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CHEMISTRY NSEC 2016-2017 EXAMINATION

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-5151200Website : www.careerpoint.ac.in, Email: [email protected]

CAREER POINT

ASSOCIATION OF CHEMISTRY TEACHERSNATIONAL STANDARD EXAMINATION IN CHEMISTRY 2016-2017

Date of Examination 27th November 2016[Ques. Paper Code : C323]

1. The kinetic energy of an electron that has a wavelength of 10 nm is

(a) 2.4 × 10–21 J (b) 4.8 × 10–21 J (c) 2.4 × 10–29 J (d) 4.8 × 10–29 J

Ans. [a]

Sol. According to debroglie

mv

h

m

h

Kinetic energy = 2mv2

1

22

2

m

hm

2

1

2

2

m

h

2

1

227

234

)10(101.9

)10626.6(

2

1

= 2.4 × 10–21 Joules

2. Which of the following compounds contain 3-centered 2-electron bonding ?

(i) [BeF2]n (ii) [Be(CH3)3]n (iii) [BeCl2]n (iv) [BeH2]n

(a) (i) and (ii) (b) (ii) and (iii) (c) (ii) and (iv) (d) (iii) and (iv)

Ans. [c]

Sol. (i) [BeF2]n (ii) [Be(CH3)3]n (iii) [BeCl2]n (iv) [BeH2]n

(i) [F – Be – F]n

F

Be

FBe

F :

F F

BeF

(3c–4e–)

:

: :

[Be(CH3)2]n BeCH3

BeCH3

BeCH3 CH3

(3c – 2e–)

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CAREER POINT

[BeCl2]n

Cl

Be

ClBe

Cl

Cl Cl

BeCl

(3c–4e–)

[BeH2]n

BeH

BeH

Banana bonding

(3c–2e–)

3. 3-Methypentane on monochlorination gives four possible products. The reaction follows free radical

mechanism. The relative reactivities for replacement of –H are 3°:2°:1° = 6:4:1.

3-Methyl pentane

Cl2

light

Cl

A +

ClB

Cl

C +Cl D

Relative amounts of A, B, C and D formed are

(a) 6/31, 16/31, 6/31, 3/31 (b) 16/31, 6/31, 6/31, 3/31

(c) 6/31, 16/31, 3/31, 6/31 (d) 6/31, 3/31, 6/31, 16/31

Ans. [c]

Sol.

3-Methyl pentane

1°H = 92°H = 43°H = 1

C–C–C–C–C

Cl

C

(1)

C–C–C–C–C

Cl

C

(2)

C–C–C–C–C

Cl

C

(3)

C–C–C–C–C

C–Cl

(4)

Cl2

16

9

44

19

y

x

2

1

6

16

61

44

z

y

3

2

% of x = 1006169

9

% of y = 100

31

16 % of z = 100

31

6

3 / 35



CAREER POINT

Relative amounts

of A,B,C,D are31

6,

31

3,

31

16,

31

6

(A) (B) (C) (D)

4. White phosphorous on reaction with NaOH gives PH3 and

(a) Na2HPO3 (b) NaH2PO2 (c) NaH2PO3 (d) Na3PO4

Ans. [b]

Sol. P4 + NaOH NaH2PO2 + PH3

Phosphene

5. Given the E0 values for the half reactions :

Sn4+ + 2 e– Sn2+, 0.15 V

2Hg2+ + 2 e– 22Hg , 0.92 V

PbO2 + 4 H+ + 2 e– Pb2+ + 2 H2O, 1.45 V

Which of the following statements is true ?

(a) Sn2+ is a stronger oxidizing agent than Pb4+ (b) Sn2+ is a stronger reducing agent than 22Hg

(c) Hg2+ is a stronger oxidizing agent than Pb4+ (d) Pb2+ is a stronger reducing agent than Sn2+

Ans. [b]

Sol. Lower the value of Reduction Potential Higher will be it oxidizing nature i.e. strong Reducing Agent.

6 For the conversion CCl4() CCl4(g) at 1 bar and 350 K, the correct set of thermodynamic parameters is

(Boiling point of CCl4 is 77 °C)

(a) G = 0, S = +ve (b) G = 0, S = – ve (c) G = – ve, S = 0 (d) G = – ve, S = +ve

Ans. [a]

Sol. CCl4)() CCl4(g)

P = 1 bar ; T = 350 K; Boiling point of CCl4 = 77ºC = 350 KG = 0G = H – TS0 = H – TSTS = H

S =T

H

= +ve

7. How many isomers are possible for complex [Co(ox)2Cl2]+ ?

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

Ans. [b]

Sol.

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CAREER POINT

[Co(ox)2Cl2]

[M(AA)2a2]n±type

component

ClCo

Cl

O

O

O

CO

C O

(Cis)+

O C OC

O

Optically Active (2)

OCo

O

Cl

O

OC O

O C O

CO

CO (Trans) Optically Inactive (1)(3)

8. The compound that will not react with silver perchlorate under normal conditions is

(a) 3-bromocyclopropene (b) tetraethyl ammonium chloride

(c) tetramethylammonium hydroxide (d) polyvinyl chloride

Ans. [d]

Sol.(a) Br

AgClO4 + AgCl

(b)AgClO4 AgCl[(Et)4N]

+Cl–

[(Et)4N]+

+

(c) AgClO4[(Me)4N]+

OH–

[(Me)4N]ClO4 + AgOH (Acid base reaction)+

(d) CH2=CH–Cl [CH2–CH]n

AgClO4 No reaction

Cl

9. The conductivity of 0.10 M KCl solution at 298 K is 1.29 × 10–2 S cm–1. The resistance of this solution is

found to be 28.44 . Using the same cell, the resistance of 0.10 M NH4Cl solution is found to be 28.50 .

The molar conductivity of NH4Cl solution is S cm2 mol–1 is

(a) 0.130 (b) 13 (c) 130 (d) 1300

Ans. [c]

Sol. K= Ga

1.29 × 10–2 =a44.8.2

1

0.367 =a

Cell constant = 0.367

m =c

k× 1000

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CAREER POINT

10.0

367.050.28

1

× 1000

m = 130 Scm2 mol–1

10. Consider a compound CsXY2 where X and Y are halogens. Which of the following statement/s is/are correct ?

(i) X and Y have different oxidation states.

(ii) For Y with lower atomic number than X, X can assume oxidation states higher than normal.

(iii) Such compounds exist because Cs+ has a high charge to size ratio.

