1. The total number of isomers of Co(en)₂Cl₂ (en = ethylenediamine) is

(a) 4

(b) 3

(c) 6

(d) 5

Ans. (b)

Sol. (Co(en)₂Cl₂]

So in the above complex has 3 isomers.

Correct option is (b)

2. Metal – metal quadruple bonds are well – known for the metal

(a) Ni

(b) Co

(c) Fe

(d) Re

Ans. (d)

Sol.

Correct option is (d)

3. The reaction of Al₄C₃ with water leads to the formation of

(a) methane

(b) propyne

(c) propene

(d) propane

Ans. (a)

Sol. Al₄C₃ + H₂O Al(OH)₃ + CH₄

Correct option is (a)

4. The correct statement about C₆₀ is

(a) C₆₀ is soluble in benzene

(b) C₆₀ does not react with tert – butyllithium

(c) C₆₀ is made up of 10 five – membered and 15 six – membered rings

(d) two adjacent five – membered rings share a common edge.

Ans. (a)

Sol. C₆₀ is non-polar hence soluble in benezne. Ir reacts with t-butyl lithium and it is made up of 12-five membered and 20 six-membered ring which five-membered rings are surrounded by six-membered ring only.

Correct option is (a)

5. The lattice parameters for a monoclinic crystal are:

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Lattice parameters for monoclinic

Correct option is (a)

6. The magnetic moment of [Ru(H₂O)₆]²⁺ corresponds to the presence of

(a) four unpaired electrons

(b) three unpaired electrons

(c) two unpaired electrons

(d) zero unpaired electrons

Ans. (d)

Sol. [Ru(H₂O)₆]²⁺ has zero unpaired electron it is low spin d⁶ complex

Ru = 4d⁶ 5s²

Correct option is (d)

7. The compound that is NOT aromatic is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

8. The order of stability for the following cyclic olefins is

(a) I < II < III < IV

(b) II < III < IV < I

(c) II < III < I < IV

(d) IV < II < I < III

Ans. (c)

Sol. 6 membered and five membered ring

6 membered and five membered ring double bond present at bridgent head position

7 membered and five membered ring with double bond at bridged head position

6 membered with 6 membered ring.

Note : double bond at bridged head position is not feasible according to Bredt's rule.

Double bond present in larger size ring with decrease strin and increase stability.

So correct order of stability IV > I > III > II.

Correct option is (c)

9. The most acidic species is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

10. The major product of the following reaction is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Correct option is (c)

11. In the carbylamine reaction, R – X is converted to R – Y via the intermediate Z. R – X, R – Y and Z, respectively, are

(a) R – NH₂, R – NC, carbene

(b) R – NH₂, R – NC, nitrene

(c) R – NC, R – NH₂, carbene

(d) R – OH, R – NC, nitrene

Ans. (a)

Sol. Carbylamine reaction

So, X = R – NH₂

Y =

Z = Carbene

Correct option is (a)

12. The compound that is NOT oxidized by KMnO₄ is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Compound which lack benzylic hydrogen is not oxidized by KMnO₄.

Correct option is (d)

13. Cyanogen bromide (CNBr) specifically hydrolyses the peptide bond formed by the C – side of

(a) methionine

(b) glycine

(c) proline

(d) serine

Ans. (a)

Sol. The chemical cyanogen bromide (CNBr) hydrolysis the peptide chain from c-terminal side of methionine.

Correct option is (a)

14. The Hammett reaction constant ρ is based on

(a) the rates of alkaline hydrolysis of substituted ethyl benzoates

(b) the dissociation constants of substituted acetic acids

(c) the dissociation constants of substituted benzoic acids

(d) the dissociation constants of substituted phenols

Ans. (c)

Sol. The Hammett equation in organic chemistry describes a linear free energy relationship rates and equilibrium constant for many reaction involving benzoic acid derivatives with meta and para-substituents to each other.

Correct option is (c)

15. The lifetime of a molecule in an excited electronic state is 10^–10 s. The uncertainty in the energy (eV) approximately is

(a) 2 × 10^–5

(b) 3 × 10^–6

(c) 0

(d) 10^–14

Ans. (b)

Sol.

leV = 1.6 × 10^–19 J

Correct option is (b)

16. For a one component system, the maximum number of phases that can coexist at equilibrium is

(a) 3

(b) 2

(c) 1

(d) 4

Ans. (a)

Sol. Maximum number of phase that can exist in equilibrium is 3.

