KVPY-SX 2019 (CHEMISTRY) - Byju's

KVPY-SX_2019 (CHEMISTRY)

KVPY-SX_2019 (CHEMISTRY) Page | 1

Copyright © Think and Learn Pvt. Ltd.

PART-I

1. The major product of the following reaction aqueous NaOH⎯⎯⎯⎯⎯→ are:

(a) Br3C–OH and

(b) and CHBr3

(c) and NaBr (d) PhH and CBr3CO2Na

2. Among the following,

The compounds which can undergo an SN1 reaction in an aqueous solution, are

(a) I and IV only (b) II and IV only

(c) II and III only (d) II, III and IV only

3. The major product of the following reaction

3

Excess DIBAL-H

Toluene, –78°C then H O+⎯⎯⎯⎯⎯⎯⎯→ is

(a) (b)

(c) (d)

HO

O

H

O

EtO

O

H

O

H

O

CN H

O

H

O

EtO

O

CN

Br

I

Br

II

Br

III MeO Br

IV

Ph CHBr2

O

Ph ONa

O

Ph H

O

Ph CBr3

O




4. Permanent hardness of water can be removed by

(a) heating

(b) treating with sodium acetate (CH3CO2Na)

(c) treating with Ca(HCO3)2

(d) treatment with sodium hexametaphosphate (Na6P6O18)

5. Alkali metals (M) dissolve in liquid NH3 to give

(a) MNH2 (b) MH

(c) [M(NH3)x]+ + [e(NH3)y]– (d) M3N

6. The absolute configurations of the following compounds

Respectively are

(a) R and R (b) S and S (c) R and S (d) S and R

7. The diamagnetic species among the following is

(a) 2O+ (b) –2O (c) O2 (d) 2–

2O

8. Among the following transformations, the hybridization of the central atom remains

unchanged in

(a) CO2⎯→ HCOOH (b) BF3⎯→ –4BF

(c) NH3⎯→ 4NH+ (d) PCl3⎯→ PCl5

9. For an octahedral complex MX4Y2 (M = a transition metal, X and Y are monodentate

achiral ligands), the correct statement, among the following, is

(a) MX4Y2 has 2 geometrical isomers one of which is chiral

(b) MX4Y2 has 2 geometrical isomers both of which are achiral

(c) MX4Y2 has 4 geometrical isomers all of which are achiral

(d) MX4Y2 has 4 geometrical isomers two of which are chiral

CH2SH

CH2OH

H

H3C CH2SH

HO

H3C

H




10. The values of the Henry's law constant of Ar, CO2, CH4 and O2 in water at 25°C are

40.30, 1.67, 0.41 and 34.86 kbar, respectively. The order of their solubility in water at

the same temperature and pressure is

(a) Ar > O2> CO2> CH4 (b) CH4> CO2> Ar > O2

(c) CH4> CO2> O2> Ar (d) Ar > CH4> O2> CO2

11. Thermal decomposition of N2O5 occurs as per the equation below

2N2O5⎯→ 4NO2 + O2

The correct statement is

(a) O2 production rate is four times the NO2 production rate

(b) O2 production rate is the same as the rate of disappearance of N2O5

(c) rate of disappearance of N2O5 is one-fourth of NO2 production rate

(d) rate of disappearance of N2O5 is twice the O2 production rate

12. For a 1st order chemical reaction.

(a) the product formation rate is independent of reactant concentration

(b) the time taken for the completion of half of the reaction (t1/2) is 69.3% of the rate

constant (k)

(c) the dimension of Arrhenius pre-exponential factor is reciprocal of time

(d) the concentration vs time plot for the reactant should be linear with a negative

slope

13. The boiling point of 0.001 M aqueous solutions of NaCl, Na2SO4, K3PO4 and CH3COOH

should follows the order

(a) CH3COOH < NaCl < Na2SO4< K3PO4

(b) NaCl < Na2SO4< K3PO4< CH3COOH

(c) CH3COOH < K3PO4< Na2SO4< NaCl

(d) CH3COOH < K3PO4< NaCl < Na2SO4

14. An allotrope of carbon which exhibits only two types of C–C bond distance of 143.5

pm and 138.3 pm, is

(a) charcoal (b) graphite (c) diamond (d) fullerene




15. Nylon-2-nylon-6 is a co-polymer of 6-aminohexanoic acid and

(a) glycine (b) valine (c) alanine (d) leucine

16. A solid is hard and brittle. It is an insulator is solid state but conducts electricity in

molten state. The solid is a

(a) molecular solid (b) ionic solid (c) metallic solid (d) covalent solid

17. The curve that best describes the adsorption of a gas (x g) on 1.0 g of a solid

substrate as a function of pressure (p) at a fixed temperature

is:

