CLASS XII CHEMISTRY

CONTENTS

1. Solid State

2. Solutions

3. Electro Chemistry

4. Chemical Kinetics

5. Surface Chemistry

6. General Principles and Processes of Isolation Of Elements

7. p-Block elements

8. The d-and f-block elements

9. Co-ordination compounds

10.Haloalkanes and Haloarenes

11.Alcohols, Phenols and Ethers

12.Aldehydes, Ketones and Carboxylic Acids

13.Amines

14.Biomolecules

15.Polymers

16.Chemistry in Everyday life

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Chapter-1

SOLID STATE

One mark:1. Glass is called as pseudo solid (or) super cooled liquid. why? Glass, even though a solid, moves like liquid very slowly. So, it is called as super cooled liquid.2. Does plastic show cleavage property? Why? No, because plastic is an amorphous substance.3. NaCl does not conduct electricity in the solid state. why? In the solid state, the ions are tightly packed and they are unable to move.4. How does a hexagonal unit cell vary from a monoclinic cell? For Hexagonal unit cell a=b≠ c , α = β= 90˚ , δ= 120˚ For monoclinic unit cell a≠ b ≠ c, α = δ = 90˚ , β≠ 90˚ .5. A solid has very high melting point and an electrical insulator in solid as well as in Molten state. What type of solid is it? It must be a covalent net work solid like diamond.6. What type of stoichiometric defect is shown by 1) ZnS 2) AgBr? ZnS Shows Frenkel defect because its ions have large difference in size. AgBr shows both Frenkel and Schottky defect. 7. What type of defect can arise when a solid is heated? Which physical property is affected by this? On heating, vacancy defects are created in solids because some atoms leave the lattice Site. This reduces the density of the crystals.8. A group 14 element is to be converted in to n – type semi conductor by doping it with a suitable impurity. To which group should this impurity belong? As the n- type semi conductor conducts electricity by excess electrons, group 14 element should be doped with group 15 elements.9. Among paramagnetic and ferromagnetic substances, which one forms better permanent magnet. Why?

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Ferromagnetic substances. This is because, when placed in the magnetic field, the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced.

2 marks 1. In a crystal, oxygen form Fcc lattice and Al occupy 2/3 of octahedral voids present in the crystal. What is the formula of the solid? Let the no of oxygen atom is = 1. The no of octahedral voids would be =1 So, the ratio of Al and oxygen atom is = Al 2/3 O Hence the formula is Al2 O3

2. What type of magnetic property is shown by MgFeO4? What happens to its magnetic property when it is heated? Mg FeO4 shows ferrimagnetisms. On heating it changes to Para magnetism due to irregular arrangement of domains.3. ZnO is colorless in ordinary temperature, but turns yellow on heating. Why? When ZnO is heated, some of the oxygen ions move out of the crystal as oxygen gas.The excess Zn ions move to the neighboring sites while the electrons occupy the anionic site. So, there are F- center created and thus it becomes, colored.4. Among octahedral void and tetrahedral voids which one is larger in size. Why? Octahedral void is larger. This is because octahedral void is surrounded by 6 particles and tetrahedral void is surrounded by 4 particles.5. In NaCl & in CsCl what is the co ordination number of Na+ & Cs+ ions. Why? The co-ordination number of Na+ is 6 & that of Cs+ is 8. This is because NaCl has FCC structure with each Na+ & Cl- ions are surrounded by the opposite ions in octahedral manner. CsCl has BCC structure in which each Cs+ & Cl- ions are surrounded by 8 opposite charged ions.6. Al crystallizes in a cubic close- packed structure. Its metallic radius is 125 pm. 1) What is the length of the side of the unit cell? 2) How many unit cell are there in 1 cm3 of Al? 1. For ccp, that is Fcc structure, edge length is 2√2 r. r = 125 pm

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so, edge length = √2 150 pm. = 354 pm. 2. Volume of one unit cell = (354 * 10 -10cm)3

No of unit cells present = 1/ 4.44* 10 -23

= 2.25 * 10 22.

3 marks

1. Germanium is an intrinsic semiconductor. What do you mean by this? Explain how can we increase the conductivity of such conductor?

It means that the conductivity of ‘Ge’ increases with increase in temperature. By doping their conductivity can be increased. While doping, it can be converted to ‘n’ type or ‘p’ type semiconductor.

2. What are non – stoichiometric solids? Nickel oxide has the formula Ni 0.98 O 1.00. What fraction of nickel exists as Ni2+ & Ni3+ ions?

Those solids in which the ratio of atoms is not present as given by the formula is called as non-stoichiometric solids. Suppose x atoms of Ni exist as Ni2+ & (98-x) atoms as Ni3+. As the no of +ve & -ve charges must be equal in number for the electrical neutrality, ( x * 2+) + (98-x )* 3+ = 200 2x + 294 – 3x = 200 x = 94 Ni 2+ = 94 Ni 3+ = 4 % of Ni 2+ = 96% & that of Ni 3+ = 4%

3. calculate the packing efficiency of FCC lattice. Refer page 18 topic 1.7.1

3. Draw the unit cell of BCC &FCC lattice. Calculate the number of particles present in one unit cell.

For BCC & FCC unit cells refer page 11 fig 1.10 (a) & 1.12 (a). Calculation of no of particle present : FCC: particle at corners = 1/8 * 8 = 1 atom Particle at face center = ½ * 6 = 3 atoms . total = 4 atoms.

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BCC : particle at corners = 1/8 * 8 =1 Particle at body centre = 1. Total = 1+1 =2 atoms.

Chapter-2

Solution

1mark:1. What is the significance of ppm unit of expressing concentration?Ppm is used to express the very small concentration of pollutants.2. Among one molar solution of H2SO4 and 1 molal solution of H2SO4,Which one contains more percentage by mass of H2SO4?1 molar solution. Because 1 molal solution contain more amount ofwater.3.The rate of evaporation of a liquid depends on the surface area of theliquid.100 ml each of water is kept in a measuring cylinder of 2cmdiameter and in Petridish of 5cm diameter. Which one has more vapourpressure? Why?Both have same vapour pressure.since vapour pressure does not dependon rate of vapourisation.4. The Henry’s law constant for two gases A & B are 40.3 k bar and 1.67

k bar respectively. Which one has more solubility in water at one atmosphere. Why? Gas B has more solubility. Since KH is inversely proportional to solubility at aparticular pressure.5. When a liquid A is mixed with liquid B the vapour pressure of the solution decreases. What type of deviation from ideal behaviour does the solution show?The solution shows negative deviation.since decrease of vapour pressure occurs due to new force of attraction between the molecule.6. A shrinked (limped ) carrot kept in fresh water becomes firm and fresh. Why?This is due to osmosis. Water enters through the skin to the solution.

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7. The molecular mass of acetic acid determined by colligative properties measurement using benzene as solvent show abnormal value. Why?This is because acetic acid undergo association in benzene.8. Why is the freezing point of salt water lower than that of pure water?The vapour pressure of solid state of the solution and the liquid state of the solution becomes equal at lower temperature. So the freezing point decreases. 2 marks:1. With a neat diagram of vapour pressure versus temperature, show that sea water has higher boiling point than pure water?Refer page 49 of NCERT text book.2. Derive the unit of cryoscopic constant of a solvent. Why is it constant for a particular solvent at a particular temperature.∆Tf = Kf * mKf = ∆Tf /m = K /mol/kgIt is constant for a particular solvent because it is calculated for one mole of a solution. So, it is independent of concentration.3. For a solvent if the enthalpy of fusion is greater then Kf value is smaller. Is thistrue? Support your answer with reason.Yes. Kf = R * M * Tf

2 / 1000 * ∆fus H.4. What type of solid solution is formed by tungsten carbide? Why?Interstitial solid solution. Because, the size of tungsten and carbon differwidely.5. The depression in freezing point of water observed for the same amount of aceticacid, trichloro acetic acid and trifluoro acetic acid increases in the order givenabove. Explain briefly.This is because the ionization of these compound differ due to the chlorinesubstitution.

