1

Time : 120 Minutes Max. Marks : 240

INSTRUCTIONS TO CANDIDATES :

(i) Question paper has 80 multiple choice questions. Each question has four alternatives, out of which

only one is correct. Choose the correct alternative and fill the appropriate bubble, as shown.

Q. No. 22 A C D

(ii) A correct answer carries 3 marks whereas 1 mark will be deducted for each wrong answer.

Answers & Solutionsforforforforfor

INDIAN ASSOCIATION OF PHYSICS TEACHERS

NATIONAL STANDARD EXAMINATION

IN CHEMISTRY (NSEC) 2019-20

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.: 011-47623456

DATE : 24/11/2019

Choose the correct answer :

1. Myoglobin, (Mb), an oxygen storage protein, contains

0.34% Fe by mass and in each molecule of myoglobin

one ion of Fe is present. Molar mass of Mb (g mol–1)

is (Molar mass of Fe = 55.845 g mol–1)

(A) 16407

(B) 164206

(C) 16425

(D) 164250

Answer (C)

Sol. Let molar mass of myoglobin (Mb) is x, then

0.34x 55.845

100

55.845 100x 16425

0.34

2. The following Ellingham diagram depicts the oxidation

of 'C', 'CO' and 'Fe'. Which of the following is

correct?

Paper Code : 34

Ellingham DiagramT/K

G°/kJ/m

ol)

–200

–550

–500

–450

–400

–350

–300

–250

200 400 600 800 1000 1200 1400

2C+ O2

2CO

0

C + O2

CO2

2Fe + O 2

FeO

2CO + O2

2CO2

I. FeO can be reduced by C below 600 K

II. FeO can be reduced by CO below 600 K

III. FeO can be reduced by C above 1000 K

IV. FeO can be reduced by CO above 1000 K

(A) II and III (B) I and IV

(C) I and III (D) II and IV

Answer (A)

2

NSEC 2019-20

Sol. For a reaction to be spontaneous, G < 0.

Hence, reduction is preferred on above line

while oxidation on lower line at a temperature

in Ellingham diagram.

Below 600 K, for reaction

FeO + CO Fe + CO2, G < 0

Above 1000 K, for reactions

FeO + C Fe + CO, G < 0

2FeO + C 2Fe + CO2, G < 0

3. A balance having a precision of 0.001 g was used to

measure a mass of a sample of about 15 g. The number

of significant figures to be reported in this measurement

is

(A) 2 (B) 3

(C) 5 (D) 1

Answer (C)

Sol. Final reading should be reported upto 3 places of

decimal i.e. 15.001 g or 14.999 g. Hence significant

figures to be reported will be 5.

4. N3–, F–, Na+ and Mg2+ have the same number of

electrons. Which of them will have the smallest and

the largest ionic radii respectively?

(A) Mg2+ and N3– (B) Mg2+ and Na+

(C) N3– and Na+ (D) F– and N3–

Answer (A)

Sol. Among isoelectronic species, Ionic radius 1

Z

Ionic radii : N3– > F– > Na+ > Mg2+

5. The reaction of 2, 4-hexadiene with one equivalent of

bromine at 0°C gives a mixture of two compounds 'X'

and 'Y'. If 'X' is 4, 5-dibromohex-2-ene, 'Y' is

(A) 2, 5-dibromohex-2-ene

(B) 2, 5-dibromohex-3-ene

(C) 2, 3-dibromohex-3-ene

(D) 3, 4-dibromohex-3-ene

Answer (B)

Sol.2, 4-hexadiene

Br (1 eq.)2

0°C

6

Br

Br

54

3 12

6

Br

Br

54

3

2

1

+

4, 5-dibromohex-2-ene(X)

2, 5-dibromohex-3-ene(Y)

Conjugated diene undergo 1, 2-addition and

1, 4-addition on reaction with Br2

6. The major product of the following reaction is

+ CH Cl2 2

anhyd. AlCl3

heat

Excess

(A)

CH Cl2

(B)

CHCl2

(C)

CH Cl2

CH Cl2

(D)

Answer (D)

Sol. + Cl – CH – Cl + 2

+ 2HCl

anhy.AlCl3

Diphenylmethane

7. An electrochemical cell was constructed with Fe2+/Fe

and Cd2+/Cd at 25°C with initial concentrations of

[Fe2+] = 0.800 M and [Cd2+] = 0.250 M. The EMF of

the cell when [Cd2+] becomes 0.100 M is

2

2

Fe (aq) / Fe(s) 0.44

Cd (aq) / Cd(s) 0.40

0Half cell E (V)

(A) 0.013 V (B) 0.011 V

(C) 0.051 V (D) 0.022 V

Answer (B)

Sol. Cell reaction :

Cd2+(aq) + Fe(s) Cd(s) + Fe2+(aq)

Initially 0.25 M 0.8 M

Final 0.1 M 0.95 M

Ecell

= E°cell

–

2

2

0.0591 Felog

2 Cd

=0.0591 0.95

( 0.4 0.44) log2 0.1

=0.0591

0.04 log(9.5) 0.011V2

3

NSEC 2019-20

8. The kinetic energy of the photoelectrons ejected by a

metal surface increased from 0.6 eV to 0.9 eV when

the energy of the incident photons was increased by

20%. The work function of the metal is

(A) 0.66 eV (B) 0.72 eV

(C) 0.90 eV (D) 0.30 eV

Answer (C)

Sol. KE =(E ) (W)

Incident Energy Work function

0.6 eV = E – W ...(i)

0.9 eV = 1.2E – W ...(ii)

Multiply equation (i) by 1.2 and then substract from

equation (ii)

Work function (W) = 0.9 eV

9. The alkene ligand ( - C2R

4) is both a '' donor and a ''

acceptor, similar to the CO ligand in metal carbonyls,

and exhibits synergic bonding with metals. Correct

order of C-C bond length in K[PtCl3(-C

2R

4)]

complexes in which R = H, F or CN is

(A) H > F > CN (B) H > CN > F

(C) CN > F > H (D) F > H > CN

Answer (C)