(a) 1 (b) 2 (c) 3 (d) 4

18. The octahedral complex CoSO4Cl.5NH3 exists in two isomeric forms X and Y. Isomer X

reacts with AgNO3 to give a white precipitate, but does not react with BaCl2, Isomer Y

gives white precipitate with BaCl2 but does react with AgNO3.

Isomers X and Y are

(a) Ionization isomers (b) Linkage isomers

(c) Coordination isomers (d) Solvate isomers

19. The correct order of basicity of the following amines

is:

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

20. Electrolysis of a concentrated aqueous solution of NaCl results in

(a) increase in pH of the solution (b) decrease in pH of the solution

(c) O2 liberation at the cathode (d) H2 liberation at the anode

I II

NH2

III H3C

IV

NH

2

NH

2

NH2

O2N

x

p

1

2 3

4




PART-II

21. The product of which of the following reaction forms a reddish brown precipitate

when subjected to Fehling's test?

(a)3

CO, HCl

anhy. AlCl , CuCl⎯⎯⎯⎯⎯→ (b) + (CH3CH2)2Cd ⎯→

(c) 5

2 4

1. PCl

2. H , Pd-BaSO⎯⎯⎯⎯→ (d) 3

2

1. O

2. Zn/H O⎯⎯⎯→

22. The major products X, Y and Z in the following sequence of transformations

are:

(a)

(b)

(c)

(d)

H

O X = Y =

NO2

Z = N

H

O

N

NO2

NH2

NH2

O

X = Y = Z = NH2

O

O2N

NH2

O

O2N OH

H

O X = Y =

O2N

NH2 Z =

N H

O

N

O2N

NH2

O

X = NH2

O

Y = NO2

NH2

O

Z = NO2

HO

NH2 O

O

O

X conc. HNO3

conc. H2SO4

15°C

Y aq. NaOH

Z

CO2H

O

Cl




23. In the following reaction, P gives two products Q and R, each in 40% yield.

If the reaction is carried out with 420 mg of P, the reaction yields 108.8 mg of Q.

The amount of R produced in the reaction is closest to

(a) 97.6 mg (b) 108.8 mg (c) 84.8 mg (d) 121.6 mg

24. Solubility products of CuI and Ag2CrO4 have almost the same value (~4 × 10–12).

The ratio of solubilities of the two salts (CuI: Ag2CrO4) is closest to

(a) 0.01 (b) 0.02 (c) 0.03 (d) 0.10

25. Given that the molar combustion enthalpy of benzene, cyclohexane, and hydrogen

are x, y, and z, respectively, the molar enthalpy of hydrogenation of benzene to

cyclohexane is

(a) x – y + z (b) x – y + 3z (c) y – x + z (d) y – x + 3z

26. Among the following, the pair of paramagnetic complexes is

(a) K3[Fe(CN)6] and K3[CoF6] (b) K3[Fe(CN)6] and [Co(NH3)6]Cl3

(c) K4[Fe(CN)6] and K3[CoF6] (d) K4[Fe(CN)6] and [Co(NH3)6]Cl3

OMe 1. O3 Q

40%

2. Zn, H2O + R

40% P (MW = 210)




27. The major products X and Y in the following sequence of transformations

are:

(a)

(b)

(c)

(d)

28. 3.0 g of oxalic acid [(CO2H)2.2H2O] is dissolved in a solvent to prepare a 250 mL

solutions. The density of the solution is 1.9 g/mL. The molality and normality of the

solution, respectively, are closest to

(a) 0.10 and 0.38 (b) 0.10 and 0.19 (c) 0.05 and 0.19 (d) 0.05 and 0.09

29. In a titration experiment, 10 mL of a FeCl2 solution consumed 25 mL of a standard

K2Cr2O7 solution to reach the equivalent point. The standard K2Cr2O7 solution is

prepared by dissolving 1.225 g of K2Cr2O7 in 250 mL water. The concentration of the

