1. Absolute stereochemistry of the given compound is:

(a) 4aS, 8aR

(b) raR, 8aS

(c) 4aS, 8aS

(d) 4aR, 8aR

Ans. (**)

Sol.

Fix 3^rd priority and rotate 1, 2 and 3.

(anticlockwise)

So, configuration of 4a centre is S.

Similarly, here fix I^st priority and rotate 2, 3, 4

then rotate (clockwise)

So, configuration of 8a centre is R.

2. Major product formed in the following reaction is:

(a)

(b)

(c)

(d)

Ans. (**)

Sol.

3. Major product formed in the following transformation is

(a)

(b)

(c)

(d)

Ans. (**)

Sol.

4. The activity of 'm' molal CuSO₄ solution can be expressed in terms of its mean activity coefficient as:

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

5. In an NMR spectrometer operating at a magnetic field strength of 16.45 T, the resonance frequency (in MHz, rounded off to one decimal place) of ¹⁹F nucleus is _______

Ans. (**)

Sol.

6. In the following reaction sequence,

the major products P and Q are:

(a)

(b)

(c)

(d)

Ans. (**)

Sol.

7. For an enzyme catalyzed reaction, the plot of inverse of initial rate against inverse of initial substrate concentration is linear with slope 0.16 s and intercept 2.12 mol^–1 L s. The estimated value of Michaelis constant (in mol L^–1, rounded off to two decimal places) is __________

Ans. (0.075)

Sol. For enzyme catalysis:-

8. In a uranium recovery process, an aqueous solution of uranyl ion is evaporated, dried in air at 400 ºC and subsequently reduced with hydrogen at 700ºC to obtain a uranium compound (X). The oxidation state of uranium in X is __________

(Given: atomic number of U is 92)

Ans. (+4)

Sol. In uranium recovery process an aqueous solution of uranyl ion is evaporated dried in air at 400ºC and subsequently reduced with hydrogen at 200ºC to obtained uranium compound with oxidation state of +4 i.e. VO₂

x – 4 = 0

x = +4

9. At 25ºC, the emf (in volts, rounded off to three decimal places) of the cell,

(Given: The standard emf of the cell is 0.082 V; R = 8.314 JK^–1 mol^–1; F = 96500 C mol^–1)

Ans. (0.058 V)

Sol.

10. When three moles of helium is mixed with one mole of neon at constant temperature and pressure (25ºC, 1 atm), the entropy of mixing (in JK^–1, rounded off to two decimal places) is ________

(Given: R = 8.314 JK^–1 mol^–1)

Ans. (18.69 JK^–1)

Sol.

= 18.69 JK^–1

11. A solution containing a metal complex absorbs at 480 nm with molar extinction coefficient of 15,000 L mol^–1 cm^–1. If the path length of the cell is 1.0 cm and transmittance is 20.5%, the concentration (in mol L^–1) of the metal complex is:

(a) 2.29 × 10^–5

(b) 8.75 × 10^–5

(c) 4.59 × 10^–5

(d) 1.37 × 10^–5

Ans. (**)

Sol. A = ECl

C = 4.59 × 10^–5 mol/L

12. The maximum number of microstates for d² electronic configuration is _________

Ans. (45)

Sol. Number of microstate

n is twice of number of orbital r is number of e^–.

For d²; n = 10

r = 2

Correct answer is (45)

13. Fluorescence quantum yield and fluorescence lifetime of a molecule are 0.4 and 5 × 10^–9 s, respectively. If the fluorescence decay rate constant is Y × 10⁷ s^–1, the value of Y (rounded off to nearest integer) is ________

Ans. (8)

Sol.

14. In the following reaction,

the number of peaks exhibited by the major product P in its broad band proton decoupled ¹³C NMR spectrum is ________

Ans. (**)

Sol.

15. The CORRECT statement regarding the substitution of coordinated ligands in Ni(CO)₄ and Co(NO)(CO)₃ is:

(Given: Co–N–O bond is nearly linear; atomic numbers of Co and Ni are 27 and 28, respectively)

(a) Ni(CO)₄ and Co(NO)(CO)₃ follow associative and dissociative pathways, respectively.

(b) Both Ni(CO)₄ and Co(NO)(CO)₃ follow associative pathway.

(c) Both Ni(CO)₄ and Co(NO)(CO)₃ follow dissociative pathway.

(d) Ni(CO)₄ and Co(NO)(CO)₃ follow dissociative and associative pathways, respectively.

Ans. (b)

Sol. Ni(CO)₄ is 18e^– complex.

Co(NO)(CO)₃ is also 18e^– complex.

Substitution reaction in 18e^– complexes undergo by dissociative pathways but in Co(NO)(C))₃, NO in linear mode changes into bent mode and Co(NO)(CO)₃ Now is 16e^– complex because NO give 1e^– in bent mode 18e^– complex undergo by associative mechanism thats why Co(NO)(CO)₃ undergo substitution by associative pathway.

