jee chemistry cl

8/16/2019 JEE Chemistry CL

1/39

Appendix B

Mock Test 1

PAPER 1

Section A: Single Option Correct Type Questions

1. In a certain polluted atmosphere containing O 3 at a steady concentration of 2.0 ¥ 10−8 mol/L, the hourly production

of O3 by all sources was estimated as 7.2 ¥ 10−15 mol/L. The rate constant for the destruction reaction, 2O 3 Æ 3O 2 is

(a) 1.3 ¥ 10 −3 L/mol/s (b) 5 ¥ 10 −3 L/mol/s (c) 1.9 ¥ 10 −3 L/mol/s (d) 3.6 ¥ 10 −15 L/mol/s 2. A solution of 0.4 mole of KI (100% dissociated) in 1000 g water freezes at T 1°C. Now to this solution, 0.2 mole

HgI2 is added and the resulting solution freezes at T 2°C. Which of the following is correct?

(a) T 1 = T 2 (b) T 1 < T 2 (c) T 1 > T 2 (d) Cannot be predicted

3. Which of the pair of orbitals have electronic density along the axis?

(a) d xy

, d xz

(b) d x

2

– y

2, d z

2 (c) d xz

, d yz

(d) d xy

, d z

2

4. There is a mixture of Cu(II) chloride and F(II) sulphate. The best way to separate the metal ions from this mixture

in qualitative analysis is by treating it with

(a) hydrogen sulphide in mild acidic medium, where only Cu(II) sulphide will be precipitated.

(b) ammonium hydroxide buffer, where only Fe(II) hydroxide will be precipitated.

(c) hydrogen sulphide in mild acidic medium, where only Fe(II) sulphide will be precipitated.

(d) ammonium hydroxide buffer, where only Cu(II) hydroxide will be precipitated.

5. The anhydride of nitric acid is

(a) Nitric oxide (b) Nitrous oxide (c) Dinitrogen trioxide (d) Dinitorgen pentoxide

6.

O200°C, D

Product

(+) R form

The most suitable product is

(a)

(–) S form

(b)

ptically inactive

O (c)

(–) R form

O

(d) O

(+) S form

7. In aqueous solution, molecules remain in equilibrium with open chain structure:

OCH3

OCH3

CH O3

CH O3 OH

O

The open chain structure is:

(a)

OH

H

OCH3

OCH3

H

H

H

CH O3

CH OCH2 3

CHO

(b)OH

H

OCH3

OCH3

H

H

H

H O3

CH OCH2 3

CHO

(c)

H

OCH3

OCH3

H

H

H

H O3

CH OH2

CH OH2

OCH3

(d)

OH

H

OCH3

OCH3

H

H

H

CH O3

CH OCH2 3

CHO


2/39

8. Tryptophan [B-COOH; pK a = 2.4] does not migrate towards any pole, when it is placed in electrified solution of pH

5.9. The pK a ( NH )+

3- is

(a) 2.4 (b) 5.9

(c) 9.4 (d) 3.5

9. At 20°C and 1.00 atm partial pressure of hydrogen, 18 mL of hydrogen measured at STP, dissolves in 1 L of water. If

water at 20°C is exposed to a gaseous mixture having a total pressure of 1400 torr (excluding the vapour pressure of

water) and containing 68.5% H2 by volume, what volume of H 2, measured at STP, will dissolve in 1 L of water?

(a) 18 mL (b) 12 mL

(c) 23 mL (d) 121 mL

10.

CH CH CH

CH CH CH CH CH CH CH CH CH CH CH

3

3 3B

3 2 3A

3

3 3

| | | æ æ == æ ¨ æ æ æ æ æ æ æ Æ æ æ CC CH CH CH

OH

2 3== æ æ |

A and B are, respectively,

(a) A = conc. H 2SO4, 140°C, B = conc. H2SO4, 120°C

(b) A = conc. H 2SO4, 140°C, B = Al2O3, pyridine, 350°C

(c) A = Al 2O3, pyridine, 350°C, B = conc. H2SO4, 140°C

(d) A = Al 2O3, pyridine, 350°C, B = Al2O3, 350°

Section B: More Than One Correct Option

11. If 0.53° is Bohr’s radius for the first orbit. It suggests in the light of the wave mechanical model that

(a) The product of Y2 and 4 Qr 2dr remains constant till it reaches at the distance of 0.53 Å. (b) Only Y2 goes on increasing, 4Q r 2dr remains constant till it reaches at the distance of 0.53 Å. (c) Y2 goes on increasing, 4Q r 2dr goes on decreasing till it reaches at the distance of 0.53 Å. (d) Only 4Q

2dr goes on increasing, Y2 remains constant till it reaches at the distance of 0.53 Å.

12. M (alkali metal solution ( + ) NH [M(NH ) ]3 3Ammoniatedmetal c

) + Æ + x y xaation

3

Ammoniated electron

+ [ (NH ) ]e y-

Mark out the correct statement/s about the solution.

(a) Solution has high electrical conductivity due to ammoniated electron.

(b) Solution imparts blue colour due to greater polarization of metal ion.

(c) Solution is quite stable, which is considered as a dilute metal. (d) On the addition of substance like iron oxide, solution decomposes and releases H 2 gas.

13. The rate of reaction is greatly influenced by change in solvent and its polarikty. Which of the following combination/s

(S N2 reaction reactants–effect on rate) is/are correct, when solvent polarity is increased?

L = leaving group, Nu\

= nucleophile

(a) R – L + Nu \ … small decrease in rate (b) R – L + Nu … large decrease in rate

(c) R – L + + Nu \ … large decrease in rate (b) R – L + + Nu … small decrease in rate


3/39

14.

Mark out correct statements regarding these conformations:

(a) Conformation I and II both are equivalent in stability.

(b) Conformation stability: II > I

(c) Conformation II possesses pitzer strain.

(d) Conformation I possesses trans-annular strain.

15. H PO H H PO ;3 4 2 4X VX W XX + -+ K a1

H PO H HPO ;2 442

2

- + -+ K a

HPO H PO ;442 3

3

- + -

+ K a Mark out the incorrect statements:

(a) K a1 > K a2 > K a3 (b) pH(H PO ) pK pK

22 4

- = +a a1 2

(c) HPO42– is more acidic than H 2PO4

– . (d) Only HPO42– is the amphiprotic anion in the solution.

Section C: Integer Type Questions

16. In the reaction, HO–CH 2CHOdil.NaOH æ Æ æææ how many distinct products (saturated) are possible?

17. How many dichloro products (including stereoisomers) will be formed when R-2-chloropentane reacts with Cl 2 in

presence of UV radiation?

18. Vant Hoff factor, i for 100% ionized K 2HgI4 solution in water is

19. The magnetic moment of a transition metal ion is 15 B.M. Find out the number of unpaired electrons in it.

20. The number of moles of NaNH 2 required to convert CH CH

Br

CH

Br

to CH C C H3 3 æ |

æ |

æ ∫∫ æ 2 in liquid NH 3 are.

PAPER 2

Section A: More than One Correct Option

1. Which of the following will give a yellow precipitate with iodine and alkali?

(a) 2-hydroxy butane (b) C 6H5 – CO – CH 2 – CH3

(c) C 6H5 – CH2 – CO – CH 3 (d) C H NH C

O

CH6 5

| |

3 æ æ æ 2. Which of the following molecules has an O–O bond?

(a) H 2S2O8 (b) H 2S2O7 (c) H 2SO5 (d) H 2S2O6 3. 1 mole of Ba(OH) 2 will exactly neutralize

(a) 0.5 mole HCl (b) 1 mole of H2SO4 (c) 1 mole of H 3PO3 (d) 2 mole of H 3PO2


4/39

4. Which one is/are correct among the followings?

Given, the half cell emf’s

E E Cu |Cu

°

Cu |Cu

°+2 +1= =0 337 0 521. , .

(a) Cu +1 disproportionates

(b) Cu and Cu 2+ comproportionates (reverse of dislproportionation into Cu +).

(c) E E Cu|Cu

°

Cu |Cu

°+2 +1+ is positive

(d) All of these

5. Phenol and benzoic acid may be distinguished by their reaction with

(a) Aqueous NaOH (b) Aqueous NaHCO 3 (c) Neutral FeCl 3 (d) Aqueous NH 3.