3 marks:1. How many ml of a 0.1M HCL are required to react completely 1g mixture ofNa2CO3 and NaHCO3 containing equimolar amounts of two.Let the amount of Na2CO3 be x in the mixture .Amount of NaHCO3 = 1 – xMoles of Na2CO3 = x/ 106

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Moles of NaHCO3 = 1 – x / 84.Since number of moles of both are equal.x/ 106 = 1-x / 84.84x = 106 – 106 xx = 0.558Moles of Na2CO3 = 0.558/ 106 = 0.00526Moles of Na2CO3 = 1 – 0.558 / 84 = 0.00526According to the reactions,1 mol of Na2CO3 will react with 2 mol of HCl0.00526 mol of Na2CO3 will react with 2 x 0.00526 mol of HClSimilarly, 1 mole of NaHCO3 will react with 1 mole of HCl0.00526 mol of NaHCO3 will react with 0.00526 mol of HClThere fore the total moles of HCl required is 2 x 0.00526 + 0.00526= 0.01578 mol.So, the volume of 0.1M HCl required = 1000 x 0.01578 / 0.1= 157.8 ml2. A solution containing 0.5g of KCl dissolved in 100g of water freezes at – 0.24˚. Calculate the degree of dissociation of salt. (Kf for water = 1.86˚C)

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Chapter-3

Electrochemistry

1 MARK

1. Can you store copper sulphate solutions in a zinc pot? Justify.Ans. Copper sulphate cannot be stored in a zinc pot zinc is more reactive than copper in electrochemical series. So a redox reaction will immediately take place when Cuso4 solution is placed in zinc pot

Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s)

2. Arrange the following metals in the order in which they displace each other from the solution of their salts.Ans . Al,Cu,Fe,Mg and ZnA metal is able to displace other metals from their salt solutions: If the metal has a lower electrode potential than the others. The above metals are, therefore, arranged in the increasing order of their Eº values, so the order is given as

Mg > Al > Zn >Fe > Cu

3. Out of Zn and Sn which one protects iron better even after cracks and why?

Ans. Zn protects better because oxidation potential of Zn is greater but that of tin is less than that of iron

4.Account why alkaline medium inhibits the rusting of iron?Ans. Rusting takes place in the presence of H+. An alkaline medium

inhibits the

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rusting of iron by neutralizing H+ ions. Thus the presence of OH-

furnished by alkaline solution removes H+ from the reaction and retards

the oxidation of Fe to Fe2+

2 MARKS

(Q1) Calculate the potential of hydrogen electrode in contact with a solution whose pH is10.

(A1) If pH of solution is 10, that means [H+] = 10-10 M. According to Nernst Equation,

EH+/H2 = EoH+/H2 – log [ ]

EH+/H2 = 0 - log [ ]

EH+/H2 = 0 – 0.0591 x 10

EH+/H2 = - 0.0591 V.

(Q2) Using the standard electric potentials below, predict if the reaction between the following is feasible.

(a) Ag+(aq) & Cu(s) solution

(b)Fe3+(aq) & Br-

(aq)

EoFe3+/Fe2+ Eo

Cu+/Cu EoAg+/Ag Eo

Br2/Br-

Eo 0.77 V 0.34 V 0.80 V 1.09 V

(A2) (a) The two half cells are Cu2+/Cu, Ag+/Ag and net reaction is

Eocell = Eo

cathode – Eocathode

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= EoAg+/Ag – Eo

Cu2+/Cu

= 0.80 – 0.34 = 0.46 V

The reaction is feasible as Eocell is positive.

(b)2Fe3+ + 2e- 2Fe2+

2Br- + 2e- Br2___________________Br- + Fe3+ Fe2+ + ½Br2

Eocell = Eo

cathode – Eocathode

= EoFe3+/Fe2+ – Eo

Cu2+/Cu

= 0.77 – 1.09 = -0.32 V

The reaction is not feasible as Eocell is negative.

(Q3) Predict the products of electrolysis in each of the following:(i) An aqueous solution of AgNO3 with silver electrodes.(ii) An aqueous solution of AgNO3 with platinum electrodes.

(A3) (i) On electrolysis of an aqueous solution of AgNO3 with silver electrodes, silver will get deposited on cathode and the silver anode will be dissolved slowly.

Cathode: Ag+ + e- AgAnode: Ag Ag+ + e-

(ii) On electrolysis of an aqueous solution of AgNO3 with platinum electrodes, silver will be obtained at the cathode and oxygen will be evolved at the anode.

Cathode: Ag+ + e- Ag Anode: 2H2O O2 + 4H+ + 4e-

(Q4) The Eo values for two metal electrodes are given below. i) Cr3+/Cr2+ = - 0.4 V ii) Fe3+/Fe2+ = 0.8 VComment on the result of treating a solution of Cr (II) with a solution containing Fe (III) ions.

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(A4) On the basis of above value, we can say the value of reduction electrode potential of Cr3+/Cr2+ is – 0.4 V. It is less than the reduction electrode potential of Fe3+/Fe2+ is + 0.8 V. In both cases, Cr2+ will be oxidized to Cr3+ and Fe3+ will be reduced to Fe2+.Cr2+ + Fe3+ Cr3+ + Fe2+

Thus, when Fe3+ ion is added, to Cr2+ solution, Fe3+ will be reduced to Fe2+ and Cr2+ will be oxidized to Cr3+.

(Q5) Explain why electrolysis of aqueous solution of NaCl gives H2 at cathode and Cl2 at anode. Write overall reaction.

(EoNa+/Na = - 2.71 V, Eo

H2O/H2 = - 0.83 V, EoCl2/2Cl- = 1.36 V; Eo

H+ + O2 / H2O = 1.23 V)

(A5) Reaction at Cathode: Na+(aq) + e- Na(s) 2H2O(l) + 2e- H2(g) + 2OH-

(aq)

Standard reduction potential for reduction of water (- 0.83 V) is more than the standard reduction potential of Na+(aq) (- 2.71 V). Hence, reduction of water takes place in reference to Na+ ions and H2 gas will liberate at cathode.

Reactions at Anode:2Cl-(aq) Cl2(g) + 2e-

H2O(l) ½O2(g) + 2H+(aq) + 2e-

Actually, oxidation of water needs large activation energy due to slow process of transfer of electros. It creates electrical resistance at anode surface, Extra voltage is required to overcome this resistance. Due to requirement of extra voltage for oxidation of H2O, oxidation of Cl- ions takes place in preference to water. Hence, chlorine is given out at anode. NaCl(aq) +H2O Na+(aq)+ OH-(aq) + ½H2(g)+ ½ Cl2(g)

3 MARKS

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1. Using the standard electrode potentials given below, predict if the reactions are

feasible:(a) Fe3+(aq) and I- (aq)

EoI2/I- =0.541V, Eo

Cu2+/Cu =0.34V(b)Ag+(aq) and Cu(s)

EoBr

2-/Br- =1.09V, Eo

Ag+/Ag=0.80(c) Fe3+(aq) and Br- (aq) Eo

Fe3+

/Fe2+ =0.77V

Ans:(a) In this reaction, Fe3+ ions are reduced by I- ions. As Eo

Fe3+/Fe2+ is greater than Eo

I2/I-, the reaction is feasible.

(b) In this reaction, Ag+ ions are being reduced. This is feasible, as Eo

Ag+/Ag is greater than EoCu2+/Cu

(c) This reaction is not feasible, as EoFe3+/Fe2+ is less than Eo

Br2/Br- Hence Br- ions can’t reduce Fe3+ ions.

2. Predict the products of electrolysis in the following:(i) An aqueous solution of AgNO3 with silver electrodes(ii) An aqueous solution of AgNO3 with platinum electrodes(iii) A dilute solution of H2SO4 with platinum electrodes(iv) An aqueous solution of CuCl2 with platinum electrodes

Given that EoredH2O= -0.41V; Eo

oxi of H2O=-1.23V; Eo

Ag+/Ag =+0.80V; EoCu2+/Cu =0.34V

Ans .(i) At cathode: Ag ion will get reduced.

Ag+ + e-AgAt anode: Ag will get oxidized. Ag Ag++ e-

(ii) At cathode: Ag++ e- AgAt anode: H2O will get oxidized. H2O 2H+ + ½ O2 + 2e-

(iii) At cathode: 2H+ + 2e- H2

At anode: CuCl2 Cu2+ + 2Cl-

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(iv) At cathode: Cu2+ will get reduced Cu2+ + 2e- Cu At anode: Cl- ion will get oxidized

2Cl- Cl2 + 2e-

3.Calculate the electrode potential of the following half cell: Cr2O7

2-(aq) + 14H+ (aq) +6e- 2Cr3+(aq)+7H2O(l) Eored = +1.33 V

When [Cr2O72-] = 0.2 M; [H+] = 1M; [Cr3+] =0.1M.