Correct option is (a)

17. At T = 300 K, the thermal energy (k_BT) in cm^–1 is approximately

(a) 20000

(b) 8000

(c) 5000

(d) 200

Ans. (d)

Sol. At T = 300 k

Thermal energy = k_BT = 1.38 × 10^–23 300 k

Correct option is (d)

18. For the reaction 2X₃ = 3X₂, the rate of formation of X₂ is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct option is (d)

19. The highest occupied molecular orbital of HF is:

(a) bonding

(b) antibonding

(c) ionic

(d) nonbonding

Ans. (d)

Sol.

Correct option is (d)

20. The residual entropy of the asymmetric molecule N₂O in its crystalline state is 5.8 J K^–1 mol^–1 at absolute zero. The number of orientations that can be adopted by N₂O in its crystalline state is

(a) 4

(b) 3

(c) 2

(d) 1

Ans. (c)

Sol. Residual entropy for N₂O = 5.8 JK^–1 mol^–1

Number of orientations = ?

S = R In W; W = ?

5.8 JK^–1 mol^–1 = × 2.303 log₁₀W; log W = 0.3025 ~ 0.3010; log W = log 2

Correct option is (c)

21. The spectroscopic ground state symbol and the total number of electronic transitions of [Ti(H₂O)₆]²⁺ are

(a) ³T_1g and 2

(b) ³A_2g and 3

(c) ³T_1g and 3

(d) ³A_2g and 2

Ans. (c)

Sol. The spectroscopic ground state symbol of [Ti(H₂O)₆]²⁺ is ³T_1g. In this compound, there will be three electronic transition.

Correct option is (c)

22. The structures of the complexes [Cu(NH₃)₄] (ClO₄)₂ and [Cu(NH₃)₄](ClO₄) is solution respectively are

(a) square planar and tetrahedral

(b) octahedral and square pyramidal

(c) octahedral and trigonal bipyramidal

(d) tetrahedral and square planar

Ans. (a)

Sol. [Cu(NH₃)₄]²⁺

[Cu(NH₃)₄]²⁺

[Cu(NH₃)₄]⁺

Correct option is (a)

23. In biological systems, the metal ions involved in electron transport are

(a) Na⁺ and K⁺

(b) Zn²⁺ and Mg²⁺

(c) Ca²⁺ and Mg²⁺

(d) Cu²⁺ and Fe³⁺

Ans. (d)

Sol. In biological systems, Cu²⁺ and Fe²⁺ metal ion involved in electron transport. This is due their variable oxidation state.

Correct option is (d)

24. In a homogeneous catalytic reaction, 1.0 M of a substrate and 1.0 μM of a catalyst yields 1.0 mM of a product in 10 seconds. The turnover frequency (TOF) of the reaction (s^–1) is

(a) 10^–2

(b) 10²

(c) 10^–3

(d) 10³

Ans. (b)

Sol. TOF =

Correct option is (b)

25. The expected magnetic moments of the first – row transition metal complexes and those of the lanthanide metal complexes are usually calculated using

(a) µ_s.o equation (s.o. = spin only) for both lanthanide and transition metal complexes

(b) µ_s.o equation for lanthanide metal complexes and µ_j equation for transition metal complexes

(c) µ_s.o equation for transition metal complexes and µ_j equation for lanthanide metal complexes

(d) µ_L+S equation for transition metal complexes and µ_s.o equation for lanthanide metal complexes

Ans. (c)

Sol. Magnetic moment of transition metal complexes is given by spin only formula

Magnetic moment of Lanthanoids complexes is given by using J-value. This is due to strong spin orbit coupling in lanthanoids.

Correct option is (c)

26. The Bronsted acidity of boron hydrides follows the order

(a) B₂H₆ > B₄H₁₀ > B₅H₉ > B₁₀H₁₄

(b) B₂H₆ = B₄H₁₀ > B₅H₉ = B₁₀H₁₄

(c) B₁₀H₁₄ > B₅H₉ > B₄H₁₀ > B₂H₆

(d) B₅H₉ > B₄H₁₀ > B₂H₆ > B₁₀H₁₄

Ans. (c)

Sol. The correct order of bronsted acidity of boron hydrides follows the order

B₁₀H₁₄ > B₅H₉ > B₄H₁₀ > B₂H₆

acidity of boron hydride depends on the size of borane greater the size higher will be acidity. This is because the negative charge, formed upon deprotonation, can be better delocalized over a larger anion with many boron atoms than over a small one.