Correct option is (b)

16. Major product formed in the following reaction sequence is:

(a)

(b)

(c)

(d)

Ans. (**)

Sol.

17. Major product formed in the given reaction is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

18. Among the following, the suitable reagents for the given transformation is:

(a) NaBH₄/CeCl₃·7H₂O

(b) H₂N–NH₂/KOH,

(c) Li/Liq. NH₃

(d) H₂, Pd/C

Ans.

Sol. (**)

19. The CORRECT statement about hexagonal boron nitride is:

(a) It is reactive towards fluorine.

(b) It has same layer stacking as that of graphite.

(c) It is a good electrical conductor.

(d) It has lower thermal stability in air compared to that of graphite.

Ans. (c)

Sol. Boron nitride is reactive towards fluorine.

Correct option is (c)

20. For a cubic crystal system, the powder X-ray diffraction pattern recorded using Cu soucrce ( = 1.54 Å) shows a peak at 33.60º for (111) plane. The lattice parameter 'a' (in Å, rounded off to two decimal place) is _________

Ans. (4.61 Å)

Sol.

21. The character table for a pyramidal AB₃ molecule of C_3v point group is given below:

The reducible representation of pyramidal AB₃ is

The CORRECT option representing all the normal Raman active modes of pyramidal AB₃ is:

(a) A₁ + A₂ + 2E

(b) 2A₁ + 2E

(c) 3A₁ + A₂ + E

(d) 3E

Ans. (b)

Sol.

So, Raman active modes are 2A₁ + 2E.

22. Among the following linear combination of atomic orbitals, the CORRECT representation of the lowest unoccupied -molecular orbital of butadiene is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Lowest unoccupied -molecular orbital is

Correct option is (a)

23. Among the following

the total number of aromatic species is __________

Ans. (4)

Sol.

All the 4 are aromatic species

24. Among the following species, the one that has pentagonal shape is:

(Given: atomic numbers of O, F, S, I and Xe are 8, 9, 16, 53 and 54, respectively)

(a) [SF₅]^–

(b) XeOF₄

(c) [XeF₅]^–

(d) IF₅

Ans. (d)

Sol. XeF₅^–

Stearic number

5 bond pair + 2 lone pair.

Shape is pentagonal planar.

Correct option is (d)

25. In oxyhemocyanin, the coordiantion number, mode of oxygen binding, color and the net magnetic behavior of copper ions, respectively are:

(Given: atomic number of Cu is 29)

(a) Four, blue and diamagnetic.

(b) Five, colorless and paramagnetic.

(c) Five, blue and diamagnetic.

(d) Four, colorless and paramagnetic.

Ans. (c)

Sol.

Coordination number = 5

Mode of oxygen binding =

Blue colour due to LMCT.

d⁹, unpaired electron of 1 Cu couples with 2^nd Cu atom.

So diamagnetic due to antiferro.

Magnetic coupling.

Correct option is (c)

26. The following table lists the reaction/conversion catalyzed by metalloenzymes.

(a) P-I; Q-IV; R-III; S-II

(b) P-II; Q-III; R-IV; S-I

(c) P-IV; Q-III; R-II; S-I

(d) P-II; Q-I; R-III; S-IV

Ans. (c)

Sol.

27. Major product formed in the following synthetic sequence is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

28. The hydrogen-like radial wave function of the 3s orbital is given as

where = 2Zr/a₀; Z = atomic number; r = distance from the nucleus and a₀ = Bohr radius.

Positions of the radial nodes (in units of ) of the 3s orbital are at

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Radial mode, R_3,0 = 0

29. The frequency (in cm^–1, rounded off to two decimal places) for pure rotational line in the spectrum of NO molecule due to change in the quantum number from J = 1 to J = 2 is ______

(Given: Moment of inertia of NO = 1.6427 × 10^–46 kg m²; h = 6.626 × 10^–34 J s; c = 3 × 10⁸ m/s)

Ans. (6.81 cm^–1)

Sol. Given; change in quantum no. from

J = 1 to J = 2

So find, pure rotation a spectra of No molecule

We know that

Correct answer is 6.81 cm^–1

30. The fission reaction of with thermal neutron is represented below.

the primary fission fragment pair, which undergo series of radioactive decay to form stable nuclei X₃ and Y₄ (chain enders). The X₃ and Y₄, respectively are:

1. 
2. 
3. 
4. 

Ans. (b)

Sol.