Sol. Due to more electron withdrawing nature of

–CN, the 'C = C' bond length is maximum in

case of –CN

10. The correct order of CFSE among [Zn(NH3)4]2+,

[Co(NH3)6]2+ and [Co(NH

3)6]3+ is

(A) [Co(NH3)6]3+ > [Co(NH

3)6]2+ > [Zn(NH

3)4]2+

(B) [Zn(NH3)4]2+ > [Co(NH

3)6]2+ > [Co(NH

3)6]3+

(C) [Co(NH3)6]3+ > [Zn(NH

3)4]2+ > [Co(NH

3)6]2+

(D) [Co(NH3)6]2+ > [Co(NH

3)6]3+ > [Zn(NH

3)4]2+

Answer (A)

Sol. CFSE increase with increase in charge on

metal ion

0 >

t

CFSE order :

3 2 2

3 2 23 6 3 6 3 4

(Co , C.N. 6) (Co , C.N. 6) (Zn , C.N. 4)

[Co(NH ) ] [Co(NH ) ] [Zn(NH ) ]

11. When acid 'X' is heated to 230°C, along with CO2 and

H2O, a compound 'Y' is formed. If 'X' is

HOOC(CH2)2CH(COOH)

2, the structure of 'Y' is

(A) HOOC(CH2)3COOH (B)

OOO

(C) CH3CH

2CH(COOH)

2(D)

O

O

O

Answer (D)

Sol.COOH

COOH

COOH(X)

–CO2

COOH

COOH

Glutaric acid

o

o

O

–H O2

12. Which of the following is correct about the isoelectronic

species, Li+ and H– ?

I. H– is larger in size than Li+

II. Li+ is a better reducing agent than H–

III. It requires more energy to remove an electron from

H– than from Li+

IV. The chemical properties of the two ions are the

same

(A) I only (B) II and III

(C) I, II and IV (D) I and II

Answer (A)

Sol. H Lir r

Ion of larger size (H–), will require less energy

to remove electron

Due to smaller size of Li+ than H–, release of

electron will be difficult in Li+ than H– i.e. its

oxidation will be difficult than H–.

13. Number of products formed (ignoring stereoisomerism)

in the monochlorination of ethylcyclohexane is

(A) 6 (B) 8

(C) 5 (D) 4

Answer (A)

Sol.

+

Cl

Cl

+ Cl

+

Cl

+

Cl

+

Cl

Monochlorination

(ethyl cyclohexane)

4

NSEC 2019-20

14. The number of asymmetric carbon atoms in strychnine,

whose structure given below is

N

N

O O

Strychnine

(A) 5 (B) 4

(C) 6 (D) 7

Answer (C)

Sol. N

N

O O

The number of asymmetric carbon atoms = 6

15. Molten NaCl is electrolysed for 35 minutes with a

current of 3.50 A at 40°C and 1 bar pressure.

Volume of chlorine gas evolved in this electrolysis

is

(A) 0.016 L (B) 0.98 L

(C) 9.8 L (D) 1.96 L

Answer (B)

Sol. NaCl Na Cl

Anode : 2

2Cl Cl 2e

1 mole 2 F

Number of faraday used = It

96500

= 3.5 35 60

96500

= 0.076 F

2 F charge gives Þ 1 mole Cl2

\ 0.076 F charge will give = 1

0.0762

= 0.038 mole Cl2

Now, PV = nRT

\ Volume of Cl2 obtain =

nRT

P

= 0.038 0.083 313

1

0.98 L�

16. Which of the following pairs of compounds can be

stable while retaining the identity of each compound

in the pair over a period of time?

I. FeCl3, SnCl

2II. HgCl

2, SnCl

2

III. FeCl2, SnCl

2IV. FeCl

3, Kl

(A) I only (B) I and III

(C) III only (D) II and IV

Answer (C)

Sol.3 2 2 4

2FeCl SnCl 2FeCl SnCl

2 2 42HgCl 2SnCl 2Hg SnCl

2 2FeCl SnCl No reaction

3 2 22FeCl 2KI 2FeCl 2KCl I

17. The reaction xX(g) yY(g) zZ(g)���⇀↽���

was carried

out at a certain temperature with an initial pressure of

X = 30 bar. Initially ‘Y’ and ‘Z’ were not present. If the

equilibrium partial pressures of ‘X’, ‘Y’ and ‘Z’ are 20, 5

and 10 bar respectively, x : y : z is

(A) 4 : 1 : 2 (B) 2 : 1 : 2

(C) 1 : 2 : 1 (D) 1 : 1 : 2

Answer (B)

Sol.

eg

xX yY zZ

t 0 30 0 0

y zt t 30 a a a

x x

���⇀↽���

According to Question 30 – a = 20

a 10

Nowya 5

x

y 5 1

x 10 2

And za 10

x

z 10

1x 10

So, x : y : z

= y z

1 : :x x

= 1

1 : : 12

= 2 : 1 : 2

5

NSEC 2019-20

18. The major product ‘P’ formed in the following sequence

of reactions is

NH2

O

O

P

(i) Ethylene glycol, dry HCl(ii) NaOBr

(iii) H O3

+

(A) OH

N

(B)

N

(C)

N

(D)

OHN

Answer (C)

Sol.

NH2

O

O

(P)

CH OH2

CH OH, dry HCl2

NaOBr

NH2

O

NH2

O O

O O

H O3

+H N2

O

N(C)

19. Sodium lauryl sulphate (SLS) is a surface active agent,

which is adsorbed on water surface. The number of

molecules of SLS that can be adsorbed on the surface

of a spherical water droplet of diameter 3.5 mm is

(effective area of one molecule of SLS = 4.18 nm2)

(A) 9.20 × 1012 (B) 9.20 × 1018

(C) 1.15 × 1012 (D) 3.68 × 1013

Answer (A)

Sol.molecules

S.A of spherical dropn

S.A of SLSmolecule

2

18 2

4 r

4.18 10 m

6

18

3.5 3.54 3.14 10

2 2

4.18 10

= 9.2 × 1012

20. The unit of Planck’s constant, ‘h’, is the same as that

of

(A) Angular momentum (B) Energy

(C) Wavelength (D) Frequency

Answer (A)

Sol. E = h

So h = E

v

units of h = 1

JJ s

s

Units of angular momentum = mvr

= matr [v = at]

= F × r × t

= w × t [w = work]

So, units of angular momentum = J s

Also nh

mvr2

So dimensions of h must be same as mvr

21. The set in which all the species are diamagnetic is

(A) B2, O

2, NO (B)

2 2O , O , CO

(C) – –

2 2N , O , CN (D) 2–

2 2C , O ,NO

Answer (D)

Sol.x y

2 *2 2 *2 2 22 1S 2S 2S 2p 2p1SC :

z x y

2 2 *2 2 *2 2 2 22 1S 2S 2S 2p 2p 2p1SO :

x y

*2 *22p 2p

x y z

2 *2 2 *2 2 2 21S 2S 2S 2p 2p 2p1SNO :

Hence, 2

2 2C ,O

and NO+ are diamagnetic.