Ans: n=6 Q= [Cr 3+ ] 2 [H 2O] 7 [Cr2O7

2-][H+]14 = (0.1M) 2 x 1 = 0.05 (0.2M) x (1)14

log Q = log 0.05= log 5 X 10-2 = 0.6990-2 Ered=Eo

red – 2.303 RT log Q nF At 25oC Ered = Eo

red – 0.0591 log Q n

= +1.33 V - 0.0591 x (-1.3010) n

= 1.33 V + 0.013 V = +1.343 V

_____________________________________________________________

NOTE :- While substituting the datas in the formula, the datas are to be substituted along with units only and not only the datas.

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Chapter-4

Chemical Kinetics

1. For a chemical reaction XY, the rate increases by the factor 2.25 when concentration of X is increased by 1.5 suggest the rate law and order.

( 1 mark )Ans: r=k[X] 2; order is 2.

2. For a chemical reaction AH, the rate is found to increase by the factor 1.59 when conc. of A is doubled. What is the order?

( 1 mark )Ans: 2/3

3. When does a rate of the reaction become equal to specific reaction rate? Ans: r=k[A]n. Where n=0,then r=k. ( 1 mark )

4. The rate law for the decomposition of N2O5 is rate = k[N2O5]. What is significance of k in this equation? ( 1 mark )Ans: k is known as specific rate constant.

5. Nitric oxide, NO reacts with oxygen to produce nitrogen dioxide: 2NO(g) + O2(g) 2NO2(g) What is the predicted rate law, if the mechanism is NO + O2 NO3 (fast)

NO3 + NO NO2 + NO2 (slow)? ( 2 mark )Ans: The rate law is given by slowest step i.e, R = k1 [NO3] [NO] As NO3 is obtained from the fast step, therefore its conc is calculated from Step 1 :

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k= [NO3] / [NO][ O2]. Here k is equilibrium constant or

k[NO][ O2]= [NO3]. Substituting this value in rate equation, we get

or r= k1 X k[NO][ O2][NO] = k’[NO]2[O2].

6. What would be the probable plot of conc. vs. time for zero order reaction? ( 2 mark )Ans :

7. The initial rate of the reaction A + B2 C2 + D is doubled if the initial concentration of A is doubled, but is quadrupled if the initial concentration of b2 is doubled:

i) What is the order with respect to each of the reactants?ii) What is the overall order of the reaction?iii) Could this possibly be a single step reaction? Explain. ( 2 mark )

Ans : 1st order w.r.t. A and 2nd order w.r.t B2.

8. Nitric oxide, NO reacts with oxygen to produce nitrogen dioxide: 2NO(g) + O2(g) 2NO2(g) The rate law for this reaction is Rate=k[NO]2[O2] Propose a mechanism for the above reaction. ( 2 mark )Ans : The rate law indicates that order of reaction is 2 w.r.t. NO and 1 w.r.t. O2. The possible mechanism for given reaction may be, NO+O2 NO3 …..(Fast step) NO3 +NO NO2 +NO2 …..(Slow step) Overall reaction,(by addition of two steps) 2NO+O2 2NO2

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As slowest step of mechanism of reaction determine the rate of reaction, Rate=k2[NO3][O2]

[NO2] = k1[NO][O2](Since NO3 is intermediate specie and its formation is in equilibrium state) Rate = k1k2[NO][O2][NO] = k[NO]2[O2] … (Where k is rate constant and k=k1k2) The above expression of rate law derived from proposed mechanism is same as in given data.

9. Following reaction takes place in one step, 2NO(g)+O2(g) 2NO2(g) How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one-third of its original volume? Will there be any change in the order of the reaction with the reduced volume?

( 2 mark ) Ans : Rate=k[NO2]2[O2]

Conc = M/VConc 1

Vol

Rate=k [ Mass of NO ]2 [O2] 1/3 V 1/3V

Rate = k X 32 X 3 [NO2]2[O2] = k X 27 X [NO2]2[O2]The rate increases by 33 = 27 times. There will be no change in order of the reaction. 10. First order reaction is 15% complete in 20 minutes. How long will it take to be 60% complete? ( 3 mark )Ans: Use the following relation to calculate k, rate constant of first order reaction. t = log Here, t = 20minutes, a = 100, x = 15 (for completion of 15% of reaction)

20 = 2.303 log _100_

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k 100-15

There fore Rate constant, k = 2.303 log _100_ t 85

= 2.303 X _0.0706_ 20 1

= 0.00813 min-1

Again use the relation for completion of 60% if reaction: t60%=2.303 log _a_ k a-x

Now, a=100, a-x =100-60 = 2.303 log _100_ 0.00813 40 = 2.303 X _0.3979_ = 112.7 minutes 0.00813 1 The 60% completion of reaction will take 112.7 minutes.

11. A certain reaction is 50% complete in 20 minutes at 300k and the same reaction is again 50% complete in 5 minutes at 350 K. Calculate energy of activation it is a reaction of first order. ( 3 mark )

Ans : Let k1 and k2 be the rate constants at 300k and 350k respectively. t1/2= 0.693 , we can write k k1= 0.693 min-1 and k2= 0.693 min-1

20 5

Ea is given by, log k2 =_Ea__{ T2-T1} k1 2.303 T1T2

log [ 0.693 X __20__ ]5 0.693

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= ____Ea_______ [350-300 ] 2.303 X 8.314 300X350 log 4 = ____Ea_______ [__50___ ] 2.303 X 8.314 300X350 Ea = 0.06021 X 2.303 X8.314 X 300 X 350 50 = 24210 J mol-1

= 24.21 KJ mol-1

NOTE:- The students are asked to substitute the values in the formulae along with the units.

18

Chapter-5

Surface Chemistry

Q1.Gelatine is generally added to ice creams. Why?Ice cream is an emulsion of milk or cream in water (o/w type). Gelatine is

generally added to act as emulsifiers to form the stable emulsion.

Q2. Why is chemisorption called activated absorption? For a chemisorption to take place ,reactants should acquire minimum energy of activation.

Q3. Why is desorption necessary in catalytic reaction?Desorption is necessary in catalytic reaction to maintain the flow of reaction.

Q4. In chemical reaction can a product formed act as catalyst? If so give example.Yes, Example: hydrolysis of ester. Ester hydrolysis is catalysed by acid. In the course of the reaction carboxylic acid is produced which catalyses the reaction.

Q5. What is colloidion solution?Colloidion solution is the solution which is used to impregnate the filter paper to make it an ultra filter.

Q6. Give conditions for Tyndall effect to take place?The refractive indices of dispersion medium & dispersed phase should not be same. The wavelength of the light should not be too low.

Q7 Why is the ester hydrolysis slow in the beginning & becomes faster after sometime?The ester hydrolysis slow in the beginning & becomes faster after sometime

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because the carboxylic acid produced acts as the catalyst.

Q8. What are the reasons for charge on sol particles?Frictional electrificationElectron capturePreferential adsorption of ions from solutionsDissociation of surface molecules

Q9. Why colloidal medicines are more effective?Colloidal medicines have large surface area for the reaction. Hence colloidal medicines are more effective.

Q10.Give the factors governing colour of colloidal solution?Wavelength of light scattered, the manner in which light is received by observer, particles size.

Q11. Why are Lyophylic colloids self stabilized?Due to the force of attraction between dispersion medium & dispersed phase. Q12. State one application ZSM-5. Used for the conversion of alcohol to gasoline.

Q13. Explain the role of catalyst in a chemical reaction. It lowers the activation energy & hence increases the rate of the reaction. Q14. Adsorption is always exothermic. Why? During adsorption, the bonding takes place between adsorbate & adsorbent and there is a decrease in surface area & decrease in entropy. Q15. Define weeping of gels. Give the other name of weeping?When gels are allowed to stand for a long time they give out small quantity of trapped liquid which accumulates on its surface. This action of gels is called as weeping, other name is syneresis.

Q16. What is Thixotrophy? Conversion of sol to gel is called thixotrophy.