Correct option is (c)

27. NaCl is crystallised by slow evaporation of its aqueous solution at room temperature. The correct statement is

(a) The crystals will be non – stoichiometric

(b) The crystals should have Frenkel defects

(c) The percentage of defects in the crystals will depend on the concentration of the solution and its rate of evaporation

(d) The nature of defects will depend upon the concentration of the solution and its rate of evaporation

Ans. **

Sol. We can't decide the type of defect. The amount of defect will depend on experimental conditions.

28. GaTiO₃ has a perovskite crystal structure. The coordination number of titaniumin CaTiO₃ is:

(a) 9

(b) 6

(c) 3

(d) 12

Ans. (b)

Sol.

The co-cordination number of Ti will be form edge centre = 6

29. If CIF₃ were to be stereochemically rigid, its ¹⁹F NMR spectrum (I for ) would be (assume that Cl is not NMR active)

(a) a doublet and a triplet

(b) a singlet

(c) a doublet and a singlet

(d) two singlets

Ans. (a)

Sol.

²F_a splits into a doublet with ¹F_b

¹F_b splits into a triplet with ²F_b

Correct option is (a)

30. The point group of NSF₃ is:

(a) D_3d

(b) C_3h

(c) D_3h

(d) C_3v

Ans. (d)

Sol. Point group of NSF₃

31. When NiO is heated with a small amount of Li₂O in air at 1200ºC, a non – stoichiometric compound Li_xNi_(1–x)O is formed. This compound is

(a) an n – type semiconductor containing only Ni¹⁺

(b) an n – type semiconductor containing Ni¹⁺ and Ni²⁺

(c) a p – type semiconductor containing Ni²⁺ and Ni³⁺

(d) a p – type semiconductor containing only Ni³⁺

Ans. (c)

Sol. As a result of this defect some of the Ni²⁺ ions oxidise into Ni³⁺ ions and decrease the defficiency of electrons and semiconductor will be p-type.

Correct option is (c)

32. White phosphorus, P₄, belongs to the

(a) closo system

(b) nido system

(c) arachno system

(d) hypo system

Ans. (b)

Sol.

TEC = 9 × 4 + 12 × 2 = 60 (14n + 4)

PEC 60 – 4 × 12 = 12

(n + 2) hence Nido

Correct option is (b)

33. Among the compounds Fe₃O₄, NiFe₂O₄ and Mn₃O₄

(a) NiFe₂O₄ and Mn₃O₄ are normal spinels

(b) Fe₃O₄ and Mn₃O₄ are normal spinels

(c) Fe₃O₄ and Mn₃O₄ are inverse spinels

(d) Fe₃O₄ and NiFe₂O₄ are inverse spinels

Ans. (d)

Sol. FeFe₂O₄ is inverse type spinel, because CFSE of Fe²⁺ is more than Fe³⁺ in octahedral site.

NiFe₂O₄ is also inverse type spinel. because CFSE of Ni²⁺ is more than Fe³⁺ in octahedral site thats way Ni²⁺ present in oh void.

Correct option is (d)

34. The number of M – M bonds in Ir₄ (CO)₁₂ are

(a) four

(b) six

(c) eight

(d) zero

Ans. (b)

Sol. Ir₄(CO)₁₂

TVE = 9 × 4 + 12 × 2 = 36 + 24 = 60

A = n × 18 – TVE where, n = number of metal

= 4 × 18 – 60 = 72 – 60 = 12

B = 12

B/2 Total number of m-m bond

Then, = 6

Correct option is (b)

35. Schrock carbenes are

(a) Triplets and nucleophilic

(b) Triplets and electrophilic

(c) Singlets and nucleophilic

(d) Singlets and electrophilic

Ans. (a)

Sol. Shrock's carbene

Nucleophilic nature is due presence of strong σ donor ligands.

Correct option is (a)

36. The INCORECT statement about linear dimethylopolysiloxane, [(CH₃)₂ SiO]_n is

(a) it is extremely hydrophilic

(b) it is prepared by a KOH catalysed ring – opening reaction of [Me₂SiO]₄

(c) it has a very low glass transition temperature

(d) it can be reinforced to give silicon elastomers

Ans. (a)

Sol. Due to presence of hydrophobic CH₃ group, linear dimethyl polysiloxane [(CH₃)₂ SiO]_n is hydrophobic in nature.