22. A solid comprises of three types of elements, ‘P’, ‘Q’

and ‘R’. ‘P’ forms an FCC lattice in which ‘Q’ and ‘R’

occupy all the tetrahedral voids and half the octahedral

voids respectively. The molecular formula of the solid

is

(A) P2Q

4R (B) PQ

2R

4

(C) P4Q

2R (D) P

4QR

Answer (A)

Sol. Effective no. of P per unit cell 4

Effective no. of Q per unit cell 8

Effective no. of R per unit cell 2

P : Q : R

Ratio 4 : 8 : 2

2 : 4 : 1

Hence, molecular formula of solid is P2Q

4R.

6

NSEC 2019-20

23. The following qualitative plots depict the first, second

and third ionization energies (I.E.) of Mg, Al and K.

Among the following, the correct match of I.E. and the

metal is

8000

6000

4000

2000

1 2 3

X

Y

ZI.E./kJmol–1

(A) X-Al ; Y-Mg ; Z-K

(B) X-Mg ; Y-Al ; Z-K

(C) X-Mg ; Y-K ; Z-Al

(D) X-Al ; Y-K ; Z-Mg

Answer (C)

Sol. Order of 1st IE : Mg > Al > K

Order of 2nd IE : K > Al > Mg

Order of 3rd IE : Mg > K > Al

So, X Mg, Y K, Z Al

24. The structure of compound ‘X’ (C8H

11NO) based on

the following tests and observations is

Reagent/s

Neutral FeCl3

Lucas reagent

NaNO /HCl at 273 K2

Observation

No coloration

Turbidity

Yellow oil

(A)

HO

H N2

(B)

OH

H

NH

(C)

OH

N

(D)N

HO

H

Answer (D)

Sol. Neutral FeCl3 test is for phenol which is negative

showing absence of phenolic group

Yellow oil is obtained in case of secondary amine

with NaNO2 and HCl.

So answer is (D)

CH OH2

NH

25. The number of stereoisomers is maximum for

(A) [Co(en)3]3+ (B) [Co(en)

2ClBr]+

(C) [Co(NH3)4Cl

2]+ (D) [Co(NH

3)4ClBr]+

Answer (B)

Sol. Complex Number of Stereoisomers

(A) [Co(en)3]3+ 2

(B) [Co(en)2ClBr]+ 3

(C) [Co(NH3)4Cl

2]+ 2

(D) [Co(NH3)4ClBr]+ 2

Hence, answer is (B)

26. Reaction of C6H

5MgBr with phenol gives

(A)

(B)

OH

(C)

O

(D)

Br

Answer (A)

Sol. +Acid-Base

MgBr–

+ OH H O

+

MgBr– +

27. The power and wavelength emitted by a laser pointer

commonly used in Power point presentations are

1.0 mW and 670 nm respectively. Number of photons

emitted by this pointer during a presentation of

5 minutes is

(A) 1.01 × 109 (B) 1.01 × 1021

(C) 1.6 × 1016 (D) 1.01 × 1018

Answer (D)

7

NSEC 2019-20

Sol. Power of laser pointer = 1 mW

3 J10

s

Total energy = 10–3 × 5 × 60

= 0.3 J

Number of photon × Energy of one photon = 0.3 J

hc

n 0.3

9

34 8

0.3 0.3 670 10n

hc 6.626 10 3 10

= 1.01 × 1018

28. The work done (kJ) in the irreversible isothermal

compression of 2.0 moles of an ideal gas from 1 bar

to 100 bar at 25°C at constant external pressure of

500 bar is

(A) 2452 (B) 490

(C) 2486 (D) –490

Answer (A)

Sol.ext ext 2 1

W P ( V) P (V V )

ext

2 1

nRT nRT(P )

P P

1 1(500) (2 R 298)

100 1

99500 2 8.314 298

100

2452.79 kJ 2452kJ �

29. Atropine (C17

H23

O3N) is a naturally occurring

compound used to treat certain types of poisoning.

The degree of unsaturation in atropine is

(A) 7 (B) 6

(C) 5 (D) 4

Answer (A)

Sol. Degree of unsaturation H N X

(C 1)2

23 118

2

= 7

30. MnCl2.4H

2O (molar mass = 198 g mol–1) when

dissolved in water forms a complex of Mn2+. An

aqueous solution containing 0.400 g of MnCl2.4H

2O

was passed through a column of a cation exchange

resin and the acid solution coming out was neutralized

with 10 mL of 0.20 M NaOH. The formula of the complex

formed is

(A) [Mn(H2O)

4Cl

2] (B) [Mn(H

2O)

6]Cl

2

(C) [Mn(H2O)

5Cl]Cl (D) Na[Mn(H

2O)

3Cl

3

Answer (C)

Sol. Millimoles of Base = 10 × 0.2 = 2

Millimoles of acid coming out of cation

exchanger = 2

And Millimoles of complex 400

2198

�

The complex has cationic charge equal to (+1)

Hence, the complex that must have formed is

[Mn(H2O)

5Cl]Cl when the given complex is dissolved

in water

31. Which of the folliwng is NOT correct about hydrides?

I. Saline hydrides are stoichiometric and metallic

hydrides are non-stoichiometric

II. BeH2 is monomeric whereas MgH

2 is polymeric

III. Hydrides of the elements of Group 13 are electron

deficient and those of Group 15 are electron rich

IV. NaH reacts with water and liberates H2 whereas

B2H

6 does not react with water

(A) IV only (B) I and III

(C) III only (D) II and IV

Answer (D)

Sol. I. Saline hydrides are ionic and stoichiometric

Metallic hydrides are non-stoichiometric.

