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S.No Questions Answers Q.1 What is meant by ‘doping’ in a semiconductor?

Ans.1 In semiconductor production, doping intentionally introduces impurities into an extremely pure (also referred to as intrinsic) semiconductor for the purpose of modulating its electrical properties. The impurities are dependent upon the type of semiconductor. Lightly and moderately doped semiconductors are referred to as extrinsic.

Q.2 What is the role of graphite in the electrometallurgy of aluminium? Ans.2 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.3 Which one of PCl4+ and PCl4-is not likely to exist and why?

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

Q.4

Ans.4Bromo-2 –methyl -1-Propene

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

Q.6 Arrange the following compounds in an increasing order of their reactivity in

nucleophilic addition reaction. Ethanol ,propanal,propanone,butanone. Ans.6 Ethanol>Propanal>Propanone>Butanone.

Q.7 Arrange the following in the decreasing order of their basic strength aqueous

solution.

Ans.7(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.

Q.8 Define the term homo polymerisation giving an example.

Ans.8 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.

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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 Name the two groups into which phenomenon of catalysis can be divided. Given an example of each group with the chemical equation involved.

Ans.11 Homogeneous catalysis and heterogeneous catalysis. Homogeneous catalysis is a sequence of reactions that involve a catalyst in the same phase as the reactants.Eg: CH3CO2CH3 + H2O CH3CO2H + CH3OH Heterogeneous catalysis refers to the form of catalysis where the phase of the catalyst differs from that of the reactants.eg: 2SO2(g) Fe(S) 2SO3(g)

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

Ans.12The 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 neutralised. 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 coagulation or meteral coagulation.

(iii) Dialysis: Dialysis is the artificial process of eliminating waste (diffusion) and unwanted water (ultrafiltration) from the blood. Our

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kidneys do this naturally. Q.13 Describe the principle involved in each of the following processes.

(i)Mond’s process for refining of Nickel. (ii) Column Chromatography for purification of rare elements.

Ans.13(i) Monds process-The impure nickel reacts with carbon monoxide at 50–60 °C to form the gas nickel carbonyl, leaving the impurities as solids.

Ni(s) + 4 CO(g) → Ni(CO)4(g)

The mixture containing nickel carbonyl (and synthesis gas) is heated to 220–250 °C, resulting in decomposition back to nickel and carbon monoxide:

Ni(CO)4(g) → Ni(s) + 4 CO(g (ii)Column Chromatography for purification of rare elements : Principle of this technique is that different components of a mixture are differently absorbed on an absorbent.the components of the mixture get absorbed on the column at different places.

Q.14 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.14 (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.15

Ans.15(i) Cr2O72- +14 H+ +6 I- 2Cr+3 + 7H2O + 3I2

(ii) 2MnO4- + 5NO2

- + 6H+ 2Mn+2 + 5NO3- + 3H2O

Q.16 What is meant by (i)Pepetide linkage (ii)Bio-catalysts

Ans.16(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.17 Write any two reactions of glucose which cannot be explained by open chain structure of glucose molecule.

Ans.17 (i) The penta-acetate of glucose doesn’t react with NH2OH indicating the absence of aldehydic group. (ii) Glucose doesn’t give 2,4-DNP test and Schiffs test even having free –CHO group.

Q.18 Draw the strucuture of the monomer for each of the following polymers: (i)Nylon 6 (ii) polypropene

Ans.18(i)

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(ii)

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 Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K.(Kf of water= 1.86K Kg/mol)

Ans.20 wB =? wA = 1000 g ΔTf = 2 K Kf = 1.86 K kg mol−1 MB of KCl = 39.09 + 35.45 = 74.54 mol−1

wB=80.56gm.

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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 How would you account for the following? (i) Many of the transition elements are known to form interstitial compounds. (ii)The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group member of 4-d series. (iii)Lanthanoids from primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds +4 and even +6 is being typical.

Ans.22 (i) Transition metals are large in size and contain lots of interstitial sites so transition elements form a few interstitial compounds with elements possessing small atomic radii, like hydrogen, carbon boron, and nitrogen. Small atoms of these type of elements get entrapped in between the void spaces (called as interstices) of the metal lattice.for eg-TiC, Mn4N, Fe3H, TiH2 etc. (ii) It is due to lanthanoids contraction across the lanthanide series, electrons are added to the 4f shell. This first f shell is inside the full 5s and 5p shells (as well as the 6s shell in the neutral atom); the 4f shell is well-localized near the atomic nucleus and has little

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effect on chemical bonding. (iii) The wide range of oxidation states of actinoids is attributed to the fact that the 5f, 6d and 7s energy levels are comparable therefore all these three sub shells can participate. But the most common oxidation state of actinoids is +3.