Correct option is (a)

37. Match the entries a – d with their corresponding structures P – S

(A) bridged system (P)

(B) atropisometric system (Q)

(C) spiro system (R)

(D) Fused system (S)

(a) A – S, B – R, C – Q, D – P

(b) A – P, B – S, C – Q, D – R

(c) A – Q, B – P, C – S, D – R

(d) A – S, B – R, C – P, D – Q

Ans. (d)

Sol. (A) Bridged system : In which carbon joined to each other by three different chain of carbon exist in different plane.

(B) Atropisomerism : Isolable stereoisomers resulting from restricted rotation about single bond are called atropisomers.

(C) Spiro system : When the two rings are shared by only one carbon atom called a spiro compound.

(D) Fused system : When two rings can share a pair of adjacent carbon the arrangement is called a fused bicylic system.

Correct option is (d)

38. The reaction between X and Y to give Z proceeds via

(a) – conrotatory opening of X followed by endo Diels – Alder cycloaddition.

(b) – disrotatory opening of X followed by endo Diels – Alder cycloaddition.

(c) – conrotatory opening of X followed by exo Diels – Alder cycloaddition.

(d) – disrotatory opening of X followed by exo Diels – Alder cycloaddition.

Ans. (a)

Sol.

Correct option is (a)

39. The major products P₁ and P₂, respectively, in the following reaction sequence are

1. 

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

40. The products Y and Z are formed, respectively, from X via

(a) conrotatory opening and disrotatory opening.

(b) disrotatory opening and conrotatory opening.

(c) conrotatory opening and disrotatory opening.

(d) disrotatory opening and conrotatory opening.

Ans. (a)

Sol.

41. o – Bromophenol is readily prepared from phenol using the following conditions

(a) (i) (CH₃CO)₂O; (ii) Br₂; (iii) HCl–H₂O,

(b) (ii) H₂SO₄, 100ºC (ii) Br₂ (iii) H₃O⁺, 100ºC

(c) N – Bromosuccinmide, bibenzoyl peroxide, CCl₄,

(d) Br₂ / FeBr₃

Ans. (d)

Sol.

Correct option is (d)

42. The major product of the following reaction is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

As trick, view from above side, hydroxyl group takes place on left side and reagent is L-(+)-diethyltartrate, so epoxide formed below the plane.

Correct option is (d)

43. The photochemical reaction of 2 – methylpropane with F₂ gives 2 – fluoro – 2 – methylpropane and 1 – fluoro – 2 – methylpropane in 14:86 ratio. The corresponding ratio of the bromo products in the above reaction using Br₂ is most likely to be

(a) 14 : 86

(b) 50 : 50

(c) 1 : 9

(d) 99 : 1

Ans. (d)

Sol.

Ratio between 2-bromo-2-methyl propane and 1-bromo-2-methyl propane is 99 : 1

Correct option is (d)

44. The major product P of the following reactions is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. 1-3 diols protect ketone (group selectivity)

Correct option is (b)

45. The reagent X in the following reaction is

(a) HO₂CN=NCO₂H

(b) EtO₂CN=NC–CO₂Et

(c) EtO₂CH=NCO₂Et

(d)

Ans. (d)

Sol.

Correct option is (d)

46. The major product of the following reaction is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct option is (d)

47. The major product of the following reaction is:

(a)

(b)

(c)

(d)

Ans. (c)

Sol. PtO₂ preffered the reduction of highly strained cyclopropane ring over double bond

Correct option is (c)

48. In the following compound, the hydroxy group that is most readily methylated with CH₂N₂ is:

(a) p

(b) q

(c) r

(d) s

Ans. (b)

Sol.

Correct option is (b)

49. The most appropriate sequence of reactions for carrying out the following transformation.

(a) (i) O₃/H₂O₂; (ii) excess SOCl₂/pyridine; (iii) excess NH₃; (iv) LiAlH₄

(b) (i) O₃/Me₂S; (ii) excess SOCl₂/pyridine; (iii) LiAlH₄; (iv) excess NH₃

(c) (i) O₃/H₂O₂; (ii) excess SOCl₂/pyridine; (iii) LiAlH₄; (iv) excess NH₃

(d) (i) O₃/Me₂S; (ii) excess SOCl₂/pyridine; (iii) excess NH₃; (iv) LiAlH₄

Ans. (a)

Sol.