II. BeH2 and MgH

2 are both polymeric.

III. Hydrides of elements of Group 13 are electron

deficient while those of Group 15 are electron

rich.

IV. B2H

6 reacts with water to form Boric acid and

hydrogen gas.

B2H

6 + 6H

2O 2B(OH)

3 + 6H

2

Hence, statement (II) and (IV) are not correct.

32. The compounds ‘X’ and ‘Y’ formed in the following

reaction are

OH

O

HH O3

+

X + Y

(A) Hemiacetals with identical physical and chemical

properties

(B) Acetals with identical physical and chemical

properties

(C) Hemiacetals with different physical and chemical

properties

(D) Acetals with different physical and chemical

properties

Answer (C)

8

NSEC 2019-20

Sol.

OH

O

HH

+

OH

OH

H

+

(–H )+

OH

+ O

OH

O

Hence, a pair of diastereomeric hemiacetals are

obtained which have different chemical and physical

properties.

33. Aqueous solution of slaked lime, Ca(OH)2, is

extensively used in municipal waste water treatment.

Maximum pH possible in an aqueous solution of slaked

lime is

(Ksp

of Ca(OH)2 = 5.5 × 10–6)

(A) 1.66 (B) 8.14

(C) 12.04 (D) 12.34

Answer (D)

Sol.2

2s 2s

Ca(OH) Ca 2OH �

Ksp

= 4s3 = 5.5 × 10–6

63 5.5 10s

4

s = 0.0111

[OH–] = 2s = 0.0222

pOH = –log(0.0222) = 1.653

pH = 14 – 1.653 = 12.34

34. An electron present in the third excited state of a

H atom returns to the first excited state and then to

the ground state. If 1 and

2 are the wavelengths of

light emitted in these two transitions respectively,

1 :

2 is

(A) 4 : 1 (B) 5 : 9

(C) 3 : 1 (D) 2 : 1

Answer (A)

Sol. H

1

1 1 1R

4 16

H

1

1 12R

64

...(1)

H

2

1 1 1R

1 4

H

3R

4

...(2)

H1

2H

3R

44

12 1R

64

1 :

2 = 4 : 1

35. The percentage dissociation of 0.08 M aqueous acetic

acid solution at 25°C is (Ka of acetic acid at

25°C = 1.8 × 10–5)

(A) 2.92 (B) 1.5

(C) 1.2 (D) 4.8

Answer (B)

Sol.3 3

0.080.08 0.08 0.08

CH COOH CH COO H

�

2 2

a

0.08K

0.08 0.08

2

a

0.08K

1

0.082 = 1.8 × 10–5 (since is very-very small

compaired to 1)

= 0.015

% = 1.5%

36. In which of the following, is a new C-C bond formed in

the product?

I. dil. NaOH

3CH CHO

II. heat

3 2 5CH MgCl C H OH

III. 3H O

2 3CO CH MgBr

IV. 3CH Br

2 2 2C H NaNH

(A) I, III and IV (B) II and III

(C) III only (D) III and IV

Answer (A)

Sol. I. dil. NaOH

3 3 2

OH|

CH CHO CH —CH—CH —CHO

II. 3 2 5 4 2 5CH MgCl C H OH CH C H OMgCl

III. 3H O

3 3O C O CH MgBr CH COOH

IV. 3CH Br

2 3H—C C—H NaNH HC C—CH

9

NSEC 2019-20

37. IUPAC name of the following molecule is

H C3

OH

CH3

(A) 4-hydroxyhept2-en-5-yne

(B) Hept-2-en-5-yn-4-ol

(C) Hept-5-en-2-yn-4-ol

(D) 4-hydroxyhept-5-en-2-yne

Answer (B)

Sol.H C3

OH

CH3

1

23

4

5 6 7

Hept-2-en-5-yn-4-ol

38. The product/s of the following reaction is/are

COOH

NO2

NaOH, CaO

heat

CHO

NO2

I

CH OH2

NO2

II

NO2

III

NO2

IV

CH OH2

(A) I and II (B) II

(C) III (D) IV

Answer (C)

Sol.

COOH

NO2

NaOH, CaO

NO2

Decarboxylation

39. For which of the following processes, carried out in

free space, energy will be absorbed?

I. Separating an electron from an electron

II. Removing an electron from a neutral atom

III. Separating a proton from a proton

IV. Separating an electron from a proton

(A) I only (B) II and IV

(C) I and III (D) II only

Answer (B)

Sol. Energy will be absorbed in process II and IV, ie

removing an electron from a neutral atom and

separating an electron from a proton.

40. Decay of radioisotopes follows first order kinetics.

Radioisotope U238 undergoes decay to a stable

isotope, Th234. The ratio of the number of atoms of

U238 to that of Th234 after three half lives is

(A)1

3(B)

3

4

(C)1

4(D)

1

7

Answer (D)

Sol. U238 Th234

Initially 1 mol 0

after three

half lifes 0.125 0.5 + 0.25 + 0.125

Ratio of no. of atoms of U238 to that of

Th234 0.125

0.875

1

7

41. The anhydride of HNO3 is

(A) NO (B) NO2

(C) N2O (D) N

2O

5

Answer (D)

Sol. Anhydride of HNO3 is N

2O

5.

42. Which of the following is correct?

I. Sodium (Na) is present as metal in nature

II. Na2O

2 is paramagnetic

III. NaO2 is paramagnetic

IV. Na reacts with N2 to form Na

3N

(A) III only (B) II and IV

(C) I, III and IV (D) II, III and IV

Answer (A)

Sol. Na2O

2 contains 2

2O

which is diamagnetic.

NaO2 contains

2O

and it has one unpaired electron

in 2p* orbital that's why it is paramagnetic.

43. An excess of aqueous ammonia is added to three

different flasks (F1, F

2, F

3) containing aqueous

solutions of CuSO4, Fe

2(SO

4)

3 and NiSO

4

respectively.

Which of the following is correct about this addition?