(a) Only (i) (b) (i) and (ii) (c) Only (ii) (d) (i) and (iii)

Ans. [b]

Sol. Cs XY2 where X and Y are halogens

Cs+ (XY2)– O.S of Cs is + 1

O.S of X is + 1O.S of Y is –1

XY2 = –1

x + (–1)2 = –1

x = + 2 –1

= + 1

11. Match the compounds given in list I with their characteristic reactions in list II

List I (Compound) List II (Reaction)

1 Tertbutyl amine a Liberation of ammonia on heating with aq NaOH

2 2-methyl-2-pentanol b Effervescence with NaHCO3

3 2,4,6-trinitrophenol c Foul smell with chloroform in alkaline condition

4 Cyclohexane carboxamide d Formation of an water insoluble compounds on treatment with conc. HCl and ZnCl2

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

Ans. [b]

Sol. (1) Tert. butyl amine (c)

C–C–NH2

C

CCHCl3

OH

C–C–N==C

C

C

foul smell withchloro form & base

(2) H3C–C–CH2–CH2–CH3

OH

CConc. HCl

ZnCl2

Write turbidityproduce with inseconds (lucas test)

(d)

(3)

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CAREER POINT

NaHCO3 gives effervescenceof CO2 gases

(b)

NO2

OH

O2N

NO2

Picric acid NO2

O Na

O2N

NO2

+ H2O + CO2

(4)

aq. NaOHC–NH2

O

(a)C–OH + NH3

Ammoniagas

Cyclohexane Carboxamide

O

12. Which of the following statements is not correct regarding the galvanic cells ?

(a) Oxidation occurs at the anode.

(b) Ions carry current inside the cell

(c) Electrons flow in the external circuit from cathode to anode

(d) When the cell potential is positive, the cell reaction is spontaneous

Ans. [c]

Sol. Electrons flow in the external circuit from anode to cathode

CathodeAnodeoxidationhalf cell

Reductionhalf cell

G

ne

Mn+

Anode to cathode

13. L-Fucose with the following planar representation is a sugar component of the determinants of the A, B, O

blood group typing

H

CH3

O

H

OH

OH

H

1

5

4

H

H

OH

23

HO

The open chain structure of L-Fucose can be represented as

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CAREER POINT

(a)

CHOHOHOH

HOHH

CH3

OHH(b)

CHOOHHH

HHOHO

CH3

HHO(c)

CHOHOHOH

HOHH

CH3

HHO(d)

CHOOHOHOH

HHH

CH3

HHO

Ans. [c]

Sol.H

CH3

O

H

OH

OH

H

1

5

4

H

HOH

OH

23

HO H

H OH

H OH

HO H

CHO

CH3

L-Fucose L-Fucose

14. In ammonia the bond angle is 107º 48º while in SbH3 the bond angle is about 91º 18'. The correct explanation

among the followings is /are

(a) The orbitals of Sb used for the formation of Sb-H bond are almost pure p-orbitals.

(b) Sb has larger size compared to N.

(c) Sb has more metallic character than N.

(d) All the statements are correct.

Ans. [a]

Sol. Given Bond angle of ammonia is 105°, 48' while SbH3 is 91°, 18'According to DRAGO's Rule down the group after 3rd period bond angle almost equal approx 90° surround.We know that bond angle decrease.Then % of p-character increases

cos =I–s

s cos 90° = 0

Then % of s = 0Then meaning = 91°, 18'per % of salmost negligibleHence % of p is pure

15. Equal masses of ethane and hydrogen gas are present in a container at 25ºC. The fraction of the total pressure

exerted by ethane gas is

(a) 11.2 dm3 (b) 5.6 dm3 (c) 4.48 dm3 (d) 22.4 dm3

Ans. [b]

Sol. Number of moles of C2H6 & H2 are30

W and

2

W respectively

8 / 35



CAREER POINT

Mole fraction of C2H6( 62HCX ) =

2W

30W

30W

=

16

1

If total pressure in container is P then

Fraction of the total pressure exerted by C2H6 =16

P

16. The volume of nitrogen evolved on complete reaction of 9 g of ethylamine with a mixture of NaNO2 and

HCl at 2.73ºC and 1 atm pressure is

(a) 11.2 dm3 (b) 5.6 dm3 (c) 4.48 dm3 (d) 22.4 dm3

Ans. [Bonus ]

Sol. CH3 –CH2 –NH2HCl

NaNO2 CH3–CH2OH + N2

mole45

9

45

9

45

9

mole mole

v =p

nRT =

1

5460821.2.0 = 8.96L

Ans. is wrong

correct ans is 8.96L, Not given in option

If temperature consider as 273K. Then volume liberated is 4.48 dm3

17. The electrons identified by quantum numbers n and l, (i) n = 4, l = 1, (ii) n = 4, l = 0, (iii) n = 3, l = 2, (iv) n =

3, l = 1 can be placed in order of increasing energy from lowest to highest as

(a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii)

(c) (i) < (iii) < (ii) < (vi) (d) (iii) < (i) < (iv) < (ii)

Ans. [a]

Sol. value of (n + )

(i) n = 4, = 1 4p 5

(ii) n = 4, = 0 4s 4

(iii) n = 3, = 2 3d 5

(iv) n = 3, = 1 3p 4

According to (n + ) rule lowest value of (n + ) is in 4s & 3p but 3p has lower value of 'n' so it has lowestenergy so (i) > (iii) > (ii) > (iv)

18. Spodoptol, a sex attractant, produced by a female fall armyworm moth, can be prepared as follows.

The structure of Spodoptol is (pka : terminal alkynes ~ 25, alcohols ~ 17)

HO CH2 (CH2)7 C CH

i. NaNH2 (excess)ii. n–C4H9Br (1 eq)iii. H2/Lindlar cat

Spodoptol

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CAREER POINT

(a) n – C4H9 – O – CH2 – (CH2)7 HC CH2

(b)

HH

HO–CH2–(CH2)7 n–C4H9

(c)

H

H

HO–CH2–(CH2)7

n–C4H9

(d) HO – CH2 – (CH2)12 – CH3

Ans. [b]

Sol.

HO–CH2–(CH2)7 CCHNaNH2(excess)

aNO – CH2–(CH2)7 CC Na

—

—

–Br CH3–CH2 –CH2–CH2

|Br

(1eq)

NaO–CH2–(CH2)7 CC–CH2–CH2–CH2–CH3—

H2 Lindlar catalyst

NaO–CH2 (CH2)7

C=CnC4H9

HH

Or H

HO–CH2 (CH2)7

C=CnC4H9

HH

–

–

19. Passing H2S gas into a mixture of Mn2+, Ni2+, Cu2+ and Hg2+ in an acidified aqueous solution precipitates

(a) CuS and HgS (b) MnS and CuS (c) MnS and NiS (d) NiS and HgS

Ans. [a]

Sol.