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Q17. Why is FeCl3 preferred over KCl for a cut leading to bleeding?Fecl3 has more positive charge and causes coagulation of blood quickly.

Q18. Why is silica gel used as dehumidifier?Silica gel adsorpes moisture on its surface quickly.

Q19. Give the kinetics of adsorption of gases on metal surfaces.Adsorption of gases on metal surfaces follows zero order kinetics.

Q20. G is negative for adsorption process. Why?Adsorption is exothermic & decrease in entropy occurs in adsorption.

Q21. Give the significance of Brownian movement.It gives stirring effect to the colloid. Hence colloidal particles do not aggregate. The colloid becomes stable. Q22. What happens when blood charcoal is shaken with conc. Kclin one case and its dilute KCl solution in other case? In conc. KCl positive adsorption occurs & in dilute Kcl negative adsorption occurs.

Q23.Give the principle used in gas masks?Competing adsorption

Q24. Gold sol appears red , purple & also golden in colur Why? Colour of the gold sol changes with its particle size.Its’ colour is red & becomes puple when size increases & finally becomes golden.

Q25. Give the sign of ∆H & ∆S when Br2 gas gets adsorbed on charcoal. ∆H = negative & ∆S = positive

Q26. What is Argyrol? Colloid of Ag with proteine(mild silver protein) Antiseptic & used for treating gonoria (a disease)

Q27. What happens when NaCl is added to ferrichydroxide sol?When NaCl is added to ferrichydroxide sol coagulation occurs.

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Q28.Differentiate dilute soap solution & concentrated soap solution.Dilute soap solution behaves like a true solution & concentrated soap

solution behaves like a colloid.

Q29. Comment: Colloid is not a substance but it is a state of a substance.The nature of the sub whether colloid or crystalloid depends upon the particle size.When the size of the solute particles lies between 1000 – 10000 pm , it behaves as a colloid & whenit is less than this value it behaves as a crystalloid. Thus Colloid is not a substance but it is a state of a substance which depeds upon the particle size.

Q30. What happens when Fe(OH)3 is added to Gold sol?Why?The Gold sol is negatively charged, Fe(OH)3 is positively charged& so coagulation occurs.Color of Gold sol changes from red to blue. Gold sol changes from red to blue due to the change in the particle size.

(AS THIS CHAPTER HAS NO NUMERICALS HOTS MAY INCLUDE THE QUESTIONS OF THE ABOVE TYPE)

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Chapter-6

GENERAL PRINCIPLES ANDPROCESSES OF ISOLATION OF ELEMENTS

Q1. Copper & Silver are below hydrogen in the electro chemical series, yet they are sometimes found in combined state as sulphides in nature.Why?

Cu & Ag react with sulphur at high temperature inside the earth.So they exist as sulphides in addition to the free or native state.

Q2. Predict the mode of occurance of a) highly reactive ,b)moderately reactive & c) noble metals.

a) highly reactive-- in combined state as chlorides,sulphates,carbonates.b) moderately reactive-- as their oxides.c) noble metals-- In free state.

Q3. What is the role of pineoil in metallurgy? It is a collector in froathfltation process.

Q4. What is meant by depressant?Illustrate. Depressant is a substance used to separate the sulphide ores of two metals by

suppressing the froath formation of one metallic sulphide.(eg) In a mixture of ZnS & PbS the depressant NaCN selectively prevents the

formation of the froath of ZnS.So it acts as depressant for ZnS.

Q5. Why MgO is used for the lining in steel making furnace?It facilitates the removal of impurities of Si ,P & S through slag formation.

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Q6. Why do metals not occur as nitrates in nature? Almost all metal nitrates are soluble in Water.So any nitrate formed will get

washed away by rain water.

Q7. Give the role of fluorspar which is added in small quantities in the electrolytic reduction of Alumina dissolved in fused Cryolite.

To lower the temperature of the melt.

Q8. Zn & not Cu is used for the recovery of Ag from its cyano complex.Why?

Zn is a more powerful reducing agent than Cu & it is cheaper than Cu also.

Q9. Graphite is used as anode but not Diamond.Why? There are free electrons in graphite which helps in electrical conductivity &

no free electrons in diamond.

Q10. The reduction of metal oxides by Al becomes faster just after the ignition of the mixture.Justify.

Huge amount of heat released during ignition (exothermic) helps the reaction to occur fast.

Q11. What is ‘900 fine silver’? Ag containing 90% Ag & 10%Cu.

Q12. Suggest a condition under which Al could reduce MgO. Al can reduce MgO at high temp, because in this temp range ,the line for ∆G0

(Mg,MgO) lies above the line of ∆G0(Al,Al2O3).This indicates that Al2O3 is

more stable than MgO.

Q13. The reduction of metal oxide is easier if the metal formed is in liquid state at the temp of the reaction.Justify.

The entropy of liquid metal is more tuan that of solid metal. When ∆S becomes more positive the reaction becomes spontaneous.

Q14. Although thermodynamically feasible ,in practice, Mg metal is not used for reduction of Alumina in the metallurgy of Al.Why?

The reaction between Mg & Al2O3 is exothermic & if the temp increases to2000K Al starts to reduce MgO.Mg is also a v. costly metal.

Q15. In Ellingham diagram C,CO2 graph is a straight line.Why?

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C + O 2(G) → CO2(g).

In the above reaction there is no change in entropy.Hence ∆G value remains constant at all temperatures.Hence the graph is a st.line.

Q16. Copper can be extracted by hydrometallurgy but not Zn.Why?Zinc occupies higher position in electrochemical series & has large negative reduction potential value cannot be reduced by treatment with more electropositive metal.Copper occupies quite lower position in electrochemical series & can be easily reduced.

Q17. Out of C & CO which is better reducing agent at 673K.CO is better reducing agent.(refer Ellingham diagram )

Q18. Name the common elements present in electrolytic refining of Copper.Why are they so present?

Antimony, Selenium, Au, Ag, Pt are present as anode mud.These elements are less electropositive & they do not undergo oxidation at the anode & settle down at the bottom.

Q19. What criterion is followed for the selection of the stationary phase in chromatography?

It should be immobile and it should be immiscible with mobile phase. The components of the mixture should be differently adsorbed on the stationary

phase.

Q20. Why copper matte is put in silica lined converter?Copper matte contains Cu2S and FeS. In the converter FeS gets converted into FeO. Silica helps in removal of FeO impurity as slag.

2FeS+3O2 2FeO + 2SO2

FeO + SiO2 FeSiO3

Q21 Zinc is not extracted from ZnO through reduction using CO. Why? The line for ∆G0

(CO,CO2) lies above the line of ∆G0(Zn, ZnO) at the temperature

employed for reduction of zinc oxide. Hence, reduction of ZnO with CO is not possible as the ∆rG0 for this would be positive.

ZnO + CO Zn + CO2 On the other hand, the line for ∆G0

(C, CO) vs T lies below the line for ∆G0(Zn, ZnO).

Hence, coke can reduce zinc oxide to zinc as the ∆G0 for the reaction

would be negative and hence the reaction is feasible.ZnO + C Zn + CO

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Q22. The value of ∆G 0 for the formation of Cr oxide is -540 kJ/mol and that of alumina is -827 kJ/mol. Is the reduction of Cr oxide possible with Al?

∆G0 = -827 – (- 540) kJ = 287kJ since ∆G0 is negative the reaction is possible.

Q23. The choice of the reducing agent in a particular case depends on thermodynamic factor why?

The choice of the reducing agent in a particular case depends on thermodynamic factor because for a reaction to be feasible the reaction of metal oxide with the reducing agent should have negative ∆G0 Therfore, that reducing agent is sutiable for which ∆G0 for the reduction reaction is negative Eg- Al is correct choice for the reduction of Cr oxide and not MgO Eg- for the reduction of Zno coke is the right choice

Q24. Predict the condition under which Al is expected to reduce MgO. Al is expected to reduce MgO at a temperature of about 2000 K (refer

Ellingham diagram)

Q25. Why it is advantages to roast a sulphide ore to oxide before reduction? Oxide can be easily reduced to metal using C or H as ∆G0 is negative whereas sulphide cannot be reduced to metal using C or H as ∆G0 is positive.