6. [Co(NH 3)5 NO2]SO4 shows

(a) Ionisation isomerism (b) Coordination isomerism

(c) Linkage isomerism (d) Position isomerism

7. The standard electrode potential of a metal–metal ion (Ag|Ag +) and metal-sparingly soluble salt anion (Ag|AgCl|Cl – )

are related as

(a) E E RT F

K Ag |Ag°

Cl |AgCl|Ag°

sp+ In= +- (b) E E RT F K

Cl |AgCl|Ag°

Ag |Ag°

sp+ In- = +

(c) E E RT

F K Cl |AgCl|Ag°

Ag |Ag

°

sp

+ InCl

- = --[ ]

(d) E E RT

F

K

Cl |AgCl|Ag

°

Ag |Ag

° sp+ In

Cl- = - -[ ]

8. Which of the following compounds will give haloform test?

(a)

O||

CH C CH I3 2 æ æ (b) CH CH CH

CI

3 3 æ æ |

(c) (d) CH 3CH2Br

Section B: Comprehension-Based Questions

Comprehension I

A and B are two ores of copper.

A undergoes calcinations on to form solid, CO2(g) and H 2O B undergo roasting to form solid and gas C which turns acidi-

fied orange K 2Cr 2O7 to green solution.

B also reacts with dil. HCl to form solid and gas D which turns lead acetate solution black and also reacts with C to form

colloidal in the presence of moisture.

Choose the correct answer.

9. The copper ores and B are, respectively,

(a) Carbonate and hydroxide (b) Carbonate and oxide

(c) Carbonate and sulphide (d) Sulphide and carboante

10. The gas (C) acts as

(a) Oxidising agent (b) Reducing agent (c) Both (a) and (b) (d) Fluxing agentComprehension II

Titration is nothing but a process through which we determine the unknown strength of a solution by means of standard

solution, i.e., solution of known strength by using suitable indicators. During the titration, a stage of reaction comes where

the equivalents of externally added acid (or base) in the basic (or acidic) solution become equal and undergo complete

reaction and completely converts into salt, known as equivalence point of the reaction. During the course of the reaction or

titration, not only pH of the solution, but also the amount of externally added acid (base) is calculated at pre-equivalence

point, equivalence point and post-equivalence point.


5/39

If CH3COOH is titrated with standard NaOH solution, then at different equivalence points, its pH can be calculated

by different ways.

At pre-equivalence point, i.e., 1/3rd, 1/4th, half equivalence point, its pH can be calculated by buffer solution. At equiva-

lence points its pH is calculated by salt hydrolysis. At post-equivalence point, its pH is calculated by remaining [OH

–

].Similarly, when a basic solution is titrated with standard acid solution, then three equivalence points, i.e., pre-equivalence point, equivalence point and post-equivalence point are achieved. pH at different equivalence points is determined.

pH (of acidic buffer) = pK a + log[salt]

[acid]

For basic buffer, pH = pK a + log[salt]

[base]

Hydrolysis of CH3COONa gives [ ]OH- = = ¥hC K C h

where h is the degree of hydrolysis,

K h is the hydrolysis constant and C is the concentration of salt in the solution.

Hydrolysis of NH4Cl gives

[H +] = hC = K C h ¥

Two chemistry (Hons.) students perform two different sets of experiments I and II as under :

Experiment I: One students takes 100 ml ( M /10) CH3COOH (pH = 3) into a conical flask and he titrates it with M /40

NaOH solution at different equivalence points. At half-equivalence point, he forms solution X and at equivalence point

he forms solution Y. In solution Y, he adds 400 ml of same NaOH solution and he dilutes it by adding 100 ml H 2O to

convert into solution Z. All reactions of experiments I and II occur at 25°C.Experiment II: Another student takes a certain amount of weak base BOH in a conical flask. Then he titrates it by 10

ml of HCl taken in burette and he finds a solution A whose pH is 9. Now he adds 10 ml of HCl in solution A to convert

it into solution B as well. He finds that the pH of solution B changes by 1 unit. He prepares solution C at the equivalence

point.

11. What is the total volume of solution X?

(a) 200 ml (b) 100 ml (c) 300 ml (d) 500 ml. 12. What is the pH of the solution X?

(a) 5 (b) 6.5 (c) 4.5 (d) 7.Comprehension III: E 2 reaction Æ Elimination bimolecular In the general mechanism of the E2 reaction, a strong base abstracts a proton on a carbon atom adjacent to the one of the

leaving group. As it abstracts a proton, a double bond forms and the leaving group leaves.

Mechanism:


6/39

Choose the correct answer

13. Identify the rate of reaction of given compounds in E 2 reaction.

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

14. In given pairs, which compound is more reactive toward E 2 reaction.

(P) (Q)

(R) (S)Ph CH CH Br

(VII)2 2 æ æ æ

Ph CH CH

Br (VII)

3 æ æ |

(a) P – ΙΙ, Q – ΙΙΙ, R – V, S – VII (b) P – ΙΙ, Q – ΙΙΙ, R – VI, S – VΙI

(c) P – Ι, Q – ΙΙΙ, R – VΙ, S – VΙΙ (d) P – Ι, Q – IΙΙ, R – V, S – VΙΙΙ

Comprehension-IV

Tin and lead form divalent Sn(II) and lead (II) and tetravalent i.e., Tin (IV) and lead (IV) compounds. Tetravalent and

dioxides of lead and Tin are amphoteric in nature. Lead tetra fluorite is ionic solid where as PbCl 4 is covalent. PbI4 does

not exist lead (II) halide are white where as PbI 2 is yellow in colour

15. Reaction of SnO with NaOH gives

(a) Na 2SnO2 (b) H 2 SnO2 (c) Na 3SnO3 (d) SnO2 16. PbI4 does not exist due to

(a) Inert pair effect (b) High reducing power of I –

(c) High oxidizing power of I – (d) High electronegativity of I –

Section C: Matrix Match Type Questions

17. Match the following

Column I Column II

(p) The ratio of velocity of electron in 5th and 3rd exicted level for a H+ atom is (a)4

1

(q) The ratio of wavelength of series limit of Balmer to Lyman series for a H-atom is (b) 23

(r)The ratio of wavelength of photon corresponding to C-line of Lyman series and H-line of

Paschen series for a H-atom is(c)

1

4

(s) The ratio of energy difference between 3rd and 1st orbit of H-atom and He+ ion is (d)3

32


7/39

18. Match the following columns.

Column I Column II

(p) (a) Benzoin condensation

(q) (b) Bayer-Villiger oxidation

(r) (c) Witting reaction

(s) (d) Reformatsky reaction

19. Match the following.

Column I Column II

(p) (a) S N1

(q) (b) S N2

(r) (c) E 1

(s) (d) E2

20. Match the following columns

Column I Column II

(p) NaNO3 D æ Æ æ (a) O2

(q) Ag(NO3)2 D æ Æ æ (b) NO2

(r) Mg(NO3)2 D æ Æ æ (c) SO2

(s) FeSO4. 7H2OD æ Æ æ (d) SO3


8/39

Mock Test 2

PAPER 1Section A: Single Option Correct Type Questions

1. Mark out the optically inactive molecule:

(a) (b)

(c) (d)

2. E-1-chloro-2-methyl cyclohexanealc.KOH

E2 æ Æ æææ Product

Predominant product is

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

3. A Adiponitrile BH O

”

H , Ni

, pressure

3+

2

D D¨ æ ææ æ Æ æææ

A B Polymer + æ Æ æ D

Polymer is:

(a) Polyacrylonitrile (b) Nylon-6 (c) Nylon-66 (d) Buna-N

4. The formal charge on oxygen atom of ozone is

(a) Zero, for all the three atoms

(b) 0, +1, –1, respectively, on 1st, 2nd and 3rd O-atom (c) 2, –1, –1, respectively, on 1st, 2nd and 3rd O-atom

(d) –2, +1, +1, respectively, on 1st, 2nd and 3rd O-atom

5. At 0°C, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. The molecular mass of the oxide is

(a) 70 (b) 140 (c) 28 (d) 60

6. A hydrogen electrode is placed in a buffer solution of CH 3COONa and CH3COOH in the ratio of x : y and y : x.

which has electrode potential E 1 and E 2 volts, respectively, at 298 K ( E 1 and E 2 are reduction potentials). Mark outthe correct expression for pK a of CH 3COOH.

(a) pK E E

a = - +1 20 118.

(b) pK E E

a = -2 10 118.

(c) pK E E

a = -1 20 118.

(d) pK E E

a = +1 20 118.