I. A precipitate will be formed in all three flasks

II. Ammonia acts as a base as well as a ligand

exchange reagent in F1 and F

3

III. A soluble complex of NH3 and the metal ion is

formed in F1 and F

3

IV. A precipitate will be formed only in F2

(A) I only (B) IV only

(C) II and IV (D) II, III and IV

Answer (D)

10

NSEC 2019-20

Sol. F1 : CuSO

4 + 4NH

3 [Cu(NH

3)4]2+ 2

4SO

Soluble complex

F3 : NiSO

4 + 6NH

3 [Ni(NH

3)6]2+ 2

4SO

Soluble complex

F2 : Fe

2(SO

4)3 + NH

4OH Fe(OH)

3(From NH

3) Reddish-brown ppt

44. The reagent/s that can be used to separate

norethindrone and novestrol from their mixture is/are

O

OH

Norethindrone

HO

OH

Novestrol

I. HCl II. NaOH III. NaHCO3

IV. NaNH2

(A) III (B) I and IV

(C) I, II and III (D) II

Answer (D)

Sol. � NaOH can cause abstraction of proton from

phenolic group, therefore Novestrol gets

dissolved in NaOH.

� NaNH2 is a very strong base therefore both

Norethindrone and Novestrol become soluble in

NaNH2.

45. Which of the following is/are electrophilic aromatic

substitution reaction/s?

ICl , light2

Cl

IIBF

3

+ (CH ) CHCl3 2

IIIHCHO, H O

3

+

CH OH2

IVNaNH , 2 3NH

NH2

(A) II, III and IV (B) II and III

(C) I, II and III (D) II only

Answer (B)

Sol. (CH ) CHCl 3 2

BF3

(CH ) CH + [BF Cl ] 3 2 3

– –

+

[electrophile]

+ (CH ) CH 3 2

+[EAS reaction]

+ HCHO [EAS reaction]H

3O

+

CH OH2

46. Among the halides NCl3 (I), PCl

3 (II) and AsCl

3 (III),

more than one type of acid in aqueous solution is

formed with

(A) I, II and III (B) II only

(C) I and II (D) II and III

Answer (D)

Sol. NCl3 + H

2O NH

3 + HOCl

PCl3 + H

2O H

3PO

3 + HCl

AsCl3 + H

2O As(OH)

3 + HCl

PCl3 and AsCl

3 give two acids after hydrolysis.

47. The normal boiling point and Hvap

of a liquid 'X'

are 400 K and 40 kJ mol–1 respectively. Assuming

Hvap

to be constant, which of the following is

correct?

I. Svap

> 100 J K–1 mol–1 at 400 K and 0.5 atm

II. Svap

< 100 J K–1 mol–1 at 400 K and 1 atm

III. Svap

< 100 J K–1 mol–1 at 400 K and 2 atm

IV. Svap

= 100 kJ K–1 mol–1 at 400 K and 1 atm

(A) II and IV (B) II only

(C) I and III (D) I, III and IV

Answer (C)

Sol. Normal boiling point is calculated at 1 atm.

Svap

= vap

b

H

T

(at 1 atm)

=

31 140 10

100 JK mol400

with the decreasing pressure, volume will increase

and hence entropy will increase

Svap

> 100 J K–1 mol–1 at 0.5 atm

Svap

< 100 J K–1 mol–1 at 2 atm

11

NSEC 2019-20

48. About the energy level diagram given below, which of

the following statement/s is/are correct?

Energy

Reaction coordinate

M

X Y

Q

N

P

R

I. The reaction is of two steps and 'R' is an intermediate

II. The reaction is exothermic and step 2 is rate

determining

III. 'Q' is an intermediate and 'R' is the transition state

for the reaction M Q

IV. 'P' is the transition state for the reaction Q N

(A) III and IV (B) I, III and IV

(C) I, II and IV (D) III only

Answer (A)

Sol. ∵ Activation energy of Step 1 (i.e. X) is greater

than that of step 2 (i.e. Y)

Step 1 is rate determining step.

Given reaction is endothermic

'M' is the reactant

'N' is the product

'R' is the transition state of step 1

'P' is the transition state of step 2

'Q' is the intermediate

49. The F – X – F bond angle is the smallest in (X is the

central atom)

(A) CF4

(B) NF3

(C) OF2

(D) –

5XeF

Answer (D)

Sol. In CF4, carbon is sp3 hybridized so bond angle

is 109°28. In NF

3, nitrogen is sp3 hybridized with one lone

pair hence, bond angle is less than 109°28.

In OF2

(F – O – F).

.

.

.

oxygen is having its

hybridisation sp3 with 2 p.� hence bond angle

is close to 100°

In –

5XeF ion, Xe is sp3d3 hybridized so bond

angle is nearly 72°, smallest of all four given

molecules.

50. The correct IUPAC name of the compound,

[Pt(py)4] [Pt(Br)

4] is

(A) tetrapyridineplatinum(II) tetrabromidoplatinate(II)

(B) tetrabromidoplatinum(IV) tetrapyridineplatinate(II)

(C) tetrabromidoplatinate(II) tetrapyridineplatinum(II)

(D) tetrapyridineplatinum(IV) tetrabromidoplatinate(IV)

Answer (A)

Sol. [Pt(py)4] [Pt(Br)

4]

x + 4 × 0 + x + 4 × (–1) = 0

2x – 4 = 0 x = + 2

So, oxidation number of Pt is +2 in both cation and

anion.

It's IUPAC name is

Tetrapyridineplatinum (II) tetrabromidoplatinate (II)

51. All four types of carbon 1 ,2 ,3 and 4� � � � are present

in

I II III IV

(A) I, II and III

(B) II, III and IV

(C) I, II and IV

(D) II and III

Answer (D)

Sol.