FeCl2 solution is closest to [Given: molecular weight of K2Cr2O7 = 294 g mol–1]

(a) 0.25 N (b) 0.50 N (c) 0.10 N (d) 0.04 N

30. Atoms of an element Z form hexagonal closed pack (hcp) lattice and atoms of element

X occupy all the tetrahedral voids. The formula of the compound is

(a) XZ (b) XZ2 (c) X2Z (d) X4Z3

OH X = Y =

CO2H

OH

SO3H X = Y =

OH

CO2H

OH

OH X = Y =

CO2H

SO3H X = Y =

HO SO3H HO2C

X 1. oleum

2. molten NaOH, 3. H3O+

1. NaOH

2. CO2 3. H3O+

Y




ANSWER KEY

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

6. (d) 7. (d) 8. (c) 9. (b) 10. (c)

11. (d) 12. (c) 13. (a) 14. (d) 15. (a)

16. (b) 17. (b) 18. (a) 19. (b) 20. (a)

21. (d) 22. (b) 23. (c) 24. (b) 25. (b)

26. (a) 27. (d) 28. (c) 29. (a) 30. (c)

SOLUTIONS

PART-I

1. (b)

O

Ph C Br Br

Br

NaOH

O

Ph C Br Br

Br

–

OH

O

Ph OH + – C–Br

Br

Br

O

Ph O + H – C–Br

Br

Br

– Na+

O

+ Ph O Na+ – H – C–Br

Br

Br

+

–

OH–

Nucleophile

Good leaving group

Proton transfer

Resonance stabilised

Option (b) is correct.




2. (c)

Condition for SN1 reaction is whether type of carbocation formed is stable or not.

(I)

⎯→ not stable carbocation due to –I effect of alkene.

(II) ⎯→stable carbocation +I effect of three methyl group.

(III)

MeO ⎯→stable due to resonance and delocalization of +ve charge

(IV)

⎯→according to Bredt’s rule, carbocation at bridge position is unstable.

So, option (c) is correct.

3. (a)

[DIBAL–H] → (i–Bu)2 Al–H → Di isobutyl aluminium hydride

RCN or ester 2

DIBAL–H

H O

⎯⎯⎯⎯⎯→ R–CHO

EtO

O DIBAL–H

Toulene, –78°C then H3O CN

Ester R–CN

H

O H

O

4. (d)

The permanent hardness of water can be removed by using calgon process. The

hardness of water causes by salts of Ca and Mg.

Na2[Na4(PO3)6] + 2Ca2+/Mg2+⎯→ Na2[Ca2/Mg2(PO3)6] + 4Na+

(soluble complex)

Sodium hexametaphosphate

(Calgon)




5. (c)

All alkali metals like lithium, sodium, potassium etc. dissolves in liquid ammonia to

give deep blue coloured solution. The blue colour is due to presence of solvated

electrons.

M + (x + y)NH3⎯→[M(NH3)x]+ + [e(NH3)y]–

(Ammoniated cation) (Ammoniated anion)

(Blue colour)

6. (d)

Draw the Fischer project i.e. 2–D representation and then assign priorities.

CH2SH H

H3C CH2OH

(I)

H HO

H3C CH2SH

(II)

According to Cahn Ingold Prelog (CIP) priority rules –

CH3

CH2SH

CH2OH

H (1)

(2)

(4)

(3)

Clockwise

CH3

H

CH2SH

HO (4)

(2)

(1)

(3)

Anticlockwise

So, it should be ‘R’ but here, It should be ‘S’ lower priority group is on

lower priority group is on horizontal line i.e. ‘R’

horizontal line. Therefore,

Configuration will be reversed.

i.e. correct configuration ‘S’

So, option (d), S, R is correct.