7. A pale yellow precipitate and a gas with pungent odour are formed on warming dilute hydrochloric acid with an

aqueous solution containing

(a) Sulphate ion (b) Sulphide ion (c) Thiosulphate ion (d) Sulphite ion

8. Chloroplatinic acid is

(a) Monobasic (b) Dibasic

(c) Tribasic (d) Tetrabasic

9. Borax structure contains

(a) Two BO 4 groups and two BO 3 groups (b) Four BO 4 groups only

(c) Four BO 3 groups only (d) Three BO 4 and one BO 3 groups


9/39

10. B E-2-pentene A II I ¨ æ æ æ Æ æ

In the reaction, A and B are diastereomers. Moreover A undergoes oxidative cleavage while B is comparatively

stable on treatment with HIO4. Which of the following set of reagents are true for I and II?

(a) I = HCO 3H, II = KMnO4, ∆ (b) I = alkaline H 2O2, II = SeO2 (c) I = OsO 4, pyridine; Na2SO3, H2O, II = PdCl2, H2O

(d) I = Ag,1

2O2 ; H2O, II = OsO4, pyridine; Na2SO4, H2O

Section B: More than One Option Correct

11. Methanol, a liquid fuel, can be prepared from water gas and additional H 2 at high temperature and pressure in the

presence of suitable catalyst

2H (g) + CO(g) CH OH(g), = 92.2 kJ2 3X VX W XX D H -

Assuming at equilibrium, (a) If pressure of the system is increased, availability of CH 3OH is increased.

(b) If the temperature is increased, availability of CH 3OH is increased.

(c) Addition of extra amount of catalyst increases the availability of CH 3OH.

(d) Increase in temperature increases D H . 12. Observe following reactions:

A + C H SO Cl Clear solution6 5 2OH KOH H

+- æ Æ ææ æ Æ ææ æ Æ æ Substance is soluble in acidic medium

B + C H SO Cl Insoluble substance in H /OH6 5 2OH +

- æ Æ ææ -

C + C H SO Cl Insoluble in base,6 5 2OH

- æ Æ ææ but soluble in H Clear solution+ H

+

æ Æ æ

A, B and C are amines, respectively

(a) 1°, 2° and 3° (b) 1°, 3° and 2° (c) 1°, aniline, 2° (d) 1°, 2° and aniline

13.

Which of the following is/are correct about A and B?

(a) A and B both are same optically active retention products.

(b) A and B both are optically inactive.

(c) A is optically active product and B is optically inactive product.

(d) A is retention product and B is racemic mixture product.

14. Which of the following sequences of reagents is/are a good means to furnish the conversion? RCH 2OH Æ RCH 2 NH2 (a) KMnO 4; SOCl2; NH3; OBr

− (b) PBr 3; NaCN; H2/Ni

(c) Collins reagent; NH 3, H2/Ni (d) All of these

15. Which of the following half cell reaction is/are concerned to reference electrode?

(a) 2H 2(g) + 4OH−(aq) → 4H2O(l) + 4e

− (b) H 2(g) → 2H

+(aq) + 2e

−

(c) 2H 2O(l) + 2e− → H2(g) + 2OH

−(aq) (d)1

2Hg2Cl2(s) + e

− → Hg(l) + Cl−(aq)


10/39

Section C: Integral Type Questions

16. An intimate mixture of Fe 2O3 and Al is used in solid fuel rocket. D H of Al 2O3 and Fe 2O3 are 399 kcal and 199 kcal,respectively. Calculate the fuel value in kcal/g of mixture.

17. The temperature of 1 mole helium gas is increased by 1°C. Find the increase in internal energy.

18. 3.150 g of oxalic acid [(COOH)2. xH2O] is dissolved in water and volume is made up to 500 mL. On titration,

28 mL of this solution required 35 mL of 0.08 N NaOH solution for complete neutralization. Find the value of x.

19. On acidic hydrolysis, CH 3 –CH2 –NC will give a carboxylic acid that may contain _____ number of carbons in it.

20. In the given reaction,

Products (major)

How many carbon atoms will be present in the ring?

PAPER 2

Section A: More than One Option Correct

1. Which of the following is/are correct statements?

(a) Hardy Schulz rule is related to coagulation.

(b) Brownian movement and Tyndall effect are shown by colloids.

(c) When liquid is dispersed in liquid, it is called gel.

(d) Gold number is a measure of protective power of lyophillic colloid.

2. Which of the following compounds on hydrolysis yields a carboxylic acid?

(a) (b)

(c) (d)

3. P4 can react with O 2 to form P4O6 and P4O10 according to the following reactions P4 + 3O 2 æÆ P4O6 P4 + 5O2 æÆ P4O10 If each option is written in an order like ( x, y, z , p) where x represents moles of P4 taken, y represents moles of O 2

taken, z represents moles of P4O6 formed and p represents mole of P4O10 formed, which options are correct? (a) 1, 4, 0.5, 0.5 (b) 1, 3, 1, 0 (c) 0.5, 2.5, 0.5, 0 (d) 2, 5, 2, 3 4. Which of the following option is/are correct?

(a) The vapour phase of XeF 6 is covalent and hybridisation is sp3

d 3

. (b) The solid phase of XeF 6 is ionic and hybridisation is sp

3d

2.

(c) XeF 2 is linear in shape.

(d) XeO 2F2 follows bent rule and structure is see-saw.

5. Which one of the following statement is correct in relation to ionisation enthalpy?

(a) Ionisation enthalpy increases for each successive valence shell electron.

(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core of noble gas

configuration.


11/39

(c) The end of valence electrons is marked by a big jump in ionization enthalpy.

(d) The removal of electron from orbitals bearing lower n value is easier than from orbit having higher n value.

6. If the rms velocity of nitrogen and oxygen molecule are same at two different temperature and same pressure, then

(a) Average speed of molecules is also same. (b) Density (gm/lt) of nitrogen and oxygen is also equal.

(c) Number of moles of each gas is also equal.

(d) Most probable velocity of molecules is also equal.

7. Which of the following contain oxygen?

(a) Bauxite (b) Chalcopyrite

(c) Haematite (d) Calamine

8. Which of the following statements are true about formic acid?

(a) It is a stronger acid than acetic acid.

(b) It forms formyl chloride with PCl5.

(c) It gives CO and H 2O on heating with conc H2SO4.

(d) It reduces Tollen’s reagent.

Section B: Comprehension-Based Questions

Comprehension-I

RNH2 reacts with HNO 2 (nitrous acid from NaNO 2 and dil. HCl) forming alcohols. Chemist A wanted to prepare 2-propanol

and thus he worked out the synthesis from 2-aminopropane. Chemist B also wanted to prepare 2-propanol but she could

not arrange 2-aminopropane but instead she tried the synthesis from 1-aminopropane.

Choose the correct answer.

9. What was the major product of chemist B?

(a) 1-Propanol (b) 2-Propanol

(c) Propene (d) Cyclopropane

10. What are the most stable intermediates of chemists A and B?

A B

(a) CH CH CH3 2 2

≈ CH CH CH3 2 2

≈

(b) CH CHCH3 3

≈ CH CH CH3 2 2

≈

(c) CH CHCH3 3

≈ CH CHCH3 3

≈

(d) None of the above is correct

Comprehension-II

The equivalent weight of a species if acts as oxidant or reduction should be derived by :

Eq. weight of oxidant or reductant =Mol.wt. of oxidant or reductant

Number of electrons lost or gained or r reductant

During chemical reactions, equal equivalents of one species react with the same number of equivalents of other species

giving same number of equivalent of products. However, this is not true for reactants if they react in terms of moles. Also,

molarity can be converted to normality by multiplying the molarity with valence or n factor.

11. Equivalent weight of Fe 2O3 in term of its mol. weight in the change Fe 3O4 Æ Fe 2O3 is (a) M (b) M /2 (c) M /3 (d) 3 M /2

12. Equivalent weight of N 2 and NH 3 in the change N 2 Æ NH 3, respectively, is (a) 4.67, 12.4 (b) 9.3, 12.4 (c) 4.67, 5.34 (d) 5.34, 4.67


12/39

Comprehension-III

Hydrogen atom in its ground state is excited by means of monochromatic radiations of wavelength 970.6 Å. The ionization

energy for hydrogen atom is 13.6 eV. (h = 6.63 ¥ 10 –34 J s)

Choose the correct answer 13. How many lines are possible in the resulting emission spectrum?

(a) 6 (b) 2 (c) 3 (d) 4

14. How many of these lines are in the visible region of the spectrum?

(a) 3 (b) 2 (c) 4 (d) 5

Comprehension-IV

Sodium metal can be extracted by either Castner’s process or Down’s process.