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CAREER POINT

H2S gasMn2+ or Ni2+ or Cu2+ or Hg2+

eu dks tkuk ughaMnS

Puff

MnS

Blue

CuS

white

NiS

Blue

But H2S in acidic aq solution Pb+2 Cu2+ Cd2+ Bi+3 Hg+2 Cd+2 Hence ans. is CuS and HgS

OH–

Basic medium

20. Battery acid (H2SO4) has density 1.285 g cm–3. 10.0 cm3 of this acid is diluted to 1 L. 25.0 cm3 of this diluted

solution requires 25.0 cm3 of 0.1 N sodium hydroxide solution for neutralization. The percentage of sulphuric

acid by mass in the battery acid is

(a) 98 (b) 38 (c) 19 (d) 49

Ans. [b]

Sol. density = 1.285

d =vol

mass =

10

mass = 1.285

w = 12.85 gmH2SO4 + 2NaOH Na2SO4 + 2H2O25mL 25mL

0.1 NNumber of milli equivalents = Number of milli equivalents of NaOHof H2SO4

= 25 × 0.1milliequivalents = 2.5millimoles × 2 = 2.5

millimoles of H2SO4 =2

5.2

= 1.25

42SOHW = 1.25 × 10–3 × 10–3 × 98 × 40

= 4.9g

% H2SO4 =85.12

9.4 × 100

= 38.13 %

21. The compound that reacts fastest with methylamine is

(a)

CH2Br

Br

(b)

Br

Br

(c)

Br

(d)Br

Br

Ans. [b]

Sol. React faster with methyl amine

|NH2 – CH3

Br

|Br

|

NHCH3

|NHCH3

SN1

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CAREER POINT

22. HgO is prepared by two different methods : one shows yellow colour while the other shows red colour. The

difference in colour is due to difference in

(a) electronic d–d transitions (b) particle size

(c) Frenkel defect (d) Schottkey defect

Ans. [b]

Sol.

23. The pH of a 1.0 × 10–3 mol L–1 solution of a weak acid HA is 3.60. The dissociation constant of the acid is

(a) 8.4 × 10–8 (b) 8.4 × 10–6 (c) 8.4 × 10–5 (d) 8.4 × 10–2

Ans. [c]

Sol. HA H+ + A–

t = 0 C 0 0teq C–C C C

pH = 3.60

[H+] =4

10 3

[H+] = C

33

1014

10

= 0.25

Ka =

1

C 2

25.01

)25.0(101 23

= 8.4 × 10–5

24. The best sequence of reactions for preparation of the following compound from benzene is

(H3C)2HC COCH3

HO3S

(a) (i) CH3COCl/AlCl3 (ii) Oleum (iii) (CH3)2CH-Cl (1 mole)/AlCl3

(b) (i) (CH3)2CH-Cl (1 mole)/AlCl3 (ii) CH3COCl/AlCl3 (iii) Oleum

(c) (i) Oleum (ii) CH3COCl/AlCl3 (iii) (CH3)2CH-Cl (1 mole)/AlCl3

(d) (i) (CH3)2CH-Cl (1 mole)/AlCl3 (ii) Oleum (iii) CH3COCl/AlCl3

Ans. [b]

Sol.

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CAREER POINT

(1)(CH3)2CH–Cl/AlCl3

(2) CH3–C–Cl/AlCl3

O(3) Oleum

COCH3

SO3H

1Mole (CH3)2CH–Cl/AlCl3

CH3 – C – Cl

O

C – CH3

O

H2S2O7/(Oleum)

AlCl3

25. Which reaction is spontaneous at all temperatures at standard pressure and concentration?

(a) exothermic reaction with a decrease in entropy.

(b) exothermic reaction with an increase in entropy

(c) endothermic reaction with a decrease in entropy

(d) endothermic reaction with an increase in entropy

Ans. [b]

Sol. G = H – TS

Reaction is spontaneous at all temperature when

G < 0

G is always negative at

H < 0 Exothermic Reaction

S > 0 Increase in Entropy

26. The IUPAC name of the following compound is

CONH2

(a) 3-Aminocarbonylpent-1-en-4-yne

(b) 2-Ethenylbut-3-yn-1-amide

(c) 2-Ethynylbut-3-en-1-amide

(d) 3-Aminocarbonylpent-4-en-1-yne

Ans. [c]

Sol.

13 / 35



CAREER POINT

H – C C – CH – C – NH2

CH = CH2

3

2

4

1

O2- Ethynyl but – 3 – en- 1 - amide

27. NaOH solution is added dropwise to HCl solution and the conductance of the mixture is measured after

addition of each drop. The variation of conductance with volume of NaOH added is as shown below.

The statement that is not true for the above is

ZX

Con

duct

ance

YVolume of base added

(a) decrease in conductance from XY is due to decrease in [H+].

(b) point Y represents the equivalence point of titration.

(c) Na+ has the higher equivalence conductance than H3O+.

(d) segment YZ represents the conductance due to ions from NaCl and NaOH in solution.

Ans. [c]

Sol.

28. A colorless water-soluble compound on strong heating liberates a brown colored gas and leaves a yellow

residue that turns white on cooling. An aqueous solution of the original solid gives a white precipitate with

(NH4)2S. The original solid is

(a) Zn(NO3)2 (b) Ca(NO3)2 (c) Al(NO3)3 (d) NaNO3

Ans. [a]

Sol. NaNO3 NaNO2 + O2

Ca(NO3)2NO2+ CaO + O2

Zn(NO3)2ZnO + 2NO2O2

(NH4)2SZnSWhite ppt

Al(NO3)2 Al2O3 + O2

29. The following compounds are heated (i) KNO3, (ii) Cu(NO3)2, (iii) Pb(NO3)2, (iv) NH4NO3

Which of the following statement/s is/are correct ?

(a) (ii) and (iii) liberate NO2 (b) (iv) liberates N2O

(c) (i), (ii) and (iii) liberate O2 (d) All statements are correct.