7. (d)

According to MOT –

(a) ( ) ( )( )+ = = = 2 z x y x y

– 2 *2 2 *2 2 2 2 *1 *01s 1s 2s 2s 2p 2p 2p 2p 2pO 15e

Number of unpaired e–s = 1 Paramagnetic

(b) ( ) ( )( )= = = 2 z x y x y

– – 2 *2 2 *2 2 2 2 *2 *11s 1s 2s 2s 2p 2p 2p 2p 2pO 17e

Number of unpaired e–s = 1

(c) ( ) ( )( )= = = z x y x y

– 2 *2 2 *2 2 2 2 *1 *12 1s 1s 2s 2s 2p 2p 2p 2p 2pO 16e

Number of unpaired e–s = 2 paramagnetic

(d) ( ) ( )( )= = = z x y x y

2– – 2 *2 2 *2 2 2 2 *2 *22 1s 1s 2s 2s 2p 2p 2p 2p 2pO 18e

Number of unpaired e–s = 0 diamagnetic

Option (d) is correct.

8. (c)

(a)

O=C=O H–C–O–H

O

sp(2B.P.) sp2(3B.P.)

(b)

F–B–F F–B–F

sp2 (3B.P.)

sp3(4B.P.)

F

F

– F

(c)

N

sp3(3 B.P. + 1 .p.) sp3(4 B.P.)

+

H H H

N

H H H

H

(d)

P

sp3(3 B.P. + 1 .p.) sp3d(5 B.P.)

Cl Cl Cl

P Cl

Cl Cl

Cl

Cl




9. (b)

MX4Y2⎯→ Octahedral complex

Note → If there is plane of symmetry (POS) or centre of symmetry (COS) in a

complex. Then it will be achiral and optically inactive. Geometrical isomers exist is cis

and trans forms.

M

X X

X X

Y

Y

POS & COS present M

Y X

Y X

trans (Achiral)

POS present

Cis (Achiral)

X

X

i.e. two geometrical isomers both of which are achiral.

So, option (b) is correct.

10. (c)

According to Henry’s law, the partial pressure of a gas in vapour phase (P) is

proportional to the mole fraction of the gas (X) in the solution.

P X

P = KHX

KH = Henry’s constant

P = Partial pressure of gas above liquid surface

so, according to Henry’s law solubility of gas in liquid partial pressure of gas above

liquid surface.

KHP solubility

KH Ar O2 CO2 CH4 (given)

40.30 34.86 1.67 0.41

Solubility order Ar < O2< CO2< CH4

11. (d)

Reaction: 2N2O5⎯→ 4NO2 + O2

Rate =

= =2 5 2 2d N O d NO d O1 1

–2 dt 4 dt dt

Rate of disappearance of N2O5 =

= =2 5 2 2d N O 2d O d NO1

–dt dt 2 dt

Rate of disappearance of N2O5= twice the O2 production rate

So, option (d) is correct.




12. (c)

For Ist order reaction –

t =

o

10

t

A2.303log

K A

And =1

2

0.693t

K

Arrhenius equation: a–E /RTK Ae=

[A = pre-exponential factor]

In a first order reaction, unit of rate constant is s–1. So pre-exponential factor are also

reciprocal seconds. Because exponential factor depends on frequency of collision.Its

related to collision theory and transition statetheory.

13. (a)

We know that, Tb = i ×Kb × m

molarity or molality = 0.001 m {at very diluted solution}

So, Tb i ;{molality is constant for all}

i Boiling point

For NaCl strong electrolyte i = 2 {Na+ + Cl–}

For Na2SO4 strong electrolyte i = 3 + + 2–42Na SO

For K3PO4 strong electrolyte i = 4 + + 3–43K PO

For CH3COOH weak acid i < 2 (calculated from i= 1+ (n-1)α , where α<1 )

So, order of Boiling Point: K3PO4> Na2SO4> NaCl > CH3COOH

So, option (a) is correct.

14. (d) Fullerene a soccer ball shaped molecule has 60 vertices with a carbon atom at each

vertex. It contains both single and double bond with C–C at a distance of 143.5 pm and 138.3 pm.

Where as in diamond C–C bond length 154 pm and in graphite 140 pm.




15. (a)

Nylon–2–Nylon–6 is a copolymer of glycine and amino caproic acid (6-amino

hexanoic acid)

nH2N–CH2–COOH + nNH2 –(CH2)5–COOH

Glycine Amino caproic acid

–nH2O

C–CH2–NH–C–(CH2)5–NH

O O

Nylon-2-Nylon-6

n

16. (b)

Ionic solid is hard and brittle eg. NaCl

It is an insulator in solid state but in molten state it conducts electricity as it has free

ions in molten state.