Correct option is (a)

50. The number of optically active stereoisomers possible for 1, 3– cyclohexanediol in its chair conformation is

(a) 4

(b) 3

(c) 2

(d) 1

Ans. (c)

Sol.

So, total number of opticallly active stereoisomers for 1, 3-cyclohexadiol are = 2

51. The major product of the following reaction is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

52. In the following reaction,

The absolute configurations of the chiral centres in X and Y are

(a) 2S, 3R and 2R, 3R

(c) 2R, 3R and 2R, 3S

(c) 2S, 3S and 2R, 3R

(d) 2S, 3R and 2S, 3R

Ans. (a)

Sol.

Correct option is (a)

53. The IR stretching frequencies (cm^–1) for the compound X are as follows: 3300 – 3500 (s, br); 3000 (m); 2225 (s); 1680 (s).

The correct assignment of the absorption bands is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct option is (a)

54. The T_d point group has 24 elements and 5 classes. Given that it has two 3 – dimensional irreducible representation, the number of one dimensional irreducible representation is:

(a) 1

(b) 2

(c) 0

(d) 3

Ans. (b)

Sol. For the Td point group,

Two three dimensional irreducible respectively

Two one diemensional irreducible respectively

One two dimensional irreducible respectively

Correct option is (b)

55. The total number of ways in which two nonidentical spin – ½ particles can be oriented relative to a constant magnetic field is:

(a) 1

(b) 2

(c) 3

(d) 4

Ans. (d)

Sol. Two non-identical spin-½ particles total orientation

Correct option is (d)

56. Approximately one hydrogen atom per cubic meter is present in interstellar space. Assuming that the H – atom has a diameter of 10^–10m, the mean free path (m) approximately is

(a) 10¹⁰

(b) 10¹⁹

(c) 10²⁴

(d) 10¹⁴

Ans. (b)

Sol.

N* = No. of molecules per unit volume = 1

= 2.2 × 10¹⁹

Correct option is (b)

57. The wavefunction of a diatomic molecule has the form . The chance that both electrons of the bond will be found on the same atom in 100 inspections of teh molecule approximately is

(a) 79

(b) 20

(c) 45

(d) 60

Ans. ()b

Sol.

Chance that both electron of the bond will be found on the same atom means ionic bond found

(0.45)² = 20.25

0.2025 × 100 = 20

Correct option is (b)

58. For the reaction given below, the relaxation time is 10^–6 s. Given that 10% of A remains at equilibrium, the value of k₁(s^–1) is

(a) 9 × 10⁵

(b) 10^–5

(c) 10⁵

(d) 9 × 10^–5

Ans. (a)

Sol.

…(i)

or K₁ = 9 K_–1

Substituting it in equation (i)

9K_–1 + K_–1 = 10⁶

K_–1 = 10⁵

Hence, K₁ = 9 × 10⁵ s^–1

Correct option is (a)

59. The minimum number of electrons needed to form a chemical bond between two atoms is

(a) 1

(b) 2

(c) 3

(d) 4

Ans. (a)

Sol. The minimum number of electron needed to form a chemical bond between two atom is one.

e.g. = according to MOT, the bond order of is

Bond order in =

Correct option is (a)

60. The ground state electronic energy (Hartree) of a helium atom, neglecting the inter – electron repulsion, is

(a) –1.0

(b) –0.5

(c) –2.0

(d) –4.0

Ans. (c)

Sol. Electronic energy = –0.5n²

for He z = 2 = – 0.5 × 4 = – 2.0

Correct option is (c)

61. A particle is confined to a one – dimensional box of length 1 mm. If the length is changed by 10^–9 m, the % change in the ground state energy is

(a) 2 × 10^–4

(b) 2 × 10^–7

(c) 2 × 10^–2

(d) 0

Ans. (a)

Sol.