Castner’s process

Electrolyte æÆ fused NaOHCathode and anode are made up of a steel rod and Ni cylinder, respectively. The reaction takes place as follows At cathode

Na + Na+ e- Æ

At anode OH1

2H O +

1

4O +2 2

- -Æ ≠ e

Down’s process

Electrolyte æÆ fused NaCl and CaCl 2Cathode and anode are ring shaped iron metal and graphite respectively. The takes place as follows:

Na Na+ -+ Æe

Cl Cl- -Æ ≠ +1

22 e

15. In Castner’s process, some sodium metal is lost due to

(a) Reaction with dissolved O 2 is molten NaOH

(b) It’s high reactivity, it attacks other substance present in the system

(c) High temperature at which some of it vapourizes

(d) All of these

16. CaCl 2 is used along with NaCl due to

(a) Increase in electrical conductivity on NaCl

(b) Lowering in m.pt in the mixture compared to pure NaCl

(c) CaCl 2 protects the Na from getting vapourized

(d) Viscosity of the mixture of NaCl or CaCl 2 is lower than pure NaCl.

Section C: Match Matrix Type Questions


Column-I

Atomic masses

Column-II

% composition of lighter isotope

Isotope-I Isotope-II Average

(p) (a + 4) ( a – 1) a (a) 66.67% by moles

(q) a 5a 2a (b) 50% by moles

(r) (a + 3) ( a + 1) ( a + 2) (c) % by mass independent of a

(s) (a + 2) ( a – 1) a (d) 80% by moles


13/39


Column-I Column-II

(p) (a)

(q) (b)

(r) (c)

(s) (d)


Column I Column II

(p) Diffusion of gas (a) High when molecular mass is low

(q) Compressibility of gas (Z) < 1 (b) Negative deviation

(r) Vrms (c) Increases with increase in temperature

(s) When a gas is liquefied of gas (d) Attractive force dominates


Column I Column II

(p) ClF ,BrF ,IF5 4+

6

-(a) All molecule/ions are polar in nature

(q) ClF ,BrF ,ICl3 2+

4- (b) All molecules/ions have same number of lone

pair(s) and same shape

(r) XeF , ICl , I2 2 3- - (c) All molecules/ions have same oxidation state of

central atoms

(s) ClOF , ClF , IO F3 4+

2 2- (d) All molecules/ions have same hybridization of

central atoms


14/39

Mock Test 3

PAPER 1Section A: Single Option Correct Type Questions

1. Mark out the correct decreasing order of entropy.

I. 1 mole of H 2O(l) at 4°C and 1 mtm pressure II. 1 mole of H2O(s) at 0°C and 1 atm pressure

III. 1 mole of H 2O(g) at 100°C and 1 atm pressure IV. 1 mole of H2O(l) at 0°C and 1 atm pressure

(a) II > I > IV > III (b) II > IV > I > III

(c) III > IV > I > II (d) III > I > IV > II

2. A is an organic compound which can reduce Tollen’s reagent, reduces mercuric chloride giving black precipitate,

decolourises acidified KMnO4 and can reduce Fehling solution also. Compound A is

(a) HCOOH (b) CH = CH

(c) CH 3CHO (d) CH2 = CH 2

3.

Predominant product is

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

4.

The major product is

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

5.

Predominant product is:

(a) Polyacrylate (b) Polymethylmethacrylate

(c) Latex (d) Perspex

6. The correct increasing order of extent of hydrolysis in the following is

(a) CCl 4 < MgCl 2 < AlCl 3 < SiCl4 < PCl5 (b) CCl 4 < AlCl 3 < MgCl 2 < PCl5 < SiCl4 (c) AlCl 3 < MgCl 2 < CCl 4 < PCl5 < SiCl4 (d) SiCl4 < MgCl 2 < AlCl 3 < PCl5 < CCl 4 7. Auto-oxidation of bleaching powder gives

(a) Only calcium chlorate (b) Only calcium chloride

(c) Only calcium hypochlorite (d) Both (a) and (b)


15/39

8. When AgNO 3 is treated with excess of KI, AgI precipitates out, which forms colloidal solution with

(a) Positive charge

(b) Negative charge

(c) Charge neutralization occurs, not charge acquisition (d) First positive, later negative

9. The correct name for the complex ion [Co(en) 2Cl(ONO]+ is

(a) Cobalt diethylenediaminechloronitrate (b) Chlorodiethyldiaminenitrito cobalt (III)

(c) Chloronitritodiethyldiamine cobaltate (III) (d) Chlorobis (ethylenediamine) nitrite cobalt (III)

10. For the reaction, 3BrO \ æÆ BrO 3 \ + 2Br \ , in alkaline solution, the value of the second order (in BrO−) rate

constant at 80°C in the rate law for −D[BrO−]/Dt was found to be 0.056 L/mol/s. Then D[BrO3−]/Dt is

(a) 0.019 L/mol/s (b) 0.112 L/mol/s

(c) 0.168 L/mol/s (d) 0.056 L/mol/s

Section B: More than One Correct Option

11. Mark out the salt solution which may have pH less than 7:

(a) BaI 2 (b) [(CH 3)3 NH]Cl (c) Cr(NO3)3 (d) NH 4F

12. Following the graph (Fig. B.1), which of the statement/(s) is/(are) correct?

(a) The absolute conversion of N 2O4 into NO 2 (N 2O4 Æ 2NO 2) is spontaneous. (b) The dynamic conversion (N 2O4 S2NO 2) at equilibrium point is spontaneous.

(c) Formation of N 2O4 is more spontaneous than dissociation.

(d) Both process dissociation as well as formation are equally spontaneous.

13. Which of the following statement(s) is/are correct?

(a) Chlorine dioxide (ClO 2) is powerful oxidizing agent but bleaching action is lower than Cl 2.

(b) ClO2 in alkaline solution undergoes disproportionanation.

(c) ClO 2 is diamagnetic in nature.

(d) ClO 2 is a yellow gas but deep red liquid.

14. Which of the following pairs show co-ordination isomerism?

(a) [Co(NH 3)6] [Cr(CN)6] and [Cr(NH3)6][Co(CN)6]

(b) [Co(NH 3)3(H2O)2]Br 2 and [Co(NH 3)3(H2O)Cl.Br]Br.H2O

(c) [Pt(NH3)4Cl2]Br 2 and [Pt(NH3)3Br 2]Cl2 (d) [Co(NH 3)6][C2O4)3] and [Cr(NH3)6][Co(C2O4)3]


16/39

15. Some alkanes may be prepared by adding an alkyl iodide (iodoalkane) dropwise to sodium in a suitable inert solvent,

e.g.,

2CH 3I + 2Na Æ C 2H6 + 2NaI

If a mixture of iodomethane and 2-iodopropane is used as starting material, the product will contain (a) 2-methyl propane (b) Methane, propene and propane

(c) Ethane and 2, 3-dimethyl butane (d) Methane and propane only

16. CH CH + 2CH COOH Pr oducts. Product/31.Hg

2. ,300-400°C

2+

∫∫ æ Æ æææææ D ss is/are

(a) CH 3CHO (b) CH3 – CH(OCOCH 3)2

(c) (d)

Section C: Integral Type Questions

17. In the given oxides, how many of them are purely basic in nature: Na 2O2, KO2, BeO, CaO, CO2, BaO, RbO2?

18. How many structures for [F] are possible?

19. When chlorine reacts with hot and conc. NaOH, what will be the oxidation state of chlorine in oxidised form?

20. For a reaction dx/dt = k [H+]n, if the pH of reaction medium changes from two to one, the rate becomes 100 times of

the value of at pH = 2. What is the order of reaction?

21. For the first order reaction, the rate constant is 7.7 ¥ 10 –2 s –1. Calculate the time required for the initial concentration1.5 mole of the reactant to be reduced to 0.75 mole.

PAPER 2

Section A: More than One Option Correct

1. Which of the following is/are correct ?

(a) D H = DU + D( PV ) when P and V both changes (b) D H = DU + P DV when pressure is constant (c) D H = DU + V D P when volume is constant (d) D H = DU + P DV + V D P when P and V both changes 2. Consider the reaction 2CO(g)+O 2(g) 2CO2(g) + Heat. Under what conditions shift is undeterminable?

(a) Addition of O 2 and decrease in volume

(b) Addition of CO and removal of CO 2 at constant volume

(c) Increase in temperature and decrease in volume

(d) Addition of CO and increase in temperature at constant volume

3. The ether when treated with HI produces:

(a) (b)

(c) (d)


17/39

4. The following statement(s) is correct:

(a) A plot of log K P versus 1/ T is linear.

(b) A plot of log [ X ] versus time is linear for a first order reaction, X æÆ A.

(c) A plot of log P versus 1/ T is linear at constant volume. (d) A plot of P versus 1/ V is linear at constant temperature.