31. At 30ºC, the vapor pressure and density of a 1.0 M aqueous solution of sucrose are 31.207 mm Hg and 1.1256 g/mL, respectively. If the vapor pressure of pure water at 30ºC is 31.824 mm Hg, the activity coefficient (rounded off to three decimal places) of water in the given solution is ________

(Given: The molar mass of sucrose = 342.3 g mol^–1)

Ans. 1.003

Sol. Given;

Icrution sucrose Pure water

T = 30ºC T = 30ºC

Vap. Press = 31.207 wittig V.P. = 31.824 wittig

S = 1.1256 g/mol activity coff = ?

To find: activity coff. of water in given for we know that

Observed colligative prop (vapour press.) = 31.207 wittig and

Calculated colligative prop = (1 – x) P₀

Where x is mole fraction =

and P₀ is achiral vepour press. = 31.824 wittig

Now

given that 1M, icrution of sucrose.

i.e. 1 mole in (aqueous) it so n₁ = 1 mole

and n₂ i.e. no. of moles of water

Also density

i.e. 1 ml contains 1.1256g water

1000 ml contains 1125.6 g water

and molar mass of sucrose is 340.3 g/mol

So achiral weight of water in (1 mole g) polution is given as 1125.6 – 340.3 g.

Pressing values we get

Calculating this we get the answer as 1.003.

32. A compound with molecular formula C₁₀H₁₂O₂ showed a strong IR band at ~ 1720 cm^–1, a peak at m/z 122 in the mass spectrum and the following ¹H NMR signals: 8.1–8.0 (2H, m), 7.6–7.5 (1H, m), 7.5–7.3 (2H, m), 4.3 (2H, t), 1.8 (2H, sextet) and 1.0 (3H, t). The structure of the compound is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Molecular formula = C₁₀H₁₂O₂

IR ~ 1720 cm^–1

w/z = 122

H¹ NMR signals

33. In a reaction, reactant X is converted to products Y and Z consecutively with rate constants 6.0 × 10^–2 min^–1 and 9.0 × 10^–3 min^–1, respectively. If the initial amount of X is 12.5 moles, the number of moles (rounded off to one decimal place) of Y formed after 10 minutes is ________

Ans. (5.4 moles)

Sol.

[X]₀ = 12.5 moles

t = 10 min.

34. For the ring opening reaction of cyclopropane to propene at 25ºC, the pre-exponential factor is 4.3 × 10¹⁵ s^–1. The entropy of activation (in JK^–1 mol^–1, rounded off to two decimal places) is _______

(Given: h = 6.626 × 10^–34 Js; k_B = 1.38 × 10^–23 JK^–1; R = 8.314 JK^–1 mol^–1)

Ans. (**)

Sol. Given, A = 4.3 × 10¹⁵ s^–1

T = 25ºC = 298 k

35. The CORRECT combination of L1 and L2 among H^–, NO^–, MeCH^2– and CO, that will satisfy the 18 electron rule for both metal centers in the following neutral molecule, is

(Given: atomic number of Ru is 44)

(a) H^–, NO^–

(b) H^–, CO

(c) MeCH^2–, CO

(d) MeCH^2–, NO^–

Ans. (c)

Sol.

When L₁ and L₂ are MeCH^2– and CO respectively then complex follow 18e^– rule.

Correct option is (c)

36. Adsorption of N₂ on TiO₂ was carried out at 75 K. A plot of versus z(z = p/p⁰) gives a straight line with an intercept, 4.0 × 10^–6 mm^–3 and slope, 1.0 × 10^–3 mm^–3. The volume (rounded off to the nearest integer) corresponding to the monolayer coverage is:

(a) 555 mm³

(b) 690 mm³

(c) 785 mm³

(d) 996 mm³

Ans. (d)

Sol.

according to BET equation, we have

v_m = 996 mm³.

37. In the following reaction sequence,

the structure of B is

(Given: atomic number of Mo is 42)

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

B is thermodynamic product which is stable and follow 18e^– rule.

38. for Fe(g) are 370.7 kJ mol^–1 and 416.3 kJ mol^–1 at 298 K, respectively. Assuming is constant in the interval 250 K to 375 K, (rounded off to the nearest integer) for Fe(g) at 375 K is:

(a) 338 kJ mol^–1

(b) 310 kJ mol^–1

(c) 325 kJ mol^–1

(d) 359 kJ mol^–1

Ans. (**)

Sol.

39. Among the following,

[B₁₂H₁₂]^2–, [Ni₅(CO)₁₂]^2–, [C₂B₉H₁₁]^2–, Rh₆(CO)₁₆, Os₆(CO)₂₀, B₅H₁₁, B₆H₁₀

the total number of species having nido structure is ______

(Given: atomic numbers of H, B, C, O, Ni, Rh and Os are 1, 5, 6, 8, 28, 45 and 76 respectively)

Ans. (3)

Sol.

Number of species which are nido is 3.