Correct option is (a)

62. A certain molecule can be treated as having only a doubly degenerate state lying at 360 cm^–1 above the nondegenerate ground state. The approximate temperature (K) at which 15% of the molecule will be in the upper state is

(a) 500

(b) 150

(c) 200

(d) 300

Ans. (c)

Sol. Let total molecules = 100

i is the ground state and j is excited state

n_i = 85, g_i = 1; n_j = 15, g_j = 2

We know that,

Correct option is (c)

63. A box of volume V contains one mole of an ideal gas. The probability that all N particles will be found occupying one half of the volume leaving the other half empty is

(a) 1/2

(b) 2/N

(c) (1/2)^N

(d) (1/2)^6N

Ans. (c)

Sol. Total valume = V

Probability to be in volume =

For N number of particles,

Probability that all N particle will be found in volume =

Correct option is (c)

64. According to the Debye – Huckel limiting law, the mean activity coefficient of 5 × 10^–4 mol kg^–1 aqueous solution of CaCl₂ at 25ºC is (the Debye – Huckel constant 'A' can be takento be 0.509)

(a) 0.63

(b) 0.72

(c) 0.80

(d) 0.91

Ans. (d)

Sol.

I = 15 × 10^–4

Correct option is (d)

65. The operation of the commutator [x, d / dx] on a function f(x) is equal to

(a) 0

(b) f(x)

(c) –f(x)

(d) x df/dx

Ans. (c)

Sol.

Correct option is (c)

66. If a gas obeys the equation of state the ratio is:

(a) >1

(b) <1

(c) 1

(d) (1 – b)

Ans. (c)

Sol. P(V – nb) = nRT

C_p – C_v = nR;

(C_p – C_v)_ideal = nR

So,

Correct option is (c)

67. Physisorbed particles undergo desorption at 27ºC with an activation energy of 16.628 kJ mol^–1. Assuming first – order process and a frequency factor of 10¹² Hz, the average residence time (in seconds) of the particles on the surface is

(a) 8 × 10^–10

(b) 8 × 10^–11

(c) 2 × 10^–9

(d) 1 × 10^–12

Ans. (a)

Sol.

Correct option is (a)

68. The rotational constant for CO in the ground and the first excited vibrational states are 1.9 and 1.6 cm^–1, respectively. The % change in the internuclear distance due to vibrational excitation is

(a) 9

(b) 30

(c) 16

(d) 0

Ans. (a)

Sol.

Correct option is (a)

69. The mechanism of enzyme (E) catalysed reaction of a substrate (S) to yield product (P) is:

If a small amount of S is converted to P, the maximum rate for the reaction will be observed for

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

r_max = K₂[E]₀

If K_–1 + K₂ << K₁[S]₀

Since it is given that small amount of S is converted to P

K₁[S]₀ >> K_–2 [P] and K₁K₂ [S]₀ >> K₂K_–2 [P]

Correct option is (b)

70. The lowest energy state of the (1s)² (2s)¹ (3s)¹ configuration of Be is

(a) ¹S₀

(b) ¹D₂

(c) ³S₁

(d) ³P₁

Ans. (c)

Sol. (1s)² (2s)¹ (3s)¹

As we know closed shell 1s² will not contribute in calculation of states for given configuration.

For 2s and 3s orbitals values of l₁ and l₂ will be zero

SM = 2S + 1 = 2(1) + 1 = 3

J = |L + S|

J = |0 + 1| = 1

term symbol = ^SML_J = ³S₁

Correct option is (c)

Common Data for Q. 71, Q. 72. and Q. 73

An electron accelerated through a potential difference of volts impinges on a nickel surface, whose (100) planes have a spacing d = 351.5 × 10^–12 m (351.5 pm)

71. The de – Broglie wavelength of the electron is . The value of 'a' in volts is

(a) 1.5 × 10^–18

(b) 1.5 × 10⁶

(c) 6.63 × 10^–5

(d) 2.5 × 10¹⁸

Ans. (b)

Sol.

Correct option is (b)

72. The condition for observing diffraction from the nickel surface is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. We know

Case I : For the first order diffraction, n = 1

squation will become

Here, we know the range of sin is [–1, 1] and –ve values are not possible for

Hence, here you will observe sin range [0, 1]

Let take sin = 1

= 2d ...(1)

where sin = 1 (max.)

If sin = 0.5 =

then

= d ...(2)

If we further dec the value of sin , will also decrease. ...(3)

Hence, from (1), (2) and (3)

Case II : If we increase n value,

Let n = 2 and sin is max. i.e. equal to 1.

n = 2d sin

2 × = 2 × d × 1

again is less than 2d and if n increase than λ decrease

Hence, the correct answer is

Option (b) is correct.

73. The minimum value of for the electron to diffract from the (100) planes is

(a) 3000

(b) 300

(c) 30

(d) 3

Ans. (d)

Sol.