5. Which of the following is/are correct statements regarding these molecules (I, II and III)?

(a) I and II are diastereomers (b) I and II are enantiomers

(c) II and III are diastereomers (d) II and III are enantiomers

6. Which of the following is correct statement(s)?

(a) CH 3 – F is more reactive than CH 3 – I in S N2 reaction.

(b) DMF is polar aprotic solvent.

(c) CH 3 – F is less reactive than CH 3 – I in S N2 reaction.

(d) DMF is polar protic solvent.

7. In 0.1 M solution, K a for dissociation of H 2S is 4.0 ¥ 10 –3. Select the correct statements.

(a) Concentration of H + is 0.018 M. (b) The degree of dissociation of H2

S is 18%.

(c) pH of solution is 1.7447. (d) Concentration of [H +] is 0.18 M.

8. What is the degeneracy of the level of the hydrogen atom that has the energy- R H

9?

(a) 0 (b) 3 (c) 7 (d) 9

Section B: Comprehension Based Questions

Comprehension-I

If the boundary of system moves by an infinitesimal amount, the work involved is given by dw = – P ext dV .

For irreversible process, w = – P ext ∆V (where ∆V = V f – V i)

For reversible process, P ext = P int ± dP = P intSo for reversible isothermal process, w = – nRT in V f /V iTwo moles of an ideal gas undergoe isothermal comprehension along three different paths:

(i) reversible comprehension from P i = 2 bar and V i = 8 L to P f = 20 bar

(ii) a single stage comprehension against a constant external pressure of 20 bat, and

(iii) a two-stage compression consisting initially of compression against a constant external pressure of 10 bar until

P gas = P ext, followed by compression against a constant pressure of 20 bar until P gas = P ext.

9. Work done (in bar-L) on the gas in reversible isothermal compression is:

(a) 9.212 (b) 36.848 (c) 18.424 (d) none of these


18/39

10. Work done on the gas in single stage compression is:

(a) 36 (b) 72 (c) 144 (d) none of these

Comprehension-II

Pseudo halides are anions having resemblance with halide ions. Group I metals can form salts with pseudo halogens canact as ligands and form co-ordinate complexes. Their hydrides are weakly acidic and can be prepared in analogous way

as halogen hydrides are prepared. Azides, cyanides, selonocyanides are example of pseudo halides.

11. Cyanide, CN – is a pseudo halide. When cyanogen is heated with alkali solution, the products are

(a) HCN, H 2O (b) NH3, (NH4)2CO3 (c) NaCN, NaOCN (d) HCOONa, NH 3 12. When NaCN reacts with H 2SO4, the products are

(a) HCN and Na 2SO4 (b) HCN and NaHSO4 (c) (CN) 2 and Na 2SO4◊H2O (d) None of these

Comprehension-III

Electrophilic addition in this mechanism a positive species approaches the double or triple bond and in the first step forms

a bond by converting then Q-pair of electrons into a T pairs

In the reaction X need not actually be a positive ion but can be the positive end of a dipole or an induced dipole, with the

negative part breaking off either during the first step or shortly after.

Not all electrophilic additions follow the simple mechanism. In many cases, if (1) forms, it very rapidly cyclises giving

cyclic intermediate (2)

Whether intermediate is (1) or (2), the mechanism is called electrophilic addition bimolecular.

Stability of intermediate (1) or the formation of (2) controls the stereochemistry of the reaction. As the stability of (1)

increases rotation along C – C is possible, reaction may be non-stereospecific.

13. CH 3 – CH = CH – CH 3 X , CCl2 4 æ Æ æææ Product (X2 = Cl 2, Br 2, I2). In the reaction , which X is having maximum

tendency to form intermediate like (2)?

(a) Cl (b) Br (c) I (d) Cl = Br = I

14.

W2 is an electrophilic reagent. The addition of W 2 to the alkene A proceeds through intermediate like (2). Which

of the following is correct about the reaction?


19/39

(a) Reaction is non-steoreospecific (b) Products are meso compounds

(c) Products are threo dl pairs (d) Products are erythro dl pairs

Comprehension-IV

The most problematic part in stoichiometry is to establish the molar ratio of reactant and the product. In order to solve,

these problems, n-factor, is an important tool, which is defined according to the kinds of reaction. In order to calculate

n-factor, first of all we categories the reactions in two parts : (i) acid-base reaction and (ii) redox reaction.

For acid and base reactions, it is the number of moles of H+ and OH

– replaced per mole of acid and base respectively.

For salt, n-factor is nothing but it is no. of moles of cation charge replaced per mole of the salt. So, n-factor of acids, bases

and salt depend upon the nature of the reactions. For redox reactions, n-factor is nothing but it is the number o moles of

electrons lost or gained in a given number of moles of active element in 1-moles of reducing agent or oxidizing agent

respectively. For redox reaction, n-factor may be a fractional quantity, but the no. of electron will not be a fractional quan-

tity. Thereby the n-factor of a redox reaction is not the number of electron lost or gained. The molar ratio of the reacting

species, a problematic part, is determine by the help of n-factor.

The reciprocal of n-factor’s ratio of the reacting species is the molar ratio of the reacting species, and similarly for the

product. Through the n-factor, we calculate equivalent weight as under:

Equivalent wt. =Molecular wt.

-factor n

15. For the reaction,

H 3PO4 + NaOH æÆ NaH 2PO4 + H 2O, What will be the equivalent wt. of H 3PO4 (H – 1, P – 31, O – 16)?

(a) 96 (b) 49 (c) 32.66 (d) 98

16. For the reaction

Zn + K 4[Fe(CN)6] æÆ K 2Zn3[Fe(CN)6]2, what will be the equivalent weight of K 4[Fe(CN)6] if the molecular wt.of K 4[Fe(CN)6] is M ?

(a)

M

2 (b) M (c)

M

3 (d)

M

4

Section C: Match the Matrix


Column-I Column-II

(p) (a) Only one type of product

(q) (b) CH3COCOCH3

(r) (c) Product gives Tollen’s test

(s) CH3 –C∫C–CH3 (d) Product gives iodoform test


20/39


Column I Column II

(p) Pt|H2(1 atm)|H+(1M) (a) Wrong representation

(q) Pt|Cl – |Cl2 (b) Concentration cell

(r) Pt|H2(1 atm)|H+(C1)||H

+(C2)|H2(1 atm)|Pt (c) Standard hydrogen electrode

(s) Pt|H2( p1 atm)|H+(1M)||H+(1M)|H2( p2 atm)|Pt (d) E ° = 0


Column (I) Column (II)

Molecule Property

(p) (a) Chiral centers containing compound

(q) (b) Presence of stereocenter

(r) (c) Optically active compound

(s) (d) Compound containing plane of symmetry


Substituent in benzoic acid p a

(p) (a) 3.91

(q) (b) 3.27

(r) (c) 2.94

(s) (d) 4.94


21/39

Solutions of Mock Test 1

PAPER 1Section A

1. Ans (a): At the steady state, the rate of destruction of O 3 must be equal to the rate of its formation, 7.2 ¥ 10−15 mol/L/h

2O 3 æÆ 3O 2

Rate of destruction of OO

O33

32=

-=

DD[ ]

[ ]t

k

k =-

¥DD[ ]

[ ]

O

O3

3 2

1

t =

7 2 10

3600

1

2 0 10

15

8 2

.

( . )

¥¥

¥

-

-mol/L/h

s/h mol/L = 5 ¥ 10 −3 L/mol/s

2. Ans (b): 2KI HgI K [HgI ]0.4

20.2

2 40.2

+ æÆ

Initial number of moles of total ions = 0.8 Total number of ions after addition of HgI 2 = 0.6 [2K +[HgI4]2−]

As the final number of mole is less, that is why the magnitude of colligative property will be less. Hence, depression

in freezing point will be less.

\ T 2 > T 1 3. Ans (b): d x2 – y2, d z 2

4. Ans (a): Cu2+

is the second group radical, which gets precipitated first due to having lower solubility product

[CuS = K sp = 1 ¥ 10−44].

5. Ans (d):

6. Ans (d):

7. Ans (b):


22/39

8. Ans (c): pI (isoelectric point) = 5.9

pI pK pK

2=

+a a1 2

5.92.4 pK

2= + a 2

fi pK a2 = 9.4 9. Ans (c):

P (H2) = (1400 tor) (0.685) =959 torr

760 = 1.26 atm

According to Henry’s law,

V

V

P

P

2

1

2

1

=

orV 2

18 mL

1.26 atm

1.00 atm=

or V 2 = 23 mL

10. Ans (b):

Same path is followed in the presence of Al2O3, 350°C, but, if pyridine is taken, then following rearrangement

does not take place, because instead of double bond (Q-bond) it is the pyridine which will combine with Lewis acid

Al2O3.