Ans. [d]

Sol. KNO3 KNO2 + O2

Cu(NO3)2 NO2+ CuO + O2

Pb(NO3)2PbO + NO2 + O2

NH4NO3N2O + H2O

14 / 35



CAREER POINT

30. The diastereoisomer (stereoisomer that is not a mirror image) of 'X' is

HO H

H OH

COOCH3

COOH

'X'

(a)

COOH

HHO

HO H

COOCH3

(b)

COOH

OHHHO H

COOCH3

(c)

HOH

HOOCHO

COOCH3

H

(d) H

OH

HOOC

HO

COOCH3

H

Ans. [c]

Sol.COOCH3

COOH

H

H

OH

OH

(X)COOCH3

H

H

HO

HO

COOH

(A)

COOCH3

COOH

H

H

HO

OH

( Identical )

15 / 35



CAREER POINT

COOCH3

H

OH

HO

H

H

(B) ( Enantiomer)

OH

COOH

H

H

H3COOC

OH

(C)

COOCH3

COOH

OH

H

H

OH

( Diastereomer )

COOCH3

H

H

HO

HO

H

(D)

COOCH3

COOH

H

H

HO

OH

( Identical )

OH COOCH3

HO HCOOH H

31. Given rHº = –54.08 kJ mol–1 and rSº = 10.0 J mol–1 at 25ºC, the value of log10K for the reaction A B is

(a) 3.4 (b) 10 (c) 0.53 (d) 113

Ans. [b]

Sol. G0 = H0 – TS0 = –2.303 RT log10 k

32. Which of the complexes has the magnetic moment of 3.87 B.M.?

(a) [Co(NH3)6]3+

(b) [CoF6]3–

(c) [CoCl4]2–

(d) [Co(dmg)2] square planar complex (dmg = dimethyl glyoxime)

Ans. [c]

Sol. (a) [Co(NH3)6]3+ (b) [CoF6]

3–

(c) [CoCl4]2– (d) [Co(fmg)2] square planer cample

(a) [Co(NH3)6]3+

x + 0 × 6 = +3

x = +3

16 / 35



CAREER POINT

Co27

4s2 3d7

36C 4s0 3d6

3d6

NH3 NH3

G.S.

Eos

NH3

4s2

NH3

4p

NH3 NH3

= 0

Hybridization is d2sp3

Dia magnetic

NH3 is strong field ligand

Similar In[Co F6]3– paramagnetic (4)

Co+3 WFL n = 4 = 24and In[CoCl4]

2– Co+ 4sº3d7

Cl– in WFL3d

G.S.

3d 3d

3d

E.S.

3d 3d×

××

××

××

×

Cl– Cl– Cl– Cl–

sp3

n = 3

= 15

= 3.87 BM paramagnetic

(d) [Co(dmg)2] x + (–1) ×2 = 0 n = +2 dmg in SFL

27Co 4s23d7

Co+2 4s05d7

pairing

dsp2

n = 1

= 3 = 1.732 BM

33. For a gaseous reaction, A + B Products, the energy of activation was found to be 2.27 kJ mol–1 at 273 K.

The ratio of the rate constant (k) to the frequency factor (A) at 273 K is

(a) 0.368 (b) 3.68 (c) 4.34 (d) 0.434

Ans. [a]

Sol. K = Ae–Ea/RT

RT/EaeA

K

273314.8

2270

eA

K

17 / 35



CAREER POINT

1eA

K

e

1

A

K

718.2

1

A

K

3679.0A

K

34. In the case of dibromo derivatives of the following compound, the derivative having highest energy has the

bromo substituents in positions

213

4

56 7

8

910

(a) 1,2 (b) 2,3 (c) 4,5 (d) 1,10

Ans. [d]

Sol.2

3

4

5

6 7

8

9

10

1 For electrophillic attack (Br+)

35. The ionization energy of a certain element is 412 kJ mol–1. When the atoms of this element are in the first

excited state, however, the ionization energy is only 126 kJ mol–1. The region of the electromagnetic

spectrum in which the wavelength of light emitted in a transition from the first excited state to the ground

state is

(a) Visible (b) UV (c) IR (d) X-ray

Ans. [a]

Sol.

126 kJ/ mole

286 kJ/ mole412 kJ/mole

n =

n = 2 (1st excited level-1)

n = 1(ground stale)when electron jump from 1st excited state to ground state then 286 KJ/mole energy is released which isreleased with visible range So ans is A

18 / 35



CAREER POINT

E =hc

l = 4.15 × 10–7 m visible range.

36. The reaction of an olefin with HBr can proceed by ionic as well as radical mechanism. The reaction in the

presence of light takes place by radical mechanism, as

(a) the free energy of the reaction in radical mechanism is higher than in ionic mechanism

(b) ionic mechanism requires a catalyst while radical mechanism does not.

(c) in the presence of light the activation energy of the reaction is lower than that for ionic mechanism.

(d) a radical reaction has very low activation energy as compared to that for the corresponding ionic reaction.

Ans. [c]

Sol. A

37. The correct statement/s is/are

I. Soap is excellent for cleaning, 100 % broken down by bacteria in rivers and hence has no further

environmental damaging repercussions.

II. Soap forms an insoluble precipitate/scum when hard water containing calcium and magnesium ion is

used.

III. Soaps can be used for cleansing under acidic solutions.

(a) Only I (b) Only II (c) only III (d) I and III

Ans. [b]

Sol.

38. The kinetic data recorded at 278 K for the reaction

4NH (aq) + –

2NO (aq) N2(g) + 2H2O (l) is

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

The kinetic rate expression and the unit of rate constant (k) of the above reaction are respectively

(a) k ]NH[ 4 ]NO[ –

2 and Ms–1 (b) k ]NH[ 4 and s–1

(c) k ]NH[ 4 ]NO[ –

2 and M–1s–1 (d) k ]NO[ –2 and s–1

Ans. [c]

Sol. K [0.24]x [0.10]y = 7.2 × 10–6 –––––––(1)K [0.12] x [0.10]y = 3.6 × 10–6 –––––––(2)K [0.12] x [0.15]y = 5.4 × 10–6 –––––––(3)

Dividing 1 by 2 2x = 2

x = 1Dividing 3 by 2

y

y

10.0

)15.0( =

6

6

106.3

104.5

Set No. ]NH[ 4 /M ]NO[ –

2 /M Rate of reaction/Ms–1

1 0.24 0.10 7.2 × 10–6

2 0.12 0.10 3.6 × 10–6

3 0.12 0.15 5.4 × 10–6

19 / 35



CAREER POINT

2

3= 3/2

So order of reaction is 2unit (M sec–1) 1–2

= M–1 sec–1

39. Which of the following statements is not true for 3Ce and 3Yb ?

(a) Both absorb in UV region (b) Both show f–f transition

(c) Both show 4f to 5d transition (d) Both ions are colorless

Ans. [b]

Sol.

cerium-ion Ytterbium-ion

3+ 3+

Ce and Yb

Lanthanide-Series

La Ce Pr Nd Pm

57 58

Ac Th

89 90

Statement (a) Correct and also (d) adsorb light in the near infrared region -

58Ce1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6

Similar Yb3+ 6s2 4f2 5d0 / 6s2 4f1 5d1 similar for Yb+3 both show 4f to 5d transition but not

show f-f-transition.

