

Like ethyl benzoate Phenol does not give effervescence.

(ii)Benzaldehyde and Acetophenone By Tollen’s reagent Test R=C6H5-

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Acetophenone does not give this test.

(i)

+ ==> + HCl benzene + benzoyl chloride (AlCl3+CS2)==> diphenylmethanone + hydrogen chloride Hg2++H2SO4 (ii) CH3-C=CH CH3-CH2-CHO (H2O)