(II) (III)

1°

2°4°

3°and

1°

3°2°

2°

1°

4°

(II) and (III) contain all 4 types of carbon atoms

1 ,2 ,3 and 4� � � �

52. The mass (g) of NaCI that has to be dissolved to

reduce the vapour pressure of 100 g of water by 10%

(Molar mass of NaCI = 58.5 gmol–1) is

(A) 36.11 g

(B) 17.54 g

(C) 81.25 g

(D) 3.61 g

Answer (B)

12

NSEC 2019-20

Sol. According to Raoult's law

Relative lowering in vapour pressure = mole fraction

of the solute in solution

or

0

S

0

P –P in

in NP

10 in

100100in

18

So 5 5555 5.56n for NaCl, i 2

9i 9 2

�

Mass(g) of NaCl to be dissolved

5 5658 5

9 2

= 18.07 g

The most close option near to 18.07 is option (B)

53. The most acidic hydrogen in the following molecule

is

HO

O

IV

HO OH

OH

O

II I

III

(A) I (B) II

(C) III (D) IV

Answer (B)

Sol.

O

(IV)

O

O

(II) (I)

(III): :

O: :

O: : O: :

Most stable anion is formed by loss of H atom

labelled as (II), So H at this position will be the

most acidic.

54. Two isomeric hydrocarbons 'X' and 'Y' (C4H

6), give the

same product (C4H

8O) on catalytic hydration with

dilute acid. However, they form different products but

with same molecular formula (C4H

6Br

4) when treated

with excess bromine. 'X' and 'Y' are

(A) &

(B) &

(C) &

(D)&

Answer (C)

Sol.H O2

O

H O2

OSame Ketones

Hg /H2+ +

Hg /H2+ +

Although on bromination the two give different

tetrabromo products.

CH

Br

Br

C

Br

Br

CH2

CH3

and

CCH3

Br

Br

C

Br

Br

CH3

55. Mercury is highly hazardous and hence its

concentration is expressed in the units of ppb

(micrograms of Hg present in 1 L of water). Permissible

level of Hg in drinking water is 0.0335 ppb. Which of

the following is an alternate representation of this

concentration?

(A) 3.35 × 10–2 mg dm–3

(B) 3.35 × 10–5 mg dm–3

(C) 3.35 × 10–5 mg m–3

(D) 3.35 × 10–4 g L–1

Answer (B)

Sol. Hg 0.0335 ppb means

0.0335 micrograms of Hg in 1 L of water

1 microgram = 10–6 g = 10–6 × 1000 mg

= 10–3 mg

Also, 1 L = 1 dm3

So [Hg] = 0.0335 ppb =

3

3

0.0335 10 mg

1dm

=

5

3

3.35 10 mg

dm

= 3.35 × 10–5 mg dm–3

13

NSEC 2019-20

56. The correct sequence of reactions which will yield 4-

nitrobenzoic acid from benzene is

(A) CH3CI; HNO

3/H

2SO

4; KMnO

4/OH–

(B) HNO3/H

2SO

4; CH

3CI/AICI

3; KMnO

4/OH–

(C) CH3CI/AICI

3; KMnO

4/OH–; HNO

3/H

2SO

4

(D) CH3CI/AICI

3; HNO

3/H

2SO

4; KMnO

4/OH–

Answer (D)

Sol. + CH Cl3

Anhydrous AlCl3

HNO +3

H SO2 4

CH3

NO2

KMnO /OH4

–

COOH

NO2

CH3

57. The volume of one drop of aqueous solution from an

eyedropper is approximately 0.05 mL. One such drop

of 0.2 M HCI is added to 100 mL of distilled water.

The pH of the resulting solution will be

(A) 4.0 (B) 7.0

(C) 3.0 (D) 5.5

Answer (A)

Sol. For dilution,

M1V

1 = M

2V

2

0.2 × 0.05 = M2 × 100

2

0.2 0.05M

100

= 1 × 10–4 M

[HCl] = [H+] = 10–4 M

pH = –log[H+]

= –log(10–4)

= 4

58. In which of the following species the octet rule is NOT

obeyed?

I. –

3I II. N

2O

III. OF2

IV. NO+

(A) I and IV (B) II and III

(C) I only (D) IV only

Answer (C)

Sol. �3I does not follow octet rule.

� As per MOT, NO

is represent as N O

in

which both ‘N’ and ‘O’ atoms have complete

octet.

59. Which atom/s will have a + charge in the following

molecule?

OI

II

III

IV

(A) I and III (B) II only

(C) II and III (D) II and IV

Answer (D)

Sol. Because of resonance, electrons shift

–OI

II

III

IV

OI

II

III

IV

–

+

+

On carbon atoms II and IV, + charge is

present.

60. 2.0 moles of an ideal gas expands isothermally (27ºC)

and reversibly from a pressure of 1 bar to 10 bar. The

heaviest mass that can be lifted through a height of

10 m by the work of this expansion is

(A) 50.8 kg (B) 50.8 g

(C) 117.1 kg (D) 117.1 g

Answer (C)

Sol. During expansion of gas, pressure should be

decrease but in problem increase of pressure given

which is wrong. Solution is given only by

considering data.

Work done(W) = – 1

2

P2.303 nRT log

P

= 1

–2.303 2 8.314 300 log10

= 11.488 kJ

For heaviest mass to lift, complete work should be

used for this

i.e. W = mgh

11.488 × 103 = m × 9.8 × 10

m = 117.1 kg

61. A commercial sample of oleum (H2S

2O

7) labeled as

'106.5% oleum' contains 6.5 g of water. The

percentage of free SO3 in this oleum sample is

(A) 2.88 (B) 28.8

(C) 0.029 (D) 0.28

Answer (B)

14

NSEC 2019-20

Sol. 106.5% oleum means 100 g of oleum sample on

dilution produces 106.5 g of H2SO

4, it implies that

100 g of oleum sample requires 6.5 g of water to

produce 106.5 g of H2SO

4. since in oleum

(SO3 + H

2SO

4), SO

3 got diluted as

3 2 2 4SO H O H SO

80 g 18 g

806.5

18 g 6.5 g

= 28.88 g

Hence, percentage of free SO3 is 28.8%

62. Which of the following species has one lone pair of

electrons on the central atom?

(A) CIF3

(B) –

3I

(C) 3I (D) SF

4

Answer (D)

Sol. In SF4, 'S' has one lone pair of electrons.

'S' has 4 bond pairs 1 lone pair.

F

S

FF

F

: Shape : See saw.