Q.23 Give the formula of each of the following coordination entities: (i) Co3+ ion is bound to one Cl−, one NH3 molecule and two bidentate ethylene diamine (en) molecules. (ii) Ni2+ ion is bound to two water molecules and two oxalate ions. Write the name and magnetic behaviour of each of the above coordination entities. (At. nos. Co = 27, Ni = 28)

Ans.23(i) [CoClNH3(en2)2+ amminebis(ethane −1, 2− diamine)chlorocobalt(III) ion

Number of unpaired electrons = 4 So the complex is paramagnetic.

Q.24 Although chlorine is an electron withdrawing group, yet it is ortho­,

para­directing in electrophilic aromatic substitution reactions. Explain why it is so?

Ans.24 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)

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Q.25 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.25 (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.26 Write chemical equation for the following conversions: (i) Nitrobenzene to benzoic acid. (ii) Benzyl chloride to 2­phenylethanamine. (iii) Aniline to benzyl alcohol.

Ans.26

(ii) Benzyl chloride to 2­phenylethanamine.

(iii) Aniline to benzyl alcohol.

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Q.27 What are the following substances? Give one example of each one of them.

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

Ans.27 (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.

Q.28 (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

Ans.28 (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+ + SO4

2– PbSO4 (electrode) + 2 H2O 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.

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solution at 298 K is 0.146 × 10−3 S cm−1? E0 = Ered - Eox E0=0.80-(0.76) =0.80 +0.76 =1.56V ΔG=-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 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

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Cell Constant =k × R = 0.146 × 10−3 × 1500 == 0.219 cm−1

Q.29 (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.29 (a) (i) P4 + 10SO2Cl2 4PCl5 + 10SO2 (ii) XeF6 + H2O XeOF4 + 2HF XeF6 +2H2O XeO2F2 +4HF (b) (i)

(ii)

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

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(b) (i) (ii) (iii) Q.30 (a) Illustrate the following name reaction giving suitable example in each case

(i) Clemmensen reduction (ii) Hell –Volhard- Zelinsky reaction (b) How the following conversions carried out ? (i) Ethylcyanide to ethanoic acid (ii) benzoic acid to m- bromobenzouc acid (iii) Butanol to butanoic acid

OR (a)Illustrate the following name reactions given a suitable example for each (i) Cross aldol condensation (ii) Decarboxilation (b) Given simple test to distinguish between the following pairs of compound s (i) Pentan-2- one and Pentan-3-one (ii)Benzaldehyde and Acetophenone (iii) Phenol and Benzoic acid

Ans.30(a) Clemmensen reduction

The Clemmensen Reduction allows the deoxygenation of aldehydes or ketones, to produce the corresponding hydrocarbon. The substrate must be stable to strong acid. The Clemmensen Reduction is complementary to the Wolff-Kishner Reduction, which is run under strongly basic conditions. Acid-labile molecules should be reduced by the Wolff-Kishner protocol. Eg-

(ii) Hell-Volhard-Zelinsky Reaction

Treatment with bromine and a catalytic amount of phosphorus leads to the selective α-bromination of carboxylic acids. Eg-

- (b) (i)

(ii) ) benzoic acid to m- bromobenzouc acid

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(iii) Butanol to butanoic acid

OR

(i) Cross aldol condensation Crossed aldol condensation is a variation of aldol condensation. aldol condensation:

Crossed Aldol condensation:

Eg.

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(ii) Decarboxilation

In decarboxylation,the –COOH or –COONa group is removed and replaced with a hydrogen atom.

(b)(i) ) Pentan-2- one and Pentan-3-one Will Give +ve Iodoform test Do not give +ve Iodoform test CH3-CH2-CH2-CO-CH3, CH3-CH2- CO-CH2-CH2(ii)Benzaldehyde and Acetophenone

By Tollen’s reagent Test R=C6H5-

Acetophenone does not give this test. (iii) ) Phenol and Benzoic acid

Like ethyl benzoate Phenol does not give effervescence.