Section B

11. Ans (a): Y2, the probability of finding 1s electron, is maximum near nucleus. On moving farther to nucleus closeto 0.53 Å (Bohr’s radius), Y2 decreases. 4 Qr 2dr , volume element, on moving farther to nucleus increases.

The product Y2.4Qr 2dr on moving farther to nucleus increases till it reaches at Bohr’s radius and then starts todecrease.

12. Ans (a, c, d): The solution consists of [M(NH 3) x]+ and [ e(NH3) y]

−.

It imparts blue colour not due to polarization but due to the excitation of free electrons to higher energy levels by

the absorption of light in visible region. On the addition of iron oxide, solution decomposes:


23/39

2Na + 2NH 2NaNH H3Fe O2 3 æ Æ ææ + ≠2 2

13. Ans (a, c):

R L Nu Nu R LSubstrate

˜

Nucleophile Transition state æ + æÆ - - - - - -

- -

R L + Nu Nu R L+- æÆ - - - - - -- +

Q

In reactants, charge is more concentrated. Polar solvent will stabilize more in comparison to transition state. Hence,

rate decreases. In (c), more stabilization causes large decrease in rate.

14. Ans (b, d):

15. Ans (c, d):

HPO H + PO42 +

43- -

Unit positive charge experiences attraction severly with trinegative PO43- . Hence, equilibrium must be in the favour

of association and higher concentration of HPO42-

and thus lowest K a3.

K K a a2 3> H PO2 4fi - is more acidic than HPO , H PO and HPO4

22 4 4

2- - - . Both are amphiprotic anions.

Section C

16. Ans (4)

17. Ans (7)

18. Ans (3)

19. Ans (3): Because magnetic moment = n n( )+ 2 BM (where n is unpaired electron)

15 2 3= + fi =n n n( ) 20. Ans (3):

CH CH

Br

CH

Br

CH C

H

CH

Br

CH C C3 NaNH

3

NaNH

32 2 æ | æ | æ Æ æææ æ

|== | æ Æ æææ æ ∫∫2 æ æ æ Æ æææ æ ∫∫ +

- +H CH C C: Na NH NaNH

3 32

Terminal alkynes are acidic in nature and as their formation will start, they will react with NaNH 2 according to

the above reaction. Hence, NaNH2 will also be consumed in the acid base reaction of terminal alkynes. Therefore,

complete conversion will require 3 moles of NaNH2.


24/39

PAPER 2

Section A

1. Ans (a, c)

2. Ans (a, c)

3. Ans (b, c, d)

4. Ans (a, c):

2Cu+1

æÆ æ Cu + Cu +2

2Cu+1

+ 2e æÆ æ 2Cu Cu – 2e æÆ æ Cu +2

2Cu Cu +Cu+2+ æÆ æ 1

\ E ° =2 0 521 2 0 337

20 184

¥ +=

. ( . ).

5. Ans (b, c): Aqueous NaHCO 3 evolves CO 2 from benzoic acid and not from phenol. Similarly, neutral FeCl3 givesviolet colouration with phenol and not with benzoic acid.

6. Ans (a, c)

7. Ans (b, c)

8. Ans (a, c)

Section B

9. Ans (c)

10. Ans (c)

11. Ans (c): At half equivalence point,

Half of meq of CH 3COOH = meq of NaOH

12

10010 40

¥ Ê Ë Á ˆ ̄ ˜ = ¥ml ml M V M

where V is the volume of NaOH consumed.

fi V = 200 ml Therefore, V total of solution X will be = 100 + 200 = 300 ml

12. Ans (a): Solution X is a buffer solution and at half equivalence point.

[Acid] = [Salt]

\ pH = pK a = – log 10 –5 = 5

13. Ans (a)

14. Ans (c)

15. Ans (a)

16. Ans (b)

Section C

17. (p Æ b); (q Æ a); (r Æ d); (s Æ c)

(p) V Z

nµ

V

V

6

4

4

6

2

3= =


25/39

(q)1 1

2

12 2 M

Ê Ë Á

ˆ ¯ ˜

= -•

Ê Ë Á

ˆ ¯ ˜

B

R

1 11

12 2 M L RÊ Ë Á ˆ ¯ ˜ = - •Ê Ë Á ˆ ̄ ˜

1

1

1

4

4

1

/

/

M

M

M

M B

L

B

L

= =

(r) C-line (second line)

1 1

1

1

3

8

92 2 M

Ê Ë Á

ˆ ¯ ˜

= -Ê Ë Á

ˆ ¯ ˜ =

L

R R

H-line (third line)

1 1

3

1

6

6 3

6 32 2

2 2

2 2 M

Ê Ë Á

ˆ ¯ ˜

= -Ê Ë Á

ˆ ¯ ˜ =

-¥

Ê

Ë Áˆ

¯ ˜ P R

= 27324

336 12

R R R= =

\ 1

1

12

8 9

9

96

3

32

/

/

/

/

M

M

P

C

R

R R R

( )( )

= = =

( s) E Z

nµ -

2

2

\ ( )

( )

E E

E E

3 1

3 1

1

4

--

=+

H

He

18. (p Æ b); (q Æ d); (r Æ c); (s Æ a) 19. (p – b); (q – d); (r – a); (s – c)

20. (p Æ a); (q Æ a, b); (r Æ a, b); (s Æ c, d)


PAPER 1

Section A

1. Ans (c):

In this molecule, the plane of symmetry is present and the plane bisects two hydrogen atoms in the same plane

extending it to the third carbon containing one H (back) one side and one H-atom (front) on another side.

If two diffderent atoms or groups will be present there, then the molecule is optically active.

For example,


26/39

Molecule is:

(a) Optically active

(b) No centre of symmetry, optically active

(d) optically active

2. Ans (a):

There are two b-hydrogen atoms. As elimination proceeds through E2 (anti-elimination) —H and —Cl must be present at the dihedral angle of 180°.

Thus, b-hydrogen is abstracted and the elimination of HCl occurs. 3. Ans (c):

4. Ans (b): Formal charge on an atom in a molecule or ion = [Total no. of valence electrons in the free atoms] – [Total

no. of electrons in lone pair] –1/2[Total no. of shared electrons, bonding electrons]

Formal charge on:

1st O-atom = 6 41

24 0- - =( )

2nd O-atom = 6 21

26 1- - = +( )

3rd

O-atom = - - = -6 61

22 1( )

5. Ans (a): PV = nRT =m

M RT

fi P m

V

RT

M

dRT

M = =

d PM

RT =

At same temperature, as densities are equal for both, therefore,

PM P M 1 1 2 2Gaseous oxide Nitrogen

=

2 ¥ M 1 = 28 ¥ 5 M 1 = 70


27/39

6. Ans (a):

CH COOH CH COO + H3 3+ X VX W XX

-

CH COO Na CH COO Na3+

3+- - æÆ æ

H O H + OH2+ X VX W XX

-

Following eq. (i),

K a =+ -[H ][CH COO ]

[CH COOH]

3

3

[H ][CH COOH]

[CH COO ]

+-= K a

3

3

Hydrogen electrode: H H+ -+ æÆe1

22

For 1st case For 2nd case,

E 1 0 0591

1= - +. log [ ]H

= 0.059 log [H+]

= 0.059 log K y

xa E K

x

ya2 0 059= . log

E E K y

x K

x

ya a1 2 0 059+ = +

È

ÎÍ

˘

˚˙. log log = 0.059 log K a

2 = 0.118 log K a

or - = - +

log.

K E E

a1 2

0 118

pK E E

a = - +1 20 118.

7. Ans (c): Sulphide and sulphite ions both react with dil. HCl on warming their aqueous solutions giving H2S and

SO2 gases, respectively, but no yellow ppt. is formed. Sulphate ion does not react. The only ion left is thiosulphate

which gives SO2 gas with a pungent odour, and pale yellow ppt. of colloidal sulphur is formed.

S O + 2H SO + S + H O2 32 +

2Gaspungent odour Colloidal sulphur

2- æ Æ æ D

8. Ans (b): Chloroplatinic acid is H 2[PtCl6], which is dibasic. 9. Ans (a): Borax Na 2B4O7.10H2O

fi {Na 2[B4O5(OH)4].8H2O}


28/39

10. Ans (d): E-2-pentene

Section B

11. Ans (a): According to Le-Chatelier’s principle, the external change (increase in pressure) must be neutralized bythe system of equilibrium.