17. (b)

Given: mass = 1g

According to freundlich adsorption isotherm –

=1

nx

KPm

(n > 1)

=1

nx KP

=1

nx

Pk

=

nx

Pk

On comparing parabola equation yn = x. If n = 2

x

So, graph (b) is correct.




18. (a)

CoSO4Cl.5NH3

Isomers of CoSO4Cl.5NH3 are (I) [Co(NH3)5SO4]Cl gives white ppt with AgNO3 but not

with BaCl2 and (II) [Co(NH3)5Cl]SO4 gives white ppt with BaCl2 but not with AgSO4.

(I) [Co(NH3)5SO4]Cl + AgNO3⎯→ AgCl + [Co(NH3)5SO4]NO3

[white ppt]

(II) [Co(NH3)5Cl]SO4 + BaCl2⎯→ BaSO4 + [Co(NH3)5Cl]Cl2

[white ppt]

So, isomer (I) and (II) are ionization isomers.

19. (b)

(I) NH2

⎯→ lone pair of ‘N’ are not delocalised so, easily donate e–.

(II) NH2

⎯→No effect, lone pair of ‘N’ are in resonance. So, e– not donate

easily.

(III)

NH2

H3C ⎯→ +I effect of CH3 and resonance of lone pair of ‘N’. So,

easily donate e– as compared to (II).

(IV)

NH2

O2N ⎯→ –I effect of –NO2 and resonance of lone pair of ‘N’. So hard

to remove electrons.

So, order of basicity – I > III > II > IV

20. (a)

Electrolysis of aqueous NaCl leads to the formation of NaOH, which is a base. SO, its

pH will be increases.

NaCl + H2O ⎯⎯⎯⎯→Electrolysis NaOH + Cl2 + H2

(base)

H2 liberation will be at cathode due to (H+⎯→ H2) gain of electron at cathode.





PART-II

21. (d)

Fehling test: Fehling solution A + Fehling solution B

(Cu2+) (OH–)

Fehling reagent react with aldehyde which has –H such aldehyde form an enolate

and thus give a positive Fehling test and a reddish brown precipitate is obtained.

Reaction: R–CHO + 2Cu2+ + 5OH–⎯→RCOO– + Cu2O + 3H2O

(reddish brown)

Benzaldehyde gives tollen’s as well as schiff’s test but not gives fehling's solution

test because benzaldehyde does not contain 𝐴𝑙𝑝ℎ𝑎 hydrogen and cannot form

intermediate enolate to proceed for further and hence it does not react with fehling's

solution test however aliphatic aldehydes gives fehling's solution test.

(a)

CO, HCl

C–H

anhy. AlCl3, CuCl

O

(Gattermann Koch Rxn) (No –H)

do not give Fehling test

(b)

O –CdCl2 2

CH2–CH3 Cl

+ (CH3CH2)2Cd O

(c)

PCl5 C–Cl

O

C–OH

O

H2, Pd-BaSO4 CH2–OH

(d) 1. O3 CH3–C–CH2–CH2–CH2–CH2–C–H give Fehling test

O 2. Zn/H2O

O aldehyde

(–H)

So, option (d) product gives Fehling test.




22. (b)

(CH3CO)2O

H–N–C–CH3

O

Conc. HNO3

NH2

(x)

Conc. H2SO4

H–N–C–CH3

O

NO2 (Acetanilide) (>51%)

(y)

aq. NaOH

NH2

NO2 (z)

Pyridine (15°C)

(Aniline)


23. (c)

1. O3 OMe 2. Zn, H2O

OMe

CHO

+

OHC

(P) (40%) (40%)

mw = 210 Given: w = 420 mg

Moles = =w 420

Mw 210 = 2 m moles =

40 42

100 5 m moles =

40 42

100 5 m moles

Molecular weight for 108.8 mg = mass

mole =

108.8 mg

4m mole

5

= 136 g/mol

So, 136 g/mol is forC8H8O2

OMe OHC Q

Then CHO

R C7H6O mol. wt = 106 g mol–1

Mass of R = mole × m.wt. = 4

5 m mole × 106 g mol–1 = 84.4 mg





24. (b)