( IN THS CHAPTER TO SCORE FULL MARKS STUDENTS ARE EXPECTED TO HAVE A THOROUGH KNOWLEDGE OF ELLINGHAM DIAGRAM)

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26

Chapter-7

P-block elements

1. Nitrogen, carbon monoxide and cyanide ion are iso electronic, yet CO and cyanide ion are reactive but N 2 is not. Explain why.

Ans: N 2 is non polar and has very high ionization energy.

2. Solid N2O5 is ionic in nature. Explain Ans: It is otherwise known as nitronium nitrate. In the solid state it exists as NO2

+ NO3 -

3. Give reason for the paramagnetic nature and purple colour of S2 molecule

Ans: It has two unpaired electron in antibonding P * orbital they get excited

4. Why is it difficult to prepare and isolate HOF?Ans: F2 + H2O -500oC HOF + HF

FOF + H2O HF + H2O2

HOF + F2 HF + OF2

5. When NaBr is heated with conc.Sulphuric acid, Br2 is produced, but when NaCl is heated with Conc.Sulphuric acid, HCl is produced.Why? Ans: In the first case

NaBr + H2 SO4 NaHSO4 + HBr

2HBr + H2 SO4 2H2O+SO2 + Br2

In the later case the HCl produced does not reduce H2 SO4

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6. When a moist blue litmus paper is dipped in a solution of hypochlorous acid, the colour change is not permanent. Explain HOCl H+ + OCl-

So it turns red, but due to the following HOCl HCl + [O]

7. Aqueous solution of ammonia is best described as NH3 (aq) not as ammonium hydroxide. Why?Ans: It is possible to isolate stable crystalline hydrate of ammonia as NH3.H2O and Kb = 1.81 x10 -5

8. Though the electron gain enthalpy for O- O2-

is 702KJ/mol , a large number of Oxides contain O2-.Why?Ans: Due to high Lattice Enthalpy.

9. What is the Covalence of phosphorus in PH4+?

Ans: four

10. Complete and balance the following:- i) NH3 + I2 ii) NH3 + B2 H6

Ans: i) 2 NH3 + 3 I2 NH3 N I3 + 3HI ii) NH3 + B2 H6 2B3 N3 H6 + 12 H2

11. Consider the formation of O2+[Pt F6]- .Is it likely that O + and N2

+ shall form the same type of complex?Ans: No, because of their high I.E

12. Give reasons for the following i) NCl3 gets readily hydrolysed while NF3 does not ii) Elemental nitrogen exists as a diatomic molecule whereas elemental phosphorus is a tetra atomic molecule.Ans: i)In NCl3 , Cl has vacant orbital to accommodate the electrons donated by water and in the latter it is not so.

ii)Because of the small size it can for pP- pP multiple bonds. Phosphorus cannot do so because of its large size.

13.i)Explain Why H3PO2 and H3PO3 act as good reducing agents while H3PO4 does not? ii) What do you observe when you pass O3 in KI solution? iii) Why is only Xenon known to form established compounds?

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Ans:- i)Both H3PO2 and H3PO3 have P-H bond, but H3PO4 does not have P-H bonds

ii) Being a strong oxidizing agent, Ozone oxidizes Iodide ions to Iodine

2I- (aq) + H2O (l) + O3(g) 2OH- (aq) + I2 (g) + O2(g)

iii) Xe has least I.E among noble gases.

* * * * * * * * * *Chapter-8

d and f block Elements

1. Silver atom has completely filled d orbitals in its ground state. How can you say that it is a transition element? Ans: It forms AgF2 a dark brown crystalline solid

2. Transition elements exhibit their highest oxidation state in their oxides not in Flourides.Why?Ans: Because oxygen can form covalent multiple bonds.

3. Explain why, Zn (II) salts are white while Mn (VII) are deep purple in colour? Ans: due to charge transfer between O2- ion to Mn7+

4. KMnO4 is used in acidic medium quite frequently than in its aqueous or alkali for oxidizing purpose. Why? Ans: In acidic medium MnO4

- involves the addition of 5e- or Eo = +1.52v

5.Inspite of low enthalpy of atomization and fairly large hydration enthalpy Zn has low reduction potential ( -O.76v). Explain why?Ans: Because of very high I.E 1 + I.E. 2

6.Give reasons: i) Zr and Hf have identical sizes ii) In the titration of FeSO4 with KMnO4 in the acidic medium dil.H2SO4 is used instead of dil HClAns: i) Due to lanthanoid contraction

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ii) dil.HCl is reducing agent and liberates chlorine on reacting with KMnO4 soln.7.Calculate the spin only magnetic moment of Iron present in the following compound. [Fe(H20)5NO] 2+

Ans: 5.92 BM

8. K2 [PtCl6] is a well known compound whereas the corresponding Ni compound is not known. State the reason for it.Ans: The oxidation state of Pt in K2 [PtCl6] is +4, which is a stable oxidation state for Pt. The same oxidation state Ni is very difficult because of (I.E1 + I.E 2+ I.E 3+I.E 4)

9. Among the ionic species Sc 3+ ,Ce 4+ and Eu2+ Which one is a good oxidizing agent?Ans: Ce 4+ is a good reducing agent because it can readily change to the most stable oxidation state by gaining one electron. Others cannot do so.

10. Why are Fe 3+ and Cu 2+ prominent in their aqueous solutions?Ans: Because of high hydration enthalpies.

11. There is a dip in the melting point curve at Mn, though the preceding element also has similar electronic configuration. Why?Ans: Due to 3d 5 half filled configuration which are tightly held by more effective nuclear charge of Mn

12. CrO3 is an acid anhydride. Explain. Ans: It dissolves in water to give chromic acid. CrO3 + H2O H2 CrO4

13. Though both Cr 2+ and Mn 3+ have d4 configuration, yet Cr2+ is reducing and Mn3+ is oxidizing. Explain Why?Ans: Because of the stable half filled configuration of t2g sets Cr3+

* * * * * * * * *

30

Chapter-9

CO-ORDINATION COMPOUNDS

1.The spin only magnetic moment of [MnBr4 ]2-is 5.9BM. Predict the geometry of the complex ion?Coordination no is 4. Hybridisation may be sp3 or dsp2. Due to the presence of 5 unpaired electrons it should be tetrahedral rather than squareplanar. 2.A solution of of [ Ni( H2O)4]2+ is green but a solution of [Ni (CN)4 ]2- is colourless.Explain.H2O is a weak ligand.In Nickel in +2 ox.state two unpaired e- pair up in presence of strong ligand CN-.

3.[ Fe (CN)6] 4- and [Fe (H2O)6]2+ are of different colours in dilute solutions why?Fe is in +2 ox. state & pairing of electrons occurs in presence of strong ligand CN- ion.

4.What are Homoleptic and Hetroleptic complexes?Complexes in which a metal is bound to only one type of ligand. eg[ Co(NH3)6] 3+

is Homoleptic complex.Complexes in which a metal is bound to more than one type of ligands.eg [ Co(NH3)4 Cl2 ] + is Hetroleptic complex.

5.Aqueous Copper sulphate solution gives a green precipitate with aqueous KF solution and a bright green solution with aqueous KCl solution.Why?CuSO4 with water exist as [ Cu( H2O)4]SO4 . Water being a weak ligand replaced by a strong ligand F- gives a green ppt. Cl- being weaker than F- gives a bright green solution.

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6.[CoF6 ]3- is a high spin complex. Why?Fluoride being a weak ligand does not pair up the electrons &so the d orbitals involved belong to the outer shell which gives a high spin complex.

7.what are metallocences?Organometallic copounds containing benzene and carbocyclic ligands.

8. Though [Co(NH3)6]3+ and [CoCl6]3- have octahedral geometry they differ from each other.How? [Co(NH3)6]3+ is an inner orbital low spin complex with d2sp3hybridization and [CoCl6]3-

is an outer orbital complex with sp3d2 hybridisation.

9.Explain the macrocyclic effect?Macrocyclic effect. When a multidendate ligand form cyclic ring with metal ion with no favourable stearic effect.

10. Explain as to how the two complexes of nickel [Ni(CN)4]2- and [Ni(CO)4] have different structure, but do not differ in their magnetic behaviour(Ni=28).[Ni(CN)4]2- has dsp2 hybridisation & has paired e-s.So it is diamagnetic. [Ni(CO)4] has sp3 hybridisation & has unpaired e-s & so it is paramagnetic.