Increase in pressure increases the number of moles per unit volume. Resultantly, equilibrium will shift in the direction

which produces lesser number of moles. Hence, favours forward direction increasing concentration of CH 3OH.

12. Ans (a): C 2H5SO2Cl Hinsberg reagent

R NH C H SO Cl R NSO C H2amine

6 5 2OH

2 2 6 52∞

+ æ Æ ææ Q

æ

Insoluble in H+ due to stable lone pair of element at nitrogen

insoluble in OH – due to absence of H

R N C H SO Cl2amine

6 5 2OH

3∞+ æ Æ ææ

- No reaction takes place. That is why it is R 3 N

which is insoluble in base and soluble in H+

13. Ans (c, d):


29/39

14. Ans (c):

(a)

(b) R CH OH R CH Br R CH CN R CH CH N2PBr

2 NaCN

2

H , Ni

2 23 2 æ æ æ æ æ æ æ æ Æ ææ æ Æ ææ æ Æ ææ HH2

(one extra C-atom)

(c) R CH OH R CHO R CH== NH2Collins reagent NH H , Ni3 2 æ æ æ æ æ Æ æææææ æ Æ ææ æ Æ ææ R R CH NH2 2 æ æ

15. Ans (b, d):

H (g) 2H (aq.)2Standard hydrogen electrode

æÆ ++ -2e

1

2Hg Cl (s) + Hg(l) + Cl (aq.) Calomel electrode2 2 e

- æÆ Q

Section C

16. Ans (1): 2Al + Fe 2O3 æÆ Al 2O3 + 2Fe; D H = ?

Given, 2Al +3

2O Al O ;2 2 3 æÆ æ D H = –399 kcal (1)

and 2Fe +3

2O Fe O ;

2 2 3

æÆ æ D H = –199 kcal (2)

Subtracting Eq. (2) from eq. (1),

2Al + Fe O 2Fe + Al O ; H = 200 kcal2 3Fuel mixture

2 3 æÆ -D

Mol. wt. of fuel mixture = 2 ¥ 27 + 2 ¥ 56 + 48 = 214 g Therefore, 214 g mixture produce = 200 kcal heat

Therefore, 1 g mixture produces =200

2140 9346 1 1= ª -. kcal g


30/39

17. Ans (3): C V =DD

U

T V

Ê Ë Á

ˆ ¯ ˜

\ ∆U = C V

¥ ∆T

Therefore, C V =3

2 R for monoatomic gas and ∆T = 1

\ ∆U =3

2

3

22 3 R = ¥ = cal

18. Ans (2): Meq. of oxalic acid in 28 mL = Meq. of NaOH = 0.08 ¥ 35

Therefore, meq. of oxalic acid in 500 mL =500

2835 0 08 50¥ ¥ =.

\ 3 150

21000 50 126

.

/mm¥ = fi =

\ H 2C2O4 ¥ xH2O = 90 + 18 x = 126

or x = 2 19. Ans (1): R NC R NH + HCOOH

H /H O

2

+2 æ æ æ Æ æææ

20. Ans (6):

PAPER 2

Section A

1. Ans (a, b, d)

2. Ans (a, b, c, d)

3. Ans (a, b):

P4 + O 2 æÆ P4O6 + P4O10 (a) P4 + 3O 2 æÆ P4O6 1 4 0

0 1 1

P4O6 + 2O 2 æÆ P4O10 1 1

0.5 0 0.5

\ Remaining (0.5 mole P4O6, 0.5 mole P4O10)


31/39

(b) P4 + 3O 2 æÆ P4O6 1 3

0 0 1

\ Remaining (1 mole P4O6, 0 mole P4O10) (c) P4 + 3O 2 æÆ P4O6 0.5 2.5

0 1.0 0.5

\ Remaining (0.5 mole P4O6, 1 mole O2) (d) P4 + 3O 2 æÆ P4O6 1 5

1

3 0

5

3

\ Remaining (5/3) mole P4O6, 1/3 mole P4) 4. Ans (a, b, c, d)

5. Ans (a, b, c): Orbital bearing lower value of n will be more closer to the nucleus and thus electron will experiencegreater attraction from nucleus and so its removal will be difficult not easier.

6. Ans (a, b, d):

( ) ( )V V rms N rms O2 = 2

3 3 RT

M

RT

M

T

M

T

M

N

N

O

O

N

N

O

O

2

2

2

2

2

2

2

2

= =

Then V av and V mps is also same.

d P M

TRd

P M

RT P P d d N

N N

N

O

O N

O

N O N O2

2 2

2

2

2 2

2

2 2 2 2if then= = = =; ,

7. Ans (a, c, d)

8. Ans (a, c, d): All the three statements are correct.

Section B

9. Ans (b):

CH CH CH NH CH CH CH N CH CH CH3 2 2 2HNO

3 2 2 2 3 2 2

1,2-hydride tr 2 æ Æ ææ æÆ æ ≈ ≈ aansfer stable(2° > 1°)

3 3(2° carbocation)

CH CHCH æ Æ ææææææ ≈

10. Ans (c):

11. Ans (d):

2 3 28 3 33

2[ ] [ ]/Fe Fe+ + æÆ æ + e

\ EFe O2 3 = M

2 3/ 12. Ans (c):

6 e + (N0)2 æÆ 2(N

3– )

\ E N2 = 28 6/ ; E NH3 = 17 3/ 13. Ans (a):

M = 970.6 Å = 970.6 × 10 –10 m

E =hc

M =

¥ ¥ ¥¥

= ¥-

--6 63 10 3 0 10

970 6 102 049 10

34 8

10

18. .

.. J =

2 049 10

1 6 1012 81

18

19

.

..

¥¥

=-

- eV


32/39

Energy of an electron in nth orbit of hydrogen atom is E =-13 6

2

.

neV

In ground state, n = 1

\ E 1 = –13.6 eV Total energy of electron after absorbing 12.81 eV energy corresponding to given wavelength

E 2 = –13.6 + 12.81 = –0.79 eV

The value of n corresponds to 4.

n2 =--

=13 6

0 7917 21

.

..

This means that the electron is excited to n = 4 state.

Total number of lines obtained =n n( ) ( )-

= ¥ -

=1

2

4 4 1

26

14. Ans (b): Out of six lines, only two are in the visible region of the spectrum ( n = 4 to n = 2 and n = 3 to n = 2).

15. Ans (c)

16. Ans (b)

Section C

17. (p Æ d), (q Æ c), (r Æ b), (s Æ a) Sol. % Mole

% of ( a – 1) =a a

a a

- ++ -

¥ = ¥ =( )

( )( )%

4

4 1100

4

5100 80

% of a =( )

( )%

2 5

5100

3

4100 75

a a

a a

a

a

--

¥ = ¥ =

% of ( a + 1) =( ) ( )

( ) ( )%

a a

a a

+ - ++ - +

¥ = ¥ =2 3

3 1100

1

2100 50

% of ( a – 1) =a a

a a

- ++ - -

¥ = ¥ =( )

( ) ( ). %

2

2 1100

2

3100 66 67

% Mass

(p)80 1

2

80 80¥ -+

= -( )a

a

a

a (q)

752

752

¥ =a

a

a

a

(r)50 1

2

5 50

2

¥ ++

= +

+( )a

a

a

a (z)

66 67 1 66 67 66 67. ( ) . .¥ -=

-aa

a

a

18. (p – b); (q – a); (r – d); (s – c)

19. (p Æ a, c); (q Æ b, d); (r Æ a, c); (s Æ b, d) 20. (p Æ a, c); (q Æ c); (r Æ b, d); (s Æ a, b, c, d)


33/39

(p) Cl F , BrF , IF5 4+

6- ; all have same oxidation state (+5).

All have one lone pair of electrons each; but different shapes; N π 0, so polar. (q) ClF , BrF , ICl ;3 2

+4- all have same oxidation state (+3).

(r) XeF2 = +2; ICI – 2 = +1; I

– 3 = +1

Al have three lone pairs each and same shape but different oxidation states. In all, µ = 0. So non-polar. (s) ClOF , ClF , IO F3 4

+2 2

-; all have same oxidation number (+5).

In all N π 0, so all polar


PAPER 1

Section A

1. Ans (c): Entropy, which is the magnitude of randomness is maximum for H 2O in gaseous phase and minimum for

H2O in solid state. As H2O density is found maximum at 4°C, that is why randomness will be less there.


34/39

2. Ans (a):

HCOOH + Ag O CO + H O + 2Ag2Tollens reagent

warm2 2

Silver mirror

æ Æ ææ

HCOOH + 2CuO CO + H O + Cu OFehling solution

2 2 2Red ppt.