63. Among the following, the complex ion/s that will have

a magnetic moment of 2.82 B.M. is/are

I. [Ni(CO)4] II. [NiCI

4]2–

III. [Ni(H2O)

6]2+ IV. [Ni(CN)

4]2–

(A) I and IV (B) II only

(C) II and III (D) II, III and IV

Answer (C)

Sol. Ni(28) = [Ar] 3d84s2

Ni2+ = 3d8

Cl– being a weak field ligand, will not be able to pair

the electrons

[NiCl4]2– has 2 unpaired electrons.

Hence, magnetic moment = 2.82 B.M.

Also, In [Ni(H2O)

6]2+, Ni has 2 unpaired electrons

because H2O is a weak field ligand, will not pair the

electrons

64. Morphine, a pain killer is basic with the molecular

formula C17

H19

NO3. The conjugate acid of morphine is

(A) 17 19 3

C H NO (B) C17

H18

NO3

(C) –

17 19 3C H NO (D)

17 20 3C H NO

Answer (D)

Sol. Conjugate acid is formed by the addition of one H+.

Bronsted base + H+ conjugate acid

C17

H19

NO3 + H+

17 20 3C H NO

Morphine Conjugate acid

65. A suboxide of carbon, C3O

2, has a linear structure.

Which of the following is correct about C3O

2?

I. Oxidation state of all three C atoms is +2

II. Oxidation state of the central C atom is zero

III. The molecule contains 4 and 4 bonds

IV. Hybridization of the central carbon atom is sp2

(A) I and IV (B) II and III

(C) II and IV (D) III only

Answer (B)

Sol. The structure of C3O

2 is given below

O=C=C=C=O

The molecule of C3O

2 has 4 and 4 bonds.

Each C-atom of C3O

2 is sp-hybridised. Oxidation

state of central atom is zero.

66. Among the following, the compounds with highest

and lowest boiling points respectively are

H N2

I

HO

II

III

F

IV

O

V

(A) I and III (B) II and III

(C) I and IV (D) II and V

Answer (B)

Sol. Out of the given compounds, alcohol (II) has the

highest boiling point due to strong intermolecular H-

bonding and hydrocarbon (III) has the lowest boiling

point due to its non-polar nature.

67. At 25ºC Ka of 2–

4HPO and –

3HSO are 4.8 × 10–13 and

6.3 × 10–8 respectively. Which of the following is

correct?

(A) 2–

4HPO is a stronger acid than –

3HSO and 3–

4PO

is a weaker base than 2–

3SO

(B) 2–

4HPO is a weaker acid than –

3HSO and 3–

4PO

is a weaker base than 2–

3SO

(C) 2–

4HPO is a weaker acid than –

3HSO and 3–

4PO

is a stronger base than 2–

3SO

(D) 2–

4HPO is a stronger acid than –

3HSO and 3–

4PO

is a stronger base than 2–

3SO

Answer (C)

15

NSEC 2019-20

Sol. The given anions are partially ionised as

2– 3–

4 4HPO H PO

� Ka = 4.8 × 10–13

– 2–

3 3HSO H SO

� Ka = 6.3 × 10–8

Since Ka of

2–

4HPO is lower than that of

–

3HSO , the

2–

4HPO is a weaker acid than

–

3HSO . The conjugate

base of a weak acid is a strong base. Therefore. 3–

4PO

is a stronger base than 2–

3SO

68. The change in internal energy (U) for the reaction

H2(g) + Br

2(g) 2HBr(l) when 2.0 moles each of Br

2(g)

and H2(g) react is

(H2(g) + Br

2(g) 2HBr (g); H

reaction = –109 kJ;

Hvap

of HBr = 213 kJ mol–1)

(A) –644 kJ (B) 644 kJ

(C) –322 kJ (D) –1070 kJ

Answer (D)

Sol. H2(g) + Br

2(g) 2HBr(g) ...(1)

H1 = –109 kJ

2HBr(g) 2HBr(l) ...(2)

H2 = –2 × 213 kJ

= –426 kJ

___________________________

(1) + (2) H2(g) + Br

2(g) 2HBr(l)

H = –109 – 426

= –535 kJ

H = U + ngRT

–535 = U + (–2) 8.314 × 10–3 × 298

U = –535 + 4.955 = –530.045 kJ

When 2 moles of H2 and 2 moles of Br

2 react,

U = –2 × 530.045

= –1060.09 kJ

This value is closest to –1070 kJ. So option (D) can

be considered.

69. The structure that represents the major intermediate

formed in the bromination of toluene is

(A)

Br+

(B)

Br

+

CH2

(C)

Br

+

(D)

BrH C3

+

Answer (C)

Sol. The methyl group of toluene is o-/p-directing. The

major product will be formed when Br+ ion attacks

the p-position to form major intermediate.

Br+

Br

+

70. About sea water, which of the following statement/s

is/are correct?

I. Frozen sea water melts at a lower temperature

than pure ice

II. Boiling point of sea water increases as it

evaporates

III. Sea water boils at a lower temperature than fresh

water

IV. Density of sea water at STP is same as that of

fresh water

(A) I only (B) I and II

(C) I, II and III (D) III only

Answer (B)

Sol. Sea water has higher boiling point than fresh water

and lower freezing point than fresh water due to

dissolved salts in it. Boiling point of sea water increases

with evaporation due to increase in concentration of

dissolved salts. Also, frozen sea water melts at a lower

temperature than pure ice.

71. Saran wrap, a polymer used in food packaging is a

copolymer of 1, 1-dichloroethene and vinyl chloride. In

the chain initiation step, 1, 1-dichloroethene generates

a free radical which reacts with vinyl chloride. Structure

of Saran wrap is

(A)

Cl

ClCl

n

(B)

Cl

Cl

Cl

n

(C)

Cl

ClCl

n

(D)

Cl

Cln

Cl

Answer (D)

Sol.

CH2

CCl2 + CH

2CH Cl

1, 1-dichloroethene Vinyl chloride

C

CH2

Cl

Cl

CH2

CH

Cl

Saranwrap

n

The polymerisation of 1, 1-dichloroethene and vinyl

chloride proceeds mostly by head-to-tail mechanism.