11.Name the metal present in vitamin B12 and plastocyanin.vitamin B12 ---Cobalt , plastocyanin --- copper.

12.Which complex is used as anticancer drug?The complex is used as anticancer drug is Cisplatin.

13.Explain why K3[Fe(CN)6] is more stable than K4[Fe(CN)6]?K3[Fe(CN)6] has more charge density as the size of Fe3+ is smaller than Fe2+ hence more stable.

14.[CoF6]3- forms outer orbital complexes while [Co(CN)6 ]3- forms inner orbital complexes?Strong ligand field of CN- forces pairing of e-s in innershell vacating inner d orbitals for d2sp3 hybridisation.

15.Give reason : [CuCl4]2- exist while [Cu I4]2- does not.

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Because of the higher size of I2 atom [Cu I4]2- does not exist.

16.[Fe(CN)6]4- and [Fe(H20)6]2+ are of different colours in dilute solutions why?The ligand CN- and H2O exert different ligand fields . CN- being the strong ligand forms low spin complex whereas H2O being a weak ligand forms a high spin complex.So they have different colours in dilute solutions.

17 Blue coloured solution of [CoCl4]2- ion changes to pink on reaction with HgCl2.why?[CoCl4

2-+HgCl2 Co(HgCl4)+2Cl-

18.Which of the two[Fe(CN)6]4- and [Fe(CN)6]3- is more stable and why?]Fe(CN)6[4- is more stable due to higher nuclear charge and smaller size of

Fe2+ ion

19.Why should [Al(H20)6] 3+ be a stronger acid than [Mg(H20)6]2+?Smaller size and greater effective nuclear charge of aluminium ion makes [Al(H20)6] 3+ a stronger acid.

20). A solution of [Ni(H20)6]2- is green but a solution of [Ni(CN)4]2- is colourless.Explain?In [ Ni(CN)4]2- strong ligand forces pairing. Hence no unpaired e-s are present making the complex colourless. -------------------------------------------------- ---------------------

33

Chapters-10, 11, 12&13

Organic Chemistry

1) Haloalkanes react with KCN to form alkylcynaide while AgCN forms isocyanide. Why?

- KCN is ionic. It gives cyanide ion in solution. The attack takes place mainly through carbon since C-C bond is stronger than C-N bond.

- AgCN is mainly covalent and N is free to donate electron pair forming isocyanide.

2) Tertiary Halides undergo SN1 reactions while primary halides undergo SN2 . Explain?

- Nucleoptile can easily attack a primary halide and indergo SN2 reaction. Attack of the nucleophile on carbon of C-X of tertiary halide is not possible due sterric effect hence it undergoes SN1 reaction.

3) Allylic and Benzylic halides undergo SN1 reactions readily. Why?

Ans- Allylic and Benzylic cations are resonance stabilized.

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4) Write the basis for Saytseff rule.Ans- Saytseff rule:- In dehydrohalogenation reactions, the preferred product is that alkene which has greater number of alkyl groups attached to the doubly bonded carbon atom.2-bromobutane on reaction with alcoholic potassium hydroxide gives 2-butene as major product since secondary carbo cation is more stable than primary carbo cation.

5) Aryl halides are extremely less reactive. Why?Reason: 1-In Aryl halides , C-X bond acquires a partial double bond character.

2- Aryl halides are resonance stabilized.

6) Can you think why NO2 group shows its effects only in ortho and para positions and not at meta positions?Ans- Presence of NO2 group at the ortho and para positions withdraws the electron density from benzene ring and facilitates the attack of the nucleophiles on haloarenes. But in meta nitro benzene none of the resonating structures bear negative charge on the C atom bearing the nitro group and so no activity is observed.

7) Chloroform is stored in coloured bottles completely filled. Why?Ans- CHCl3 is stored in coloured bottles because it is light sensitive. It is fully filled in the bottle because it forms a poisonous gas phosgene (COCl2) on reaction with air.

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8)Why is phenol acidic?Ans- Phenoxide ion is resonance stabilized since phenylgroup is electron withdrawing.

9) Acidic character of alcohol decreases as primary>secondary>tertiary- Explain?

- Acidic character of alcohol is due to polar nature of O-H bond. Presence of more electron releasing alkyl groups increases the electron density of oxygen atom tending to decrease the polarity of O-H bond.

10) Ortho nitro phenol is steam volatile while para nitro phenol is not. Why?Ans- ortho nitrophenol forms intra molecular H-bonding while para nitro phenol forms inter molecular H-bonding.

11. Write the major product formed when methoxy benzene reacts with HI and give a possible reason?Ans- When methoxy benzene reacts with HI, phenol and methyl iodide are formed .

Alkyl aryl ethers are cleaved at the alkyl-oxygen bond since aryl-oxygen bond is more stable due to partial double bond character.

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12) Anisole undergoes bromination with bromine and ethanoic acid even in the absence of Iron (III) bromide. Explain.Ans- It is due to the activation of the benzene ring by the methoxy group. The major product is para isomer since it is more symmetric than ortho isomer.

13) Preparation of ethers by the acid dehydration of secondary and tertiary alcohols is not possible. Why?Ans: Preparation of ethers by acid dehydration of alcohols requires protonation of alcohol. Due to sterric hindrance protonation of 20 and 30 alcohols is very difficult.

14) Carbonyl carbon is electrophilic while Carbonyl oxygen is nucleophilic. Explain.Ans-

Since Oxygen is more electronegative than C of carbonyl group, the carbonyl carbon develops a partial positive charge (electrophilic) and oxygen bears a partial negative charge(nucleophilic).15) What is the change in the oxidation state of copper ion in fehling’s test?Ans- RCHO+2Cu2++5OH-RCOO-+Cu2O+3H2O

So the oxidation state changes from Cu2+ to Cu+.

16) Formic acid gives fehling’s test . why?- Methanoic has aldehydic group in its structure.

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17) Experimental determination of molecular mass of ethanoic acid gives it to be 120gmol-1 and not 60 gmol-1 why?Ans- Carboxylic acids undergo dimer formation in their vapour phase by H-bonding.

18) Carboxylic acids are stronger acids than phenol. Explain.Ans-

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- Carboxylate ion is stabilized by two equivalent resonating structures in which the negative charge is on the more electronegative oxygen atom.

- Phenoxide ion has non equivalent resonance structures in which negative charge is on less electronegative carbon atoms. Thus carboxylate ion is more resonance stabilized than phenoxide ion.

19) Chloroacetic acid is a stronger acid than acetic acid. Why?

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Ans:

- Chloroacetate ion is more resonance stabilized than acetate ion. Since chlorine of chloro acetic acid is electron withdrawing

20) Among the reducing agents, iron scrap and hydrochloric acid mixture is preferred for the reduction of nitrobenzene to aniline. Why?Ans:

- Fe+2HCl FeCl2 + H2

Hydrogen is used for the reduction FeCl2 + 2H20 Fe(OH)2 +2HClFeCl2 formed is hydrolysed to give HCl. Hence only small amount of HCl is required.

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21) pKb value of methaneamine is 3.38 while that of aniline ids 9.38. Explain.

Ans:

- Methaneamine is more basic due to the electron releasing effect of methyl group. So its Kb value is more and hence its pKb value is less.

- Aniline is weaker base due to the presence of electron withdrawing benzene ring (-I effect). Kb for benzene is low hence its pKb value is high.

22) (CH3)3N is stronger base than CH3NH2 in vapour phase but it is a weaker base than CH3NH2 in aqueous solution. Explain.Ana- In vapour phase only the electron releasing effect is considered .In the aqueous form, the susbstituted amine cations get stabilised not ony by +I effect but also by solvation with water. Decresing order of extent of H-bonding in water and order of stability of ions by salvation is 10>20 >30 amines . It is due to sterric considerations.

23) Acetylation reactions are faster in presence of strong baes like pyridine. Explain with an example.Ans:

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- The HCl in the product side should be removed as and when it is formed otherwise it causes the reverse reaction. Hence a base is required.

24) The pH of the medium for the reaction of aldehydes and ketones with ammonium derivatives should be carefully controlled. Explain.Ans- The reaction is catalysed by acid. In acid medium the carbonyl oxygen is protonated . The protonated carbonyl group undergoes the nucleophilic attack of ammonium derivatives.