40. Complete catalytic hydrogenation of naphthalene gives decalin (C10H18). The number of isomer of decalin

formed and the total number of isomers of decalin possible are respectively

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

Ans. [b]

Sol.

H2/Pt

Cis decalin

H

H

trans decalin

H

H

H2/Pt

20 / 35



CAREER POINT

41. The mass of argon adsorbed per unit mass of carbon surface is plotted against pressure. Which of the

following plots is correct if x and m represent the masses of argon and carbon respectively ? (----- represents

extrapolated data)

(a)

P

x/m(b)

P

x/m

(c)

P

x/m(d)

P

x/m

Ans. [d]

Sol. Fact

42. In a process n-propyl chloride is reacted with sodium butanoate in an aqueous medium. After the reaction

diethyl ether is added and the solution is shaken. The two layer are separated. The incorrect statement with

respect to this procedure is

(a) The reaction gives a solid product which precipitates in the aqueous solutions.

(b) The reaction takes place in the aqueous medium.

(c) The product is extracted in diethyl ether and the organic layer is the upper layer.

(d) The salt formed in the reaction remains in aqueous medium.

Ans. [a]

Sol.

NaCl +

CH3 – CH2 – CH2–Cl + Na O – C – CH2 – CH2 – CH3

O

n– propyl chloride Sod. Butanoate

CH3 – CH2 – CH2 – O – C– CH2 – CH2 – CH3

Soluble in Aq .medium

n – propyl Butanoate [Ester]

organic comp present in upperlayer. Salt formed in reactionremain in aq. medium

O

43. Which of the following statements about ammonium cerium (IV) nitrate, (NH4)2[Ce(NO3)6] is false ?

(a) –3NO acts as a monodentate ligand

(b) The Ce atom has a coordination number of 12

21 / 35



CAREER POINT

(c) The shape of the complex ion is icosahedron.

(d) The solution is used as oxidizing agent.

Ans. [a]

Sol. (NH4)2 [Ce(NO3)6] Ammonium Cerium nitrate 2NH4+ + [Ce(NO3)6]

–2

+1 × 2 + x + (–1) 6 = 0

x = + 6 – 2 = +4

Ce

O

O

O

O

NO

N O

ON O

O

ONO

OO

ONO

3NO acts as bidentate ligand

The cehas co-ordination number is 12.

This is oxidizing agent inorganic synthesis

And shape is icosheedral

44. The correct order of the magnitude of bound energy (kJ/mol) of the central C-C bond in the following

compounds is

(i) CH2=CH–CH=CH2 (j) Me3C–CPh3 (k) MeCO–CO–Me (l) CH ≡C–C≡CH(a) k > i > l > j (b) j > k > l > i (c) i > j > k > 1 (d) 1 > i > k > j

Ans. [d]

Sol. Bond Energy order for C – C central BondCH C – C CH > CH2 = CH – CH = CH2 > Me – C – C – Me

|| ||OO

> Me3C – CMe3

l > i > k > j

45. Which one of the following information about the compounds is correct ?

Compounds Oxidation

state of P

No. of P–OH

bonds

No. of P–H

bonds

No. of P = O

bonds

[I] H3PO2

Hypophosphorous acid

1+ 2 1 0

[II] H4P2O5

Pyropophosphorous acid

3+ 2 2 2

[III] H4P2O6

Hypophosphoric acid

4+ 2 2 2

[IV] H4P2O`

pyrophosphoric acid

5+ 3 1 4

22 / 35



CAREER POINT

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

Ans. [d]

Sol.No. of P-OH No. of P-H No. of P = O bent

H3PO2

+1H – P – O – H

H

O

1 2 1

H4P2O5

+3H – O – P – O – P – O – H

H

O

H

O

2 2 2

H4P2O6

+4H – O – P – P – O – H

O–H

O

O–H

O

4 0 2

H4P2O7

+5H – O – P – O – P – O – H

O–H

O

O–H

O

4 0 2

46. The method of preparation of 2-benzyloxynaphthalene is a base catalysed reaction of -

(a) benzyl chloride and 1-naphthol (b) 1-chloromethylnaphthalene and phenol

(c) 1-chloronaphthalene and benzyl alcohol (d) benzyl alcohol and 1-naphthol

Ans. [NA]

Sol.

47. The pair that is isostructural (i.e. having the same shape and hybridization) is

(a) NF3 and BF3 (b) 4BF and

4NH (c) BCl3 and BrCl3 (d) NH3 and 3NO

Ans. [b]

Sol. Pair of isostructural

(a) NF3

sp3

BF3

sp2

and Not isostructural

(b) 4BF and

4NH Same hybridisation and same bond pair and no-lone pair Hence same shape (isostructural)

(c) BCl3

sp2

BrCl3

B.P. = 3L.P. = 2


plane

5 = sp3d

T-shape

23 / 35



CAREER POINT

(d) NH3

sp3

3NO

sp2


B.P. = 3L.P. = 1

O = N = O

O

48. A group which departs from the substrate in a a nucleophilic substitution reaction is called a leaving group.

The case of departure is determined by the acidity of the conjugate acid of the leaving group; higher the

acidity better is the leaving group. The correct order of the reactivity of the following compounds in a given

nucleophilic reaction is -

(a) R-Cl > R-OCOCH3 > R-OSO2CH3 > RI (b) R-OSO2CH3 > R-Cl > R-OCOCH3 > ROH

(c) R-I > RNH2 > R-OCOCH3 > R-OSO2CH3 (d) R-Br > R-OSO2CH3 > R-OCOCH3 > ROCH3

Ans. [b]

Sol. Leaving group ability

O–S–CH3 > Cl > O –C–CH3 > OH

O O

O ––––

Acidic strength :

CH3–S–OH > H Br > CH3–C–OH > H–O–H

O

O

O

49. Compound 'X' in the following reaction is -

CH3

COCH3

i. Ozonolysisii. aq. KOH, heat

X

(a)CH2CH3

CH3

(b) CH2CH3

H3C

(c)

CH3

CH3

(d)CH2CH3

Ans. [c]

Sol.