Correct option is (d)

74. Common Data for Q. 74 and Q. 75

An iron complex [FeL₆]²⁺ (L = neutral monodentate ligand) catalyses the oxidation of (CH₃)₂S by perbenzoic acid.

74. The formation of the organic product in the above reaction can be monitored by

(a) gas chromatography

(b) cyclic voltammetry

(c) electron spin resonance

(d) fluorescence spectroscopy

Ans. (c)

Sol. This reaction involves perbenzoic acid, this it can be monitored by electron spin resonance

Hence, correct option is (c)

75. The oxidation state of the metalionin the catalyst can be deducted by

(a) Atomic absorption spectrosocpy

(b) Mossbauer spectroscopy

(c) HPLC

(d) Gas Chromatography

Ans. (b)

Sol. Mossbaur spectroscopy is unique in its sensitivity to subtle changes in the chemical environment of the nucleus including oxidation state changes, the effect of different ligand oin a particular atom and the magnetic environment of the sample. It is an useful tool for identifying the composition of iron containing specimens.

Correct option is (b)

Linked Answer type Q. 76 and Q. 77:

In the reaction,

76. Compound X is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

It is possible to disssociate Cl^–, but the larger PPh₃ (i.e. more steric hindrance) ligand seem to be one factor that enhance its dissociation and this provide more space for incoming larger size acetonitrile ligand.

Correct option is (d)

77. Rh(PPh₃)₃Cl reacts very fast with a gaseous mixture of H₂ and C₂H₄ to immediatel give Z. The structure of Z is

(a) H₃C–CH₃

(b)

(c)

(d)

Ans. (d)

Sol. Correct option is (d)

Linked Answer Type for Q. 78 and Q. 79:

The reaction of PCl₃ with methanol in the presence of triethylamine affords compound X. EI mass spectrum of X shows a parent ion peak at m/z = 124. Microanalysis of X shows that it contains C, H, O and P. The ¹H NMR spectrum of X shows a doublet at 4.0 ppm. The separation between the two lines of the doublet is approximately .

78. Compound X is:

(a) (CH₃O₃)P

(b) (CH₃O)₂P(O)

(c) (CH₃O)₂P(O)(OH)

(d) (CH₃O)₃PH

Ans. (a)

Sol.

Correct option is (a)

79. Upon heating, compound X is converted to Y, which has the same molecular formula as that of X. The ¹H NMR spectrum of Y shows two doublets centered at 3.0 ppm (separation of two lines ~ 20 Hz) and 4.0 ppm (separation of two ~ 15 Hz) respectively.

Compound Y is:

(a) (CH₃O)₂P(O)(OH)

(b) (CH₃O)₃P(O)

(c) (CH₃O)₂(CH₃)P(O)

(d) (CH₃O)₂(CH₃)P(OH)

Ans. (b)

Sol.

¹HNMR = [2NI_P + 1]

For, a = = 2 (doublet) and For, b = = 2 (doublet)

Correct option is (b)

Linked Answer Type Q. 80 and Q. 81.

For butyrophenone (PhCOCH₂CH₂CH₃)

80. The most probable fragmentation observed in the electron impact ionization (EI) mass spectrometry is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct option is (d)

81. Photoirradiation leads to the following set of products:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct option is (d)

Linked Answer Q. 82 and Q. 83

In the following reaction,

82. The reactive intermediate I and the product P are

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct option is (a)

83. The product P shows 'm' and 'n' number of signals in ¹H and ¹³C NMR spectra, respectively. The values of 'm' and 'n' are

(a) m = 3 and n = 2

(b) m = 2 and n = 3

(c) m = 2 and n = 2

(d) m = 4 and n = 3

Ans. (a)

Sol.

Correct option is (a)

Linked Answer Type of Q. 84 and Q. 85

The infrared spectrum of a diatomic molecule exhibits transitions at 2144, 4262 and 6354 cm^–1 corresponding to excitations from the ground state to the first, second and, third vibration states respectively.

84. The fundamental transition (cm^–1) of the diatomic molecule is at:

(a) 2157

(b) 2170

(c) 2183

(d) 2196

Ans. (b)

Sol.

Equation (1) multiply (3) and substract from (2)

Fundamental transition at 2170 cm^–1

Correct option is (b)

85. The anharmonicity constant (cm^–1) of the diatomic molecule is:

(a) 0.018

(b) 0.012

(c) 0.006

(d) 0.003

Ans. (c)

Sol. Anharmonicity constant,

Correct option is (c)