If the medium is too acidic, the ammonium derivatives form salt and so it cannot act as a nucleophile.

G-NH2 + H+ G-NH3+

If the medium is too basic, the protonation of the carbonyl group does not occur. Hence the reaction does not take place.

Thus, for carrying out the reaction, the pH is properly controlled. The optimum pH of the medium is around 3.5.

25. A Compound is formed by substitution of two chlorine atoms for H-atom in propane. What is the number of structural isomers possible?Ans : Four structures are possible:

(i) 1, 1-dichloro propane(ii) 1, 3-dichloropropane(iii) 1, 2-dichloropropane(iv) 2 , 2-dichloropropane.

26. Which of the following is optically active and why?(i) 1-bromobutane(ii) 2-bromobutane.Ans: 2-bromo butane is optically active since it contains a chiral carbon.

27. Which is a better nucleophile Br- ion or I- ion?Ans : I- ion is a stronger acid and hence better nucleophile.

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28. Explain why thionyl chloride method is preferred for preparing alkyl chlorides from alcohols?Ans : When thionyl chloride is used, the by products of the reaction, HCl and SO2, being gases, escape into the atmosphere leaving behind almost pure alkyl halide.

29. R-OH doest not react with NaBr, but on adding some H2SO4 forms RBr. Explain.Ans : Br- ion is an extremely weak bronsted base. Hence, it cannot displace the stronger base OH-. On adding H2SO4, protonation of alcohols take place, resulting in the formation of ROH2

+. Now Br- ion can easily displace H2O, which is a very weak base and hence a good leaving group.

30. When t-butanol and n-butanol are separately treated with la few drops of dil. KMnO4 in one case only, the purple colour disappears and the brown ppt. formed .Ans: Only n-butanol would be oxidised by dil.KMnO4

solution and the brown ppt. formed is of MnO2. Tertiary alcohol is resistant to oxidation.

31. Explain why Dry gaseous hydrohalic acids and not their aqueous solution are used to prepare alkyl halides from alkenes?Ans : Dry hydrogen halides are stronger bases and better electrophiles than H3O+formed in their aqueous solution. Further H2O is a nucleophile which reacts with R+ to give alcohol.

32. P-dihalobenzene have higher melting points and lower solubilities than those of ortho- and meta- isomers?Ans: The p-isomer is more symmetrical and therefore fits better into the crystal lattice and as a result of this, intermolecular forces are greater than those in corresponding ortho- and meta-isomers.

33. Iodoform is obtained by the reaction of acetone with sodium hypoiodite but not with iodide ion.Ans: To prepare iodoform from acetone, I+ is required. As I+ can be obtained from IO- and not I-, hypoiodite is used to convert acetone into iodoform. As such I- ion cannot be used for this conversion.

34. Sodium metal can be used for drying diethyl ether but not ethanol. Explain.

43

Ans: Sodium metal does not react with diethyl ether but reacts vigorously with ethanol to form sodium ethoxide.

35. During the preparation of acetaldehyde by oxidation of ethanol, acetaldehyde should be distilled out continuously why?Ans: During oxidation of ethanol, acetaldehyde is formed in the first step which should be distilled out, otherwise further oxidation of acetaldehyde to acetic acid will take place.

36. Name the major product obtained when tert-butyl bromide is treated with sodium ethoxide?Ans: The major product obtained is 2-methyl-1-propene because in the presence of alcoholic NaOH (sodium ethoxide) alkyl halides undergo dehydrohalogenation. Ease of dehydrohalogenation of alkyl halides is tertiary>secondary>primary.

37. How does a carbonyl group differ from an ethylenic bond?Ans: (1) Carbonyl group is polar in nature while ethylenic bond is non polar .

(2) Carbonyl compounds undergo nucleophilic addition reactions while ethylenic bond undergo electrophilic addition reactions.

38.What is IUPAC name of (CN)2 ?Ans : Ethanedinitrile .

39. What is role of conc. H2SO4 in the nitration of benzene ?Ans: In the nitration of benzene, H2SO4 is used to generate nitonium ion from nitric acid.

40. What is the use of Baker –Mulliken test ?Ans: It is used for the detection of nitro groups.

41.Give an example of ambident nucleophile.Ans : CN- or NO2

- .

Distinction between pair of organic compounds

1) Chlorobenzene and phenyl chloromethane .

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phenylchloromethane curdywhite precipitate

chlorobenzene no reaction

it is because chlorobezene is aryl halide(less reactive) while phenyl chloromethane is alkyl halide(more reactive).

2) CH3NH2 AND C6H5NH2

Aniline decolourises bromine water whereas methanamine does not.

3) Methanoic acid and ethanoic acid

Methanoic acid when heated with fehling’s solution I and II, forms reddish brown ppt of cuprous oxide while ethanoic acid does not give. Methanoic acid has –CHO in its structure.

Questions For Practice

1. Write the differences between aldol condensation and Cannizaro’s reaction?

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2. What are the differences between Clemmensen’s reduction and Rosenmund’s reduction?3. How can CCl4 and CHCl3 be distinguished?4. How can HCHO and CH3CHO be distinguished?5. Distinguish between ethyl nitrite and nitro ethane?6. Distinguish between Ethyl cyanide and ethyl isocyanide?7. How are the following conversions effected?

a) ethane to butane(2 steps)b) methanol to ethanol (4 steps)c) acetaldehyde to lactic acid (3 steps)d) nitromethane to trimethylamine (3 steps)e) propan-2-ol to 2 methyl propan-2-ol (2 steps)

f) Lime to benzene (4 steps)

g) benzene to meta chloro aniline (4 steps)

Word Problems Involving Organic Conversions for 5 marks in Board Examination

1. A compound ‘A’ dissolves in water and its 40% aqueous solution is used as a preservations for zoological specimen. When its aq. Solution is allowed to stand, it forms trimer ‘B’. 6 moles of it react to give ‘C’ in presence of Ca(OH)2. ‘A’ undergoes disproportionation reaction in presence of conc. Alkali to form ‘D’ and ‘E’. ‘D’ liberates H2(g) with Na Metal. Identify ‘A’ to E writing the chemical equations involvedAns: A is formaldehyde (HCHO) (aq soln formalin is used as

preservative)

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2. Complete the following and identify from A to E.

Solution:

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3. An organic compound (A) having MF C8H8O forms an orange red precipitate(ppt). (B) with 2,4-DNP. Compound (A) gives yellow ppt (‘C’). When heated in presence of I2 and NaOH along with a colorless compound ‘D’. ‘A’ does not reduce Fehling’s solution and does not decolorize Bromine water. On drastic oxidation of ‘A’ with chromic acid, a carboxylic acid ‘E’ with mF C7H6O2 is formed. Deduce the structures of A to E. Also, write the reactions involved.Ans:

4. Complete the following and identify the compounds A to E

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Ans.

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5. An organic compound A with mF C5H8O2 is reduced to n pentane on treatment with Zn-Hg/HCl. ‘A’ forms di-oxine with NH2OH and gives a positive iodoform test and Tollen’s reagent test. Identify the compound ‘A’ and deduce its structure

Ans: Contains CH3CO- group.• Gives positive Tollen’s tests. Has –CHO group• Gives n pentane on reduction. Should contain all 5 carbons in the

same chain.

6. An organic compound ‘A’ with MF C8H8O gives positive iodoform and 2,4-DNP tests. It does not reduce Fehling’s solution and does not decolorize Br2/H2O. On oxidation with chromic oxide, it gives a carboxylic acid ‘B’ with MF C7H6O2. Deduce the structures of A and BAns: A is a carbonyl compound (gives 2,4 DNP test)

• Has CH3CO group (Iodoform test)• Is saturated (not decolorising Br2 water).