24 / 35



CAREER POINT

CH3

CH3

O3|Zn|HOOzonolysis

O

O

KOH

O

–

Intra molecularAldol

CH3

OH

–H2O

OO O

50. A toxic element is to be removed from drinking water by adsorption on activated charcoal. At low

concentrations, the rate constant for adsorption is 1.8 × 10–5 s–1. The time required to reduce the

concentration of the toxic element to 10% of its initial concentration is

(a) 1.28 × 105 s (b) 5.85 × 103 s

(c) 1.28 × 106 s (d) cannot be calculated from the given data.

Ans. [a]

Sol. K = t

2.303 log

t

0

C

C

k =t

2.303 log

0.1a

a

t =5–101.8

2.303

log 10

t = 1.28 × 105

51. Assuming that Hund's rule is violated by the diatomic molecule B2, its bond order and magnetic nature will

be respectively

(a) 1, diamagnetic (b) 1, paramagnetic

(c) 2, diamagnetic (d) 2, paramagnetic

Ans. [a]

Sol. B2 5 × 2 = 10 e–

Case – IN 14 * 1s2 1s2 * 2s2 2p1

x = 2p1y 2pz

But in given question Hund's Rule is voletedThen configuration is 1s2 * 1s2 2s2 * 2s2 2p2

x = 2p0y 2p0

z

Hence B2 is Diamagnetic

But B.O. is =21

(Nb – Na)

=21

(6 – 4)

= 1

25 / 35



CAREER POINT

52. In a cubic crystal structure, divalent metal-ion is located at the body-centered position, the smaller tetravalent

metal ions are located at each corner and the O2– ions are located half way along each of the edges of the

cube. The number of nearest neighbors for oxygen is

(a) 4 (b) 6 (c) 2 (d) 8

Ans. [b]

Sol. Since the nearest distance is a/2 so the No. of nearest atom will be six

53. Organic compounds sometimes adjust their electronic as well as steric structures to attain stability. Among

the following, the compound having highest dipole dipole moment is

(a)Br

(b)

Br

(c)

Br

(d)

Br

Ans. [d]

Sol.

Br

ionisation

+

+ Br

6 e–

Aromaticcyclic delocalisationof (4n +2) e–.So more ionic nature

HigherDipole moment

54. Cyanide ion is a very good complexing agent and also functions as a reducing agent. Hence many cyanide

complexes of metals are known. Addition of an aqueous solution of KCN to a solution of copper sulphate

yields a white precipitate which is soluble in excess of aqueous of aqueous KCN to form the complex :

(a) [Cu(CN)4]1– (b) [Cu(CN)4]

2– (c) [Cu(CN)4]3– (d) [Cu(CN)4]

4–

Ans. [c]

Sol. CuSO4 + 2 KCN K2SO4 + 2

2

)CN(Cu

Cu2 (CN)2 + (CN)2

or or

])CN(Cu[K 4

1

3

KCN3 Cu CN NC – CN

55. When a certain metal was irradiated with light of frequency 3.2 × 1016 Hz, the photoelectrons emitted had

twice the kinetic energy as did the photoelectrons emitted when the same metal was irradiated with light of

frequency 2.0 × 1016 Hz. The 0 of the metal is

(a) 2.4 × 1016 Hz (b) 8.0 × 1016 Hz (c) 8.0 × 1015 Hz (d) 7.2 × 1016 Hz

Ans. [c]

Sol. h1 = h0 – K.E1 1 = 3.2 × 1016

h2 = h0 – K.E2 2 = 2 × 1016

K.E1 = 2 × K.E1

)–(hE.K

)–(hE.K

022

011

26 / 35



CAREER POINT

1

2 =

)–(

)–(

02

01

22 – 20 = 1 – 0

22 –1 = 0

2 × 2 × 1016 – 3.2 × 1016 = 0

0 = 8 × 1015 Hz

56. The major product 'S' of the following reaction sequence is

Br2/FeP

Sn/HClQ

excess Br2/H2O

i. NaNO2-H2SO4/0ºCii. CuBr,HBr

R S

NO2

(a)

Br

Br

BrBr

Br

Br

(b)

Br

BrBr

Br

Br

(c)

Br

Br

BrBr

(d)

Br

BrBr

Br

Ans. [b]

Sol.

Br2/Fe

NO2 NO2

Br

Sn./HCl

NH2

Br

Br

Br

BrNH2

Br

(i) NaNO2–H2SO4/0°C(ii) Cu–Br,HBr

Br

Br

BrBr

Br

57. 1.250 g of metal carbonate (MCO3) was treated with 500 mL of 0.1 M HCl solution. The unreacted HCl

required 50.0 mL of 0.500 M NaOH solution for neutralization. Identify the metal M

(a) Mg (b) Ca (c) Sr (d) Br

Ans. [b]

Sol. MCO3 + HCl 1.250 gm 0.1 m

Excess

mL500

HCl + NaOH O + NaClRemaining 0.5 50 mLNo. of equivalents of HCl = No. of equaivalents of NaOHn × 1 = 0.5 × 1 × 50

nHCl = 25 milli moles

27 / 35



CAREER POINT

Remaining.nHCl (Actual) = 0.1 × 500

= 50 millimolnHCl reacted with MCO3 = (50 – 25) millimoles

= 25 millimolesNumber of equilivalents of MCO3 = Number of equivalents of HCl

E

250.1 = 25 × 10–3 × 1

E = 100025

25.1

= 50Mol. wt = E × Af = E × 2 = 50 × 2 = 100 (So CaCO3)

58. An electron beam can undergo diffraction by crystals which proves the wave nature of electrons. The

potential required for a beam of electrons to be accelerated so that its wavelength becomes equal to 0.154 nm

is

(a) 63.5 V (b) 31.75 V (c) 635 V (d) 127 V

Ans. [a]

Sol. =V

150 in Aº

1.54 × 1.54 =V

150

V = 63.5V

59. A biodegradable alternating coploymer of L-alanine and glycolic acid (HO–CH2-COOH) is

(a)N

o

o

H3C

nH

OH

(b)o

n

o

N

CH3

H

H

o

(c) N

o

o

H3C

n

H

Ho (d)

o

n

o

N

H3C

H

H

o o

Ans. [a]

Sol.