D æ Æ æ

HCOOH + 2HgCl CO + 2HCl + Hg Cl2 2 2 æÆ æ -

2

2KMnO 4 + 3H 2SO4 + 5HCOOH → K 2SO4 + 2MnSO4 + 8H2O + 5CO2 3. Ans (c):

4. Ans (b):

5. Ans (b):

6. Ans (a): CCl 4 is non-polar compound. That is why it does not undergo hydrolysis. For Si and P, hydrolysis is com-

ing up favourable as hydroxides are more stable due to pQ-dQ back bonding.

Hence, order is: CCl 4 < MgCl 2 < AlCl 3 < SiCl4 < PCl5


35/39

7. Ans (d): 6CaOCl Ca(ClO ) + 5CaCl2Auto

oxidation 3 2 2 æ Æ æææ

8. Ans (b): AgNO + KI AgI + KNO3excess

3 æÆ æ

Out of excess KI, I \ which is the common constituent to solution gets adsorbed over AgI and imparts

negative charge.

9. Ans (d): The correct name is: Chloro bis-(ethylenediamine) nitrite cobalt (III)

10. Ans (a):1

3 rate of disappearance of BrO − = rate of appearance of BrO 3

−

1

3

1

30 056 0 019- = ¥ =

-DD

[ ]. .

BrO

t

Hence,D

D[ ]

. .BrO3 1

30 056 0 019

-

= ¥ =t

Section B

11. Ans (b, c, d):

BaI + 2H O Ba(OH) + 2HI2 2 2Str ong base Str ong acid

æÆ

[(CH ) NH]Cl + H O (CH ) NHOH HCl3 3 2 3 3Weak base Str ongacid

æÆ ++

Cr(NO ) + 3H O Cr(OH) + 3HNO3 3 2 3Weak base

3Str ongacid

æÆ

NH F + H O NH OH + HF } If > then only4 2 4Weak base Weak acid

æÆ K K a b,

12. Ans (b, c): As DG° for N2O4 → 2NO2 is +5.0 kJ mol−1 (positive), that is why direct conversion is non-spontane-

ous.

But at equilibrium point B, N2

O4

is dissociated upto equilibrium point (dissociation of that extent) spontaneous as

DG° = −5.40 + (−0.84) = −6.24 kJ Hence, dynamic conversion N 2O4 into equilibrium mixture and 2NO 2 into equilibrium mixture both are spontaneous.

But, as for the reverse process, DG° is −6.24 kJ (more than forward), that is why more favourable. 13. Ans (b, d): ClO 2 is a powerful oxidizing agent and also a strong chlorinating agent. Its bleaching power is almost

30 times stronger than Cl2.

In alkaline, solution undergoes disproportionation

2ClO 2 + 2NaOH → NaClO + NaClO3 + H 2O

unpaired electron that is why paramagnetic.

14. Ans (a, d): Co-ordination isomerism occurs when both cation and anion are complex. It is caused due to the inter-

change of ligands between two complex ions of the same complex.

15. Ans (a, b, c):

2CH I + 2CH CH I Products

CH

3 3 Na- - æ Æ æ

|

3

2Na 2Na + 2+ æÆ æ -e

CH I + 2 CH + I3 3e- - - æÆ

CH I + CH CH CH + I3 3 3 3- - æÆ -


36/39

2Na + 2I − → 2NaI

CH + CH CH I CH CH CH + I

CH CH

3 3 3 3- -- - æÆ - -

| |

3 3

[CH 3 \ acts as nucleophile]

CH + CH CH I CH + CH =CH CH

CH

3 3 4 2 3- - - æÆ æ -

|

3

[CH 3 \

acts as base]

Isopropyl iodide can also accept electron.

CH CH I + 2 CH CH + I

CH CH

3 3- - æÆ -- - -e

| |

3 3

CH CH CH CH I CH CH CH CH

CH CH CH CH

3 3 3 3- + - - æÆ - - --

| | | |

3 3 3 3

(nucleophilic action)

+CH CH +CH =CH

| |CH CH

3 2- 2

3 3 (basic action)

The reaction is called Wurtz reaction. It is successful for only 1° and symmetrical molecule, otherwise gives a

mixture of products.

16. Ans (b, d):

CH CH+CH COOH CH =CH O C

O||

CH3Hg

2 3Vinyl acetate

2+

∫ æ Æ ææ - - =

O

CH =CH O C CH +CH COOH CH CH( OCOCH )2 3 3Hg

3 3 2ethyliden

2+||

- - - æ Æ ææ - -ee diacetate

Section C

17. Ans (5): Na 2O2 , KO2, CaO, BaO and RbO2 are basic oxides.

BeO is amphoteric and CO 2 is an acidic oxide.

18. Ans (3)

19. Ans (5): 3Cl 2 + 6NaOH æÆ 5NaCl + Na ClO5

3

++ 3H2O

20. Ans (2):

r = K [H+]n and pH = 2


37/39

\ [H +] = 10 – 2

r 0 = K [10 – 2]n (1)

At pH = 1, [H+] = 10

– 1

\ r 1 = K [10 – 1]n (2)

r

r

n1

0

100 10= = [ ] fi n = 2

21. Ans (9):

t K

A

A=

2 303 0. log[ ]

[ ]

=2 303

7 7 10

1 5

0 752.

.log

.

.s s

=2 303

7 7 102

2

.

.log

¥ ¥- = 9

PAPER 2

Section A

1. Ans (a, b, c): D H = DU + P .DV + V .D P + D P .DV is the correct relation.

2. Ans (c, d): Le Chatelier’s principle is not quantitative. If both stress would cause the same direction of shift, theshift is determinable. If the two stress would cause shifts in opposite direction, no deduction is possible.

3. Ans (a, d)

4. Ans (a, b, d)

5. Ans (b, c):

I, II – Enantiomers

II, III – Diastereomers

6. Ans (b, c):

(a) is incorrect because CH 3 – F is less reactive than CH 3 – I because F− is poor leaving group.

(b) is correct because DMF is


38/39

(c) is correct because I − is better leaving group than F −

(d) is incorrect answer.

7. Ans (a, b, c):

H S1(1 )

H0 + HS02

+

C C C -

-

K C

a1

2

1=

-

( ) ( &C = 0.1 M)

4.0 ¥ 10 –3 =0 1

1

2.

( )

¥-

B = 0.18 = 18%

Now, [H +] = C .B = 0.1 ¥ 0.18 = 0.018 M pH = – log [H +] = – log 1.8 ¥ 10 –2 = 1.7447 8. Ans (d):

Energy of an electron of the atom (atomic no. Z) in the nth orbit is

E R Z

nn =

- H .2

2

For hydrogen atom, Z = 1

\ E n =-

= - R

n

RH H.1

9

2

2 (given)

Thus n = 3

When n = 3, l can be

l = 0 and m = 0 (one 3 p orbital)

l = 1 and m = – 1, 0, + 1 (three 3 p orbitals)

l = 2 and m = – 2, – 1, 0, + 1, + 2 (five 3d orbitals)

There are 1 + 3 + 5 = 9 states and therefore degeneracy is nine.

Section B

9. Ans (b):

PV = nRT fi 2 ¥ 8 = 2 ¥ 0.080 ¥ T T = 100 K

W rev = –2.303 ¥ n ¥ R ¥ T log P

P 1

2

= – 2.303 ¥ 2 ¥ 0.08 ¥ 100 ¥ log1

10

Ê Ë Á

ˆ ¯ ˜

= 36.848 bar-L

10. Ans (c):

wirr = – P ext ( V 2 – V 1)

= – 20 nRT

P

nRT

P 2 1-

Ê

Ë Áˆ

¯ ˜

= 144 bar-L

11. Ans (c)

12. Ans (a)

13. Ans (c): I 2 is more polarisable.


39/39

14. Ans (d):

15. Ans (d):

& n-factor of H3PO4 is 1

\ Eq. wt. =mol wt. .

.1

16. Ans (c):

& n-factor of K 4 [Fe(CN) 6] = 3.

Section C

17. (p Æ a, c, d); (q Æ b, c, d); (r Æ a, c); (s Æ a , b, d)

18. (p Æ c, d); (q Æ a); (r Æ b, d); (s Æ b, d) 19. (p Æ b, c); (q Æ b, d); (r Æ a, b, c); (s Æ b, d) 20. (p Æ b); (q Æ c); (r Æ d); (s Æ a) Ortho effect and polar nature of the group determine the acidic strength.

jee chemistry cl

Documents