Given: ( ) ( )= = 2 4

–12sp spCuI Ag CrO

K K 4 10

CuI(aq) ⎯→ Cu+(aq) + I–(aq)

S1 S1

1spK = S1 × S1 4 × 10–12 = (S1)2

S1 = 2 × 10–6 mol/L

Ag2CrO4⎯→ 2Ag+ + 2–4CrO

2S2 S2

2spK = (2S2)2 (S2) 4 × 10–12 = 4(S2)3

(S2)3 = 10–12 S2 = 10–4 mol/L

= = = –2

–61

–42 4 2

Ssolubility of CuI 2 102 10

solubility of Ag CrO S 10

=1

2

S0.02

S


25. (b)

6 6Benzene

C H + 15

2O2⎯→ 6CO2 + 3H2O; H1 = x .....(1)

6 12Cyclohexane

C H + 9O2⎯→ 6CO2 + 6H2O; H2 = y .....(2)

H2 + 1

2O2⎯→ H2O; H3 = z .....(3)

C6H6 + 3H2⎯→ C6H12; H =? .....(4)

So, equation (1) + equation (3) × 3 – equation (2); to get equation (4)

C6H6 + 3H2⎯→ C6H12

x + 3z – y = x – y + 3z

So, molar enthalpy of hydrogenation of benzene to cyclo hexane = x – y + 3z

Option (b) is correct.




26. (a)

(a) K3[Fe(CN)6] CN– is a strong field ligand so, pairing possible

Fe3+

26Fe [Ar]3d64s2

Fe3+ [Ar]3d54s04p0

3d 4s 4p

Fe3+

K2[Fe(CN)6] xx

CN–

d2sp3

xx xx xx xx xx

CN– CN– CN– CN– CN–

Number of unpaired e–s = 1 paramagnetic

K3[CoF6] F– weak field ligand, so no pairing

Co3+

27Co [Ar]3d74s2

Co3+ [Ar]3d64s04p04d0

3d 4s 4p

Co3+

4d

K3[CoF6] xx xx xx xx xx xx

F– F– F– F– F– F–

sp3d2 Number of unpaired e–s = 4 paramagnetic


27. (d)

1. oleum 2. NaOH

3. H3O+

SO3H

1. NaOH

2. CO2 3. H3O+

OH OH COOH

(x) (y)

(salicylic acid) (Kolbe schmidt reaction)

Option (d) is correct.




28. (c)

Given: Mass of oxalic acid = 3.0 g

Mol. wt. of oxalic acid (COOH)2.2H2O = 126 g mol–1

Density = 1.9 g/mL and volume of solution = 250 mL

d = m

v m = d × v

Molality = ( )

mass of solute

m.mass of solute mass of solvent kg

( )

=

mmolality

m.m d v –3

3

126 10.9 250 10=

Molality = 0.05

We know that, Normality = Molarity × nfactor( )

molemolarity

volume L

=

nfactorof oxalic acid = 2 (Because it lose 2 H+)

( ) factornn

NV L

=

Normality = –3

3126 2 0.19

250 10 =


29. (a)

Reaction: FeCl2 + K2Cr2O7

+6 Fe3+ + Cr3+

nf = 6

So, by law of equivalence,

Number of eq. of FeCl2 = number of eq. of K2Cr2O7

N1V1 = N2V2

Given: volume of FeCl2 = 10 mL

Volume of K2Cr2O7 = 25 mL

Wt. of K2Cr2O7 = 1.225 g

Volume of solution = 250 mL

m. wt. of K2Cr2O7 = 294 g mol–1

N1 × 10 = M × nf × 25

=

1 –3

1.225/ 294N 10 25 6

250 10

( )mole

molarityvolume L

=

=

1 –3

1.225 25 6N

294 250 10 10

N1 = 0.25 N




30. (c)

Let, number of atom in HCP = a

We know, the number of tetrahedral voids will be twice as that of number of atoms

present in a unit cell.

So, no. of tetrahedral void = 2a

Z = a, X = 2a

Z : X

a : 2a

1 : 1

So, formula of compound = X2Z

Therefore, option (c) is correct.

KVPY-SX 2019 (CHEMISTRY) - Byju's

Documents

Transcript of KVPY-SX 2019 (CHEMISTRY) - Byju's