O

HO

CH3H

NH2 + HO – C

O

CH2

O H

O

CH3H

NH – C

O

CH2

O

n

60. In which of the following complexes the metal ion has the lowest ionic radius ?

(a) [Ti(H2O)6]2+ (b) [V(H2O)6]

2+ (c) [Cr(H2O)6]2+ (d) [Mn(H2O)6]

2+

Ans. [b]

28 / 35



CAREER POINT

Sol. In all four complex 3d metals are in +2 state like Ti2+ , V2+, Cr2+ and Mn2+ So there ionic radius trends asfollow

110

80

Ioni

c ra

dius

(pm

)

70

100

90

60

Ca2+

Ti,2+

V2+Cr2+

Mn2+

Co2+

Fe2+

Ni2+

Zn2+

+Cu2+

increasing atomic NoLowest ionic radius = V2+

complex is 262 )OH(V

61. In cold climate, the water in a radiator of a car gets frozen causing damage to the radiator. Ethylene glycol is

used as an antifreezing agent. The amount of ethylene glycol that should be added to 5 kg of water to prevent

it from freezing at –7 ºC is

(Given) Kf for water = 1.86 K mol–1 kg ; Molar mass of ethylene glycol = 62 g mol–1)

(a) 1165 g (b) 46.7 g (c) 116.7 g (d) 93.4 g

Ans. [a]

Sol. Tf = i × kf × m

7 = 1× 1.865

n

nethylene glycol =86.1

57

62

w =

86.1

57

w = 1165 gm

62. The ratio of the energy of the electron in ground state of hydrogen atom to that of the electron in the first

exited state of Be3+ is

(a) 1:4 (b) 1 : 8 (c) 1: 16 (d) 4 :1

Ans. [a]

Sol. E =2

2

0n

zE

2

2

0Be 2

)4(EE 3

2

2

0H1

)1(EE = – 4E0

= – E0

4

1

E4

E

E

E

0

0

Be

H

3

29 / 35



CAREER POINT

63. Water insoluble, but organic solvent soluble, dye is dissolved in three organic solvents and taken in three

separating funnels, a, b and c. To each solution, water is added, shaken, and kept undisturbed. The solvents in

separating funnels a, b and c form the following figures are respectively

Organic solvent

a b c

(a) a: EtOH; b: CCl4 ; c: EtOAc (b) a: CCl4 ; b: EtOH ; c EtOAc;

(c) a : EtOAc; b: CCl4; c: EtOH (d) a : CCl4; EtOAc; c ; EtOH

Ans. [b]

Sol. (i) In picture a both are Immiscible and organic solvent is heavier So it should be CCl4

(ii) In picture b organic solvent is soluble So organic solvent should be alcohol(iii) In picture c ester is present because it is insoluble and less dense than water

64. P,Q R and S are four metals whose typical reactions are given below ;

(I) Only Q and R react with dilute HCl to give H2 gas.

(II) When Q is added to a solution containing the ions of the other metals, metallic P,R and S are formed.

(III) P reacts with concentrated HNO3 but S does not

The correct order of their reducing character is -

(a) S< P< R<Q (b) S< R <P < Q (c) R< Q < P < S (d) Q < P < S < R

Ans. [a]

Sol. According of obove given statements Q is bestReducing Agent then that of P.R and S

Hence [Ans is S<P<R<Q]

65. If a dilute solution of aqueous NH3 is saturated with H2S then the product formed is

(a) (NH4)2S (b) NH4HS (c) (NH4)2Sx (d) NH4OH + S

Ans. [b]

Sol. Dilute solution of aqueous NH3 is saturated with H2SNH3 + H2O NH4OHNH4OH + H2S NH4HS + H2O

66. Three Faradays of electricity are passed through aqueous solution of AgNO3, NiSO4 and CrCl3 kept in three

vessels using inert electrodes. The ratio (in moles) in which the metals Ag, Ni and Cr are deposited is

(a) 1: 2 : 3 (b) 3: 2 : 1 (c) 6: 3 : 2 (d) 2: 3 : 6

Ans. [c]

30 / 35



CAREER POINT

Sol. Number of moles =fnF

Q

pn

Qn

So.3AgNOn :

4NiSOn :3CrCln :

)CrCl(f)NiSO(f)AgNO(f 343n

1:

n

1:

n

1

3AgNOn :4NiSOn :

3CrCln =3

1:

2

1:

1

1

= 6 : 3 : 2

67. Nepetalactone (X) is isolated as an oil from Catnip.

O

O

CH3

(X)

The number of chiral carbon atoms and the amount of KOH consumed by 83 mg of Nepatalactone are

respectively

(a) 3, 50 mg (b) 2, 56mg (c) 3, 56 mg (d) 3, 28 mg

Ans. [d]

Sol.

O

O

**

*

CH3

3 chiral centreOH

O CH3

HO

O

HO

HC

O

KOH

166

1083 3– = 0.5 × 10–3 mole

5 × 10–4 mole of molecule require 5 × 10–4 mole of KOH

So WKOH = 5 × 10–4 × 56 = 28 × 10–3 g

moles mole wt = 28 mg

So Ans. is (d) 3. chiral centre and 28 mg of KOH

68. Number of P-S single bonds and P-S double bonds (P=S) in P4S10 are respectively

31 / 35



CAREER POINT

(a) 10, 6 (b) 16, 0 (c) 14, 2 (d) 12, 4

Ans. [d]

Sol. P4S10

PS S

S S

SS=PS

P=S

P||S

||S

Number of P-S Single bond is = (12)

and P-S Double bond is = (4)

69. If the solubility product of iron(III) hydroxide is 1.8 × 10–37, the pH of a saturated solution of iron(III)

hydroxide in distilled water is close to

(a) 4 (b) 5 (c) 7 (d) 9

Ans. [b]

Sol. Fe(OH)3 Fe3+ + 3OHs 3s

Ksp = 11 × 33 s s1 + 3

Ksp = 27s4

1.8 × 10–37 = 27 s4

S4 = 66.66 × 10– 40

S = 2.85 × 10–10

[OH–] = 3s = 3× 2.85× 10–10

= 8.55 × 10–10

pOH= 10– log 8.55= 10– 0.9319pOH = 9pH = 14 – 9 = 5

70. An alkyl halide (X) on reaction with ethanolic sodium hydroxide forms an alkene (Y) which on further

reaction with HBr gives the same alkyl halide . The alkene (Y) on reaction with HBr / peroxide followed by

reaction with Mg metal; followed by reaction with HCN produces an aldehyde(Z). Z is.

(a)CHO

(b)CHO

(c)

CHO

(d) CHO

Ans. [b,c]

Sol.