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7. A compound ‘X’ with MF C2H4O on oxidation gives Y with MF C2H4O2. X undergoes haloform reaction. On treatment with HCN, X gives Z which on hydrolysis gives 2-hydroxy propanoic acid. Write down the structures of X, Y and Z. Name the product when X reacts with diluted NaOH.Ans: Y Has 2 carbon atoms . Hence not Ketone

• X is CH3CHO (gives haloform reaction)

8. An unknown aldehyde A on reaction with alkali gives B-hydroxy aldehyde which loses H2O forming an unsaturated aldehyde 2-butenal. Another aldehyde B undergoes disproportionation reaction in the presence of con. Alkali to produce C and D. C is an aryl alcohol with MF C7H8O. Write the sequence of reactions involved. Identify A to D. Name the product when B reacts with Zn-Hg/HCl

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Ans:

9. An organic compound A (C8H6) on treatment with H2SO4 and HgSO4 gives B which also can be obtained from the reaction b/w benzene and an acid chloride in presence of AlCl3 (anhydrous). B on treatment with I2/NaOH gives C and a yellow compound D. Identify from A to D. Write the sequence of reactionsAns: B has CH3CO group (iodoform test)

• B is obtained from benzene and acid chloride in presence of anhyd. AlCl3 (FC reaction)

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10. Two moles of an organic compound A on treatment with strong base gives compounds B and C. Compound B on dehydrogenation with Cu gives A while on acidification gives a carboxylic acid D with MF CH2O2. Identify A to D and write the sequence of reactionsAns:

11. An organic compd A (C8H16O2) was hydrolysed with dil H2SO4 to yield a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid gives compound B. C on dehydration gives D which reacts with HBr to give optically active compound E. Identify A to E with the help of equations. Ans: A is an ester (on hydrolysis gives Carboxylic acid and alcohol)

• C and B have same number of Carbon atoms (C on oxidation gives B)

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Chapter-14

BIOMOLECULES

1. Justify a. Sucrose is a non reducing sugar, even though it is made up of glucose

and fructose, out of which glucose is a reducing sugar.b. Sucrose is dextrorotatory, but after hydrolysis the mixture is

laevorotatory.c. Both amylose and amylopectin have glycosidic linkage at C1- C4 of α

D (+) glucose units, amylopectin is insoluble in water where as amylose is soluble.

Ans: a. The reducing group of glucose(C1) and C2 of fructose are involved in glycosidic bond formation in sucrose, therefore sucrose is not reducing.b. Sucrose on hydrolysis gives glucose with dextrorotation(+52.5) and fructose with laevorotation(-92.40), which is higher than the dextrorotation.c. In amylopectin, extra branching occurs by C1-C6 glycosidic linkage, therefore it is insoluble.

2. Why can’t we digest cellulose, even though both starch and cellulose are made up of glucose units?

Ans: Starch is made up of α glucose while cellulose is made up of β glucose. We do not have enzymes which can digest β glucose.

3. Account for the following :-a. Glucose and sucrose are soluble in water but cyclohexane and

benzene are insoluble in water.b. Amino acids are polar in nature.Ans: a. Glucose and sucrose can form hydrogen bonds with water due to the presence of OH group whereas cyclohexane and benzene cannot.b. Due to the formation of zwittor ion.

4. How many peptide bonds will be formed when ten amino acids polymerise?

Ans: 9

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5. What makes hair, wool, and silk insoluble in water?Ans: Fibrous structure. ie. Polypeptide chains run parallel and are held together by hydrogen bonds and disulphide bonds, which makes it insoluble in water.

6. Account for a. the higher melting point and higher solubility of amino acids

than the corresponding haloacids.b. Vitamin B and C only, should be taken regularly and

frequently in our diet, than other vitamins.Ans: a. Amino acids have dipoles which account for the high solubility and strong bonds between the molecules.

b. B and C vitamins are water soluble, cannot be stored and are excreted in urine.

7. What structural difference is there between α glucose and β glucose?Ans: They differ in the orientation of OH group on C1 carbon atom.

8. If a fragment of one strand in DNA molecule has the base sequence CCATGCATG, what is the base sequence in the complementary strand?Ans: GGTACGTAC

3 marks

1. An organic compound A of formula C12 (H2O)11 on hydrolysis gives a compound B and C, both of formula C6(H2O)6. B on oxidation with bromine water forms a pentahydroxy monocarboxylic acid, while C is not. Identify A, B and C.

Ans: C12(H2O)11= C12H22O11 which is sucrose. A is sucrose. A on hydrolysis gives B and C. B with the mild oxidizing agent gives pentahydroxyacid. ie. gluconic acid, therefore B is glucose and C is fructose.

.

2. An organic compound X with formula C5H6(OH)5CHO forms hexane on prolonged heating with HI. X also forms an oxime Z with hydroxyl amine, and gets oxidized with a mild oxidizing agent, Br2/H2O to Y. write equations and identify X,Y and Z.

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Ans: X with HI forms hexane, this shows X contains 6 carbons in a straight line.

X gives oxime, this indicates the presence of carbonyl group. X is oxidized by bromine water, indicates the presence of aldehyde group. Therefore X should be glucose, Y should be gluconic acid and Z should be an oxime of glucose,

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Chapter-15

POLYMERS

1.Give the examples of semisynthetic polymers.Guncotton, vulcanized rubber.

2.Name the polymer used for making aircraft windowsPMMA (Polymethyl metha acrylate).

3.What is the trade name of polyacrilonitrile?The trade name of polyacrilonitrile is Acrilan.

4.Classify _(NH_CHR_CO)n_ as a homopolymer or copolymer?

_(NH_CHR_CO)n_ is a homopoymer derived from NH2CHRCOOH.

5.What is a biopolymer?Biopolymer is a natural polymer (eg) carbohydrate and proteins.

6.Which rubber is found in bubble gums?SBR (Styrene butadiene rubber) is found in bubble gums.

7. Name a synthetic polymer which contains an ester group. Terylene or Dacron contains an ester group.8. What is a plasticizer?Organic compounds which are added to plastic to make them soft is called as a plasticizer.(eg) cresyl tri phosphate.9. Arrange the following polymers in the increasing order of their molecular forces. i)Nylon 66, Buna S, Polyethene, Bakelite ii) Nylon 6, Neoprene, PVC, Melamine

Answer : i) Buna S < Polyethene < Nylon 66 < Bakelite ii) Neoprene < PVC < Nylon 6 < Melamine

10. In recent days PHBV is a preferred polymer.Justify. PHBV is a preferred polymer because it is a biodegradable polymer.

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Chapter-16

CHEMISTRY IN EVERYDAY LIFE

1. Account for the following:a. Metal hydroxides are better antacids than metal

hydrogencarbonates.b. Antihistamines and antacids work on different receptors.Ans: a)Metal hydrogen carbonates makes the stomach alkaline and triggers the production of more acid whereas metal hydroxides are insoluble and do not increase the pH above 7.

b)They prevent the interaction of histamine with the receptors present in the cell wall which results in release of lesser amount of acid.

2. What is the role of Cimetidine or Ranitidine in curbing acidity?Ans: Anti histamines do not affect the secretion of acid in stomach whereas antacids do. This shows that they work on different receptors.

3. How can we prevent pollution ,while using detergents?Ans: Unbranched degertants can be biodegraded and hence pollution can be prevented.

4. To which class of synthetic detergents is the detergent used in hair conditioner belong?

Ans: Cationic detergents. Example- Cetyltrimethylammoniumbromide

5. Classify the following :- a) dish washing detergent b) detergent in tooth paste. c)detergents in hair conditioner d)detergents having germicidal properties

Ans: a)Non ionic detergentsb)Anionic detergentsc)Cationic detergentsd)Cationic detergents

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6. What is the ingredient used in shaving soap to prevent rapid drying?

Ans: Glycerol

7. Identify the following:- a) A sweetner which can be used in cool drinks and soft foods only. b) The control of sweetness of food is difficult with this sweetner. c) that which is stable at cooking temperature.

Ans a) Aspartameb) Alitamec) Sucrolose

8. Out of Aspartame ,Alitame and Sucrolose, which is the preffed and why?

Ans: Sucrolose because it is stable at cooking at temperature and does not produce calories.

9. Name the component in the birth control pill which suppresses ovulation?

Ans: Progeterone

10.Name a broad spectrum antibiotic which is rapidly absorbed from the gastro intestinal tract?

Ans: Chloramphenicol

11.Name the non-narcotic analgesic which is also an antpyretic used to prevent heart attacks?

Ans: Aspirin (prevents blood clot)

12.Pick out the odd one out:a) 1) Valium 2) Serotonin 3) Veronal 4) Prontosilb) 1) erythromycin 2) Tetracyclin 3) Penicillin 4)

ChloramphenicolAns: a) Prontosil

b) Penicillin

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