40. Major products P and Q, formed in the reactions given below, are:

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

41. The total number of g_|| lines expected in the EPR spectrum of a solution of bis(salicylaldimine) copper(II) having pure ⁶³Cu and ¹⁴N at 77 K is ________

Ans. (60)

Sol. Total number of g₁₁ lines expected in the EPR spectrum of solution having pure ⁶³Cu and ¹⁴N at 77 K.

Observation of spectral line linear

= 4 × 5 × 3 = 60 lines.

Correct option is 60 lines

42. Major product formed in the following reaction sequence is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

43. Consider that AgX crystallizes in rock salt structure. The density of AgX is 6477 kg/m³ and unit cell length is 577.5 pm. Atomic weight of Ag is 107.87 g mol^–1. The atomic weight of X (in g mol^–1, rounded off to two decimal places) is ________

Ans. (**)

Sol. Density of AgX = 6477 kg/m³

a = 577.5 pm

AgX crystallizes in rock salt str.

So, Z = 4

= 186.2 g/mol = M(AgX)

As atomic weight of Ag is 107.87 g/mol

So, atomic weight of X is

44. In the following reaction sequence, the major products Q and R are:

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

45. The van der Waals constants a and b for gaseous CO are given as 1.49 L² atm mol^–2 and 0.0399 L mol^–1, respectively. The fugacity (in atm, rounded off to two decimal places) of CO at 35ºC and 95 atm is _____

(Given: R = 0.082 L atm K^–1 mol^–1)

Ans. (**)

Sol. For one mole of the gas, the van der Waals equation is

To the first approximation V_m in the correction terms may be replaced by RT/p. Thus, we have

f = 0.93 (95 atm)

f = 88.47 atm

46. The rate of solvolysis of the given compounds is in the order:

(a) Q > T > R > P > S

(b) T > Q > R > P > S

(c) T > R > Q > S > P

(d) R > T > Q > S > P

Ans. Rate of solvolysis is:

So, rate of solvolysis is T > R > Q > S > P

Sol. Rate of solvolysis is:

So, rate of solvolysis is T > R > Q > S > P

47. In the electronic absorption spectrum of an aqueous solution of [Ni(NH₃)₆]²⁺, a very weak band is observed between the bands due to the transitions . The transition responsible for the very weak band is

(Given: atomic number of Ni is 28)

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

two spin-forbidden transition.

48. Major products P and Q, in the given reaction sequence, are:

(a)

(b)

(c)

(d)

Ans. (**)

Sol.

49. Among the following sets,

the total number of set(s) of diastereomeric pair(s) is ________

Ans. (**)

Sol.

50. The % error (rounded off to two decimal places) in the ground state energy of a particle in a one dimensional box of length 'a' described by a trial variation function = x(a – x), where 0 < x < a, is ______

(Given: The true ground state energy of the above system is )

Ans. (1.32)

Sol. Given; Trial wave function

F = x (a – x) 0 < x < a

To find; ^ error in ground state energy of particle in 10 box

Also given; true ground state energy of above system is ;

Solution for given trial wave function

Percentage error = 1.32

51. The experimental magnetic moment (3.4 BM) of a hydrated salt of Eu³⁺ at 27ºC is significantly different from the calculated value. The difference is due to

(Given: atomic number of Eu is 63)

(a) strong spin–orbit coupling.

(b) pairing of electrons in f-orbitals.

(c) population of electrons at higher J level(s) via thermal excitation.

(d) strong ligand field splitting of f-orbitals.

Ans. (**)

Sol. Difference in magnetic moment of Eu³⁺ is due to population of electrons at higher J level via thermal excitation.

52. The CORRECT statement with respect to the stereochemistry of -hydroxy acids P and Q formed in the following reactions is:

(a) P is formed with inversion of configuration and Q with retention of configuration.

(b) Both P and Q are formed with retention of configuration.

(c) P is formed with retention of configuration and Q with inversion of configuration.

(d) Both P and Q are formed with inversion of configuration.

Ans. (d)

Sol.

53. Among the following,

the total number of compounds showing characteristic carbonyl stretching frequency less than 1700 cm^–1 in their IR spectra is _______

Ans. (**)

Sol.

So, one compound shows stranding frequency less than 1700 cm^–1.

54. Assuming no interaction between vibrational and rotational energy levels in HF, the frequency (in cm^–1, rounded off to the nearest integer) of the R branch line originating from J = 4 in its IR spectrum is _____

(Given: Rotational constant for HF = 19.35 cm^–1; )

Ans. (**)

Sol. Given: No interaction between vibrational and rotational energy lavels in HF.

To find: frequency of R-branch line originating from J = 4, in its IR spectra.

Also given

Rotational constant of nf = 19.35 cm^–1

We know that selection rule for R-pressure is

= 4138.5 + 10 × 19.35

55. The CORRECT 'voltage (E) versus time' excitation signal used in cyclic voltammetry is

(a)

(b)

(c)

(d)

Ans. (**)

Sol.