32 / 35



CAREER POINT

CH3 – CH – CH3

Br

X

EtO Na– +

CH3 – CH = CH2

Y

HBrCH3 – CH – CH3

Br

XHBr/R2O2

CH3 – CH2 – CH2

Br

MgCH3 – CH2 – CH2

MgBr(1) HCN(2) H3O

CH3 – CH2 – CH2 – CHO

(B)

CH3 – C – CH3

CH3

X

EtO Na– +

CH3 – C – CH3

HBrCH3 – C – CH3

CH3

Y

HBr/R2O2

H3C – CH – CH3

CH2 – Br

MgCH3 – CH – CH2

CH2MgBr

(1) HCN(2) H3O

+

CH3 – CH – CH3

(C)

Br

CH2

Br

CH2–CHO

71. HClO4 is a stronger acid than HClO. The correct statement is

(a) 4ClO ion is more stabilized than ClO

(b) 4ClO ion has higher hydration energy than ClO

(c) 4HClO is better solvated in water than HClO

(d) In HClO4, H is attached to Cl, while in HClO it is attached to O

Ans. [a]

Sol. HClO4 is a strongest acid than HClOcorrect statement are

HClO4 H+ + ClO4– O = Cl = O four R.S.

O

O–

H – O – Cl H+ + OCl–

Hence Statement (a) is correct72. For an elementary rearrangement reaction A P, the following data were recorded at 303 K, when

33 / 35



CAREER POINT

[P]0 = 0 :

Set No. [A]0/mol L–1 Rate of conversion of A / mol L–1 min–1

1. 0.340 0.100

2. 0.170 0.050

3 0.085 0.025

If the equilibrium constant of the reaction is 1.12 at 303 K, the rate constant for the reaction P A is -

(a) 0.263 min–1 (b) 0.294 min–1 (c) 0.526 min–1 (d) 0.588 min–1

Ans. [a]

Sol.Kf

KbA P

Kc =b

f

k

k

Order of reaction is 1Rate = kf [A] Rate = kf [A]0.100 = kf (0.34) 0.50 = kf (10.170)kf = 0.29 kf =0.29

kc =b

f

k

k

1.12 =bk

29.0

kb = 0.2626

73. Compound P on treatment with CH2N2 (diazomethane) produces compound Q. Compound Q on reaction

with HI produces two alkyl iodides R and S. Alkyl iodide S with higher number of carbon atoms on reaction

with KCN followed by hydrolysis gives 3-methylbutanoic acid. The compound P is

(a) 2-butanol (b) 1-butanol (c) 2-methyl-2-propanol (d) 2-methyl-1-propanol

Ans. [d]

Sol.

CH2N2

CH3–CH–CH2–O–CH3

CH3–I

CH3–CH–CH2–I

CH3

CH3–CH–CH2–OH

P

CH3

Mg / Dry ether

H3C–CH–CH2–C–OH

CH3 O

(1)HCN

(2) H3O+ H3C–CH–CH2MgI

CH3

2-methyl-1-propouolCH3

QHI

74. I2 reacts with aqueous Na2S2O3 to give Na2S4O6 and NaI. The products of reaction of Cl2 with aqueous

Na2S2O3 are

34 / 35



CAREER POINT

(a) Na2S4O6 + NaCl (b) NaHSO4 + HCl (c) NaHSO3 + HCl (d) NaHSO3 + NaCl

Ans. [b]

Sol. Na2S2O3 + I2 Na2S4O6 + 2NaI (aq)

Cl2 Water

NaHSO4 + HCl

75. The standard potentials (Eº) of 24 Mn/MnO and 2

2 Mn/MnO half cells in acidic medium are 1.51 V and

1.23 V respectively at 298 K. The standard potential of 24 MnO/MnO half-cell in acidic medium at the

same temperature is

(a) 5.09 V (b) 1.70 V (c) 0.28 V (d) 3.34 V

Ans. [b]

Sol.

MnO4

+7E1= ?

MnO2

+41.23 Mn2+

E =21

2211

nn

nEnE

1.51 =5

223.13E1

7.55 – 2.46 = 3E1

E1 = 1.69V

76. Aspartame (X) is an artificial sweetening agent and is 200 times sweeter than sugar. It is an ester of the

dipeptide of

OMe OMe

(X)

NH

H2N

COOH

OMe

(a) Alanine and phenylalanine (b) Aspartic acid and alanine

(c) Phenylalanine and glycine (d) aspartic acid and phenylalanine

Ans. [d]

Sol. Aspartame is an artificial sweetening agent formed by aspartic acid and phenylalanine

77. Which one of the following reactions is correct ?

(a) [Fe(CO)5] + 2NO [Fe(CO)2(NO)2] + 3CO (b) [Fe(CO)5] + 2NO [Fe(CO)3(NO)2] + 2CO

(c) [Fe(CO)5] + 3NO [Fe(CO)2(NO)3] + 3CO (d) [Fe(CO)5] + 3NO [Fe(CO)3(NO)3] + 2CO

Ans. [a]

Sol. Fe(CO)5 + NO [Fe(CO)2 (NO)2] + 3 CO

We know

35 / 35



CAREER POINT

that C = O

2e– Donar

N = O

3e– Donar3 × 2 6e–

Donar2 × 3e– 6e–

DonarThis meaning 3CO replaced by 2NOHence Ans. is (a)

78. Standard molar enthalpy of formation of CO2(g) is equal to

(a) Zero

(b) The standard molar enthalpy of combustion of carbon (graphite)

(c) The standard molar enthalpy of combustion of C(g)

(d) The standard molar enthalpy of combustion of CO(g)

Ans. [b]

Sol. C(s) + O2CO2(g)

Formation of CO2 = combustion of C(s) graphite

79. The reaction of 1-phenylpropane with limited amount of chlorine in the presence of light gives mainly

(a) 4-chloropropylbenzene (b) 1-chloro-1-phenylpropane

(c) 3-chloro-1-phenylpropane (d) 2-chloro-1-phenylpropane

Ans. [b]

Sol.

CH3–CH2–CH2–PhCl2 (less)

h CH3–CH2–CH–Ph

Cl

CH3–CH2–CH–Ph

It is example of photo monohalogenation by free radical mechanism. So reaction occur through benzylic freeradical.

80. An ionic solid LaI2 shows electrical conduction due to presence of

(a) La2+ and 2I– (b) La3+, 2I– and e– (c) La2+, I2 and 2e– (d) La3+, I2 and 3e–

Ans. [b]

Sol. LaI2 2La + 2ILa3+ + e–

association of chemistry teachers - career point€¦ · pbo2 + 4 h + + 2 e – pb2+ + 2 h 2o ......

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