16

NSEC 2019-20

72. The alkene 'Y' in the following reaction is

Alkene Y (i) Ozonolysis

(ii) Me S2

O

H H +

O

H

O

O

H

O

+

O

(A) (B)

(C) (D)

Answer (C)

Sol. The unknown alkene (Y) can be decided by

combining carbonyl groups of the ozonolysis

products and proceeding in the backward direction.

The ozonolysis products can be explained if option

(C) is chosen as the desired alkene

1. O3

2. Me S2

O

H H +

HC

O

O

+

O

O

H

O

(Y)

73. In solid state, PCl5 exists as [PCl

4]+[PCl

6]–. The

hybridization of P atoms in this solid is/are

(A) 2 2

3

x ysp d d d

(B) 2

3

zsp d d d

(C) sp3 and sp3d2 2 2 2x y z

d d ,d

(D) sp3d and 2

3

zdsp d d

Answer (C)

Sol.

P

ClCl

Cl

Cl

+

sp3

[PCl ]4+

P

ClCl

Cl

Cl

–

sp3 2d

[PCl ]6–

ClCl

2 2 2x y , z(d d d )

74. Which of the following compounds have chiral carbon

atom/s?

OH

OHOH

OH

Cl

I II III IV V

OH

(A) I and II (B) I, III, IV and V

(C) II, IV and V (D) II, III and IV

Answer (B)

Sol. OH

OH

H C3 H

CH2

I II IIIyesYes, chiral

carbon atompresent

No, planeof symmetryexists

H OH*

*

HOH

IVyes

OH

H C3

*CH3

Vyes

ClH

H

**

Hence, Chiral carbon atoms are present in I, III, IV

and V

75. The crystal defect indicated in the diagram below is

S+

R–

S+

R–

S+

R–

R–

R–

S+

S+

S+ R–

R–

S+ R–

R–

S+

R–

S+

S+

S+ R–

S+ R–

S+ R–

(A) Frenkel defect

(B) Schottky defect

(C) Frenkel and Schottky defects

(D) Interstitial defect

Answer (B)

Sol. In the crystal, 2 R– and 2 S+ are missing. This is

Schottky defect.

R–

S+

R–

R–

S+S+

S+ R–

R–

S+ R–

S+ R–

S+ S+

S+

S+

missing

R–

missing

S+

missing

R–

missing

17

NSEC 2019-20

76. If the standard electrode potentials of Fe3+/Fe and Fe2+/

Fe are –0.04 V and –0.44 V respectively then that of

Fe3+/Fe2+ is

(A) 0.76 V (B) –0.76 V

(C) 0.40 V (D) –0.40 V

Answer (A)

Sol. G° = –nFE°

G° = –3F(–0.04)1

Fe3+

FeFe2+

G° = –1Fy2

G° = –2F(–0.44)3

1 2 3G G G

–3F(–0.04) = –1Fy – 2F(–0.44)

0.12 = –y + 0.88

y = 0.88 – 0.12 = 0.76 V

Hence, standard electrode potential of Fe3+/Fe2+ is

0.76 V

77. Given below is the data for the reaction

2 22NO(g) N (g) O (g)� where 'k

f' and 'k

b' are rate

constants of the forward and reverse reactions

respectively

1 3 1 1 3 1

f b

6

5

Temperature (K) k (mol dm s ) k (mol dm s )

1400 0.20 1.1 10

1500 1.3 1.4 10

The reaction is

(A) exothermic and Keq

at 1400 K = 3.79 × 10–6

(B) endothermic and Keq

at 1400 K = 2.63 × 10–5

(C) exothermic and Keq

at 1400 K = 1.8 × 105

(D) endothermic and Keq

at 1500 K = 9.28 × 10–4

Answer (C)

Sol. At 1400 K,

4 4feq 6

b

k 0.20 20K 10 18.2 10

k 1.11.1 10

5

1.8 10 At 1500 K,

4 4feq 5

b

k 1.3 13K 10 9.3 10

k 1.41.4 10

1500

1400

K H 1500 1400log

K 2.303R 1500 1400

But,

4

1500

51400

K 9.3 10 9.31

K 181.8 10

1500

1400

Klog 0 H 0 Exothermic

K

78. The major product 'P' formed in the following reaction

is (*denotes radioactive carbon)

OH

*

(i) conc. H SO excess2 4

(ii) conc. HNO , conc. H S3 2 4

O

(iii) H O3

+, heat

OH

P

(A)

OH

*

NO2

OH

(B)

OH

*

NO2

HO

(C)

OH

*NO2

HO

(D)

OH

*NO

2

OH

Answer (A)

Sol.

OH

*OH

(1) Conc. H SO excess2 4

OH

* OH

SO H3

HNO3 H SO2 4

OH

* OH

SO H3

NO2

H /3O+

Ipso substitution Desulphonation

OH

*OH

NO2

HO3S

–OH activatingand o,p directing

HO3S

79. A helium cylinder in which the volume of gas = 2.24 L

at STP (1 atm, 273 K) developed a leak and when the

leak was plugged the pressure in the cylinder was

seen to have dropped to 550 mm of Hg. The number of

moles of He gas that had escaped due to this leak is

(A) 0.028 (B) 0.072

(C) 0.972 (D) 0.099

Answer (A)

Sol.

• Number of moles of He(g) escaped is due to

leakage, which will depend on pressure drop

due to leakage.

• Pressure drop = (760 – 550) mm Hg

= 210 mm Hg

• PV = nRT

He

2102.24 n 0.0821 273

760

Hen = 0.028

18

NSEC 2019-20

80. Lipoic acid with the following structure is a growth factor

required by many organisms. Percentages of 'S' and

'O' in lipoic acid respectively are (atomic masses of 'S'

and 'O' are 32.065 g mol–1 and 15.999 g mol–1

respectively)

S S

O

OH

Lipoic acid

(A) 33.03, 16.48 (B) 31.11, 18.24

(C) 31.11, 15.52 (D) 31.42, 15.68

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Answer (C)

Sol. Molecular formula of lipoic acid = C8H

14O

2S

2

molecular mass of lipoic acid

= 12 × 8 + 14 × 1 + 2 × 15.999 + 2 × 32.065

= 206.128

64.13% of S 100 31.11%

206.128

2 15.999% of O 100 15.52%

206.128