1. For a reaction of the type A + B Products, the unit of the rate constant is mol L–1s–1. The overall order of the reaction is

(a) 0 (b) 1 (c) 2 (d) 3

Ans. 1

Sol. For any nth order reaction

For zero order reaction, n = 0

A = A0 – kt

Units of k is mol L–1 s–1.

So above A → B conversion follows zero order zero order kinetics.

Correct option is (a)

2. The thermodynamic criterion for spontaneity of a process in a system under constant volume and temperature and in the absence of any work other than expansion work (if any) is

(a) change in entropy is positive

(b) change in enthalpy is negative

(c) change in Helmholtz free energy is negative

(d) change in Gibbs free energy is negative

Ans. 3

Sol. U = Q – PV … (1)

A = V – Ts … (2)

A = Q – PV – TS

taking differential form both side.

dA = dq – PdV – VdP – Tds – SdT

also

(dA) = –PdV – VdP – SdT

(dA)V,T = –VdP

Helm holtz free energy must be for spontaneity of a process

therefore dP should be greater than zero and dA is –ve.

Correct option is (c)

3. The number of vibrational mode(s) of a carbon dioxide molecule that can be detected using infrared spectroscopy is

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

Ans. 3

Sol. Total degrees of freedom 3N

Where N value for CO2 is 3

Therefore

For vibrational degrees of freedom

Out of 4 vibrational modes, one with symmetrical stretch is IR inactive, therefore remaining 3 are IR active can be detected in IR spectra.

Correct option is (c)

4. For three non-coplanar vectors a, b and c, the expression a· (b × c) can be written as

(a) (a × b) · c (b) (a × b) · (a × c) (c) (a · b) × (a · c) (d) (a · b) × c

Ans. 1

Sol. a.(b×c) = [abc]

Correct option is (a)

5. Correct trend in the bond order is

(a) (b)

(c) (d)

Ans. 4

Sol. Species No. of Bonding e– s No. of Antibonding e– B.O

O2+ 10 5 2.5

O2– 10 7 1.5

O22– 10 8 1

is correct bond order sequence.

Correct option is (d)

6. The correct option for the metal ion present in the active site of myoglobin, hemocyanin and vitamin B12 respectively is

(a) iron, iron and zinc (b) molybdenum, iron and copper

(c) iron, copper and cobalt (d) molybdenum, copper and cobalt

Ans. 3

Sol. Metallo enzyme Metal at active site

Myoglobin Fe (Iron)

Hemocyanin Cu (Copper)

Vitamin B12 Co (Cobalt)

Correct option is (c)

7. The correct order of wavelength of the halide to metal charge-transfer band of [Co(NH3)Cl]2+ (I), [Co(NH3)5Br]2+ (II) and [Co(NH3)5I]2+ (III) is

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

Ans. 2

Sol.

Charge transfer is due to bonding to eg* antibonding MO’s electronic transistion

Energy gap between is minimum and maximum for and hence value of photon to effect charge transfer follows the order

III > II > I

Correct option is (b)

8. The correct option for the major products of the following reaction is

(a) (b)

(c) (d)

Ans. 3

Sol.

Correct option is (c).

9. The major product formed in the following reaction is

(a) (b)

(c) (d)

Ans. 4

Sol.

10. The complementary strand for the following single strand of DNA is

(a)

(b)

(c)

(d)

Ans. 1

Sol. In DNA the two strands must be antiparallel (if one is 3′ to 5′ then other is 5′ to 3′) and complementary (A pairs with T and G pairs with C)

Correct option is (a)

11. The function has a minimum at

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

Ans. d

Sol.

Correct option is (c)

12. The correct option for the number of bending modes of vibration in each of H2O, CS2 and SO2 molecules, respectively, is

(a) 1, 2 and 2 (b) 2, 2 and 1 (c) 2, 1 and 2 (d) 1, 2 and 1

Ans. d

Sol. For linear molecules like CO2, Cs2 there are 2 bending modes

For non linear triatomic bent molecules like OH2, SO2

13. The total number of degrees of freedom of an HBr molecule that is constrained to translate along a straight line but does not have any constraints for its rotation and vibration is

(a) 6 (b) 5 (c) 4 (d) 3

Ans. d

Sol. Total degree of freedom in HBr = 3N = 3 × 2 = 6

Translational degree of freedom = 3

Remaining = 6 – 3 = 3

Correct option is (d)

14. According to the kinetic theory of gases, the ratio of the root mean square velocity of molecular oxygen and molecular hydrogen at 300 K is

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

Ans. c

Sol. According to KTG, root mean square speed of gas is given as

15. The half-life of the chemical reaction, for initial reactant concentrations of 0.1 and 0.4 mol L–1 are 200 and 50 s, respectively. the order of the reaction is

(a) 0 (b) 1 (c) 2 (d) 3

Ans. c

Sol.

16. The ratio of the nearest neighbor atomic distances in body-centered cubic (bcc) and face-centered cubic (fcc) crystals with the same unit cell edge length is

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

Ans. c

Sol. Nearst neighbours in body center lattice are distance apart where as in face center limit cell this distance is

Correct option is (c)

17. The correct trend in the rate of substitution of Cl– by pyridine in the following complexes is

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

Ans. a

Sol. Rate of Cl– by nucleophile is governed by kinetic trans effect of trans ligand. (opposite and leaving group) Strength of trans ligand depends on donar and tendency of ligands.

is order of ligand to exert trans effect, therefore rate of substituion of Cl– is fastest in (iv) and slowest in (iii)

Correct option is (a)

18. In qualitative inorganic analysis of metal ions, the ion which precipitates as sulfide in the presence of H2S in warm dilute HCl is

(a) Cr3+ (b) Al3+ (c) Co2+ (d) Bi3+

Ans. d

Sol.

19. The correct statement regarding the observed magnetic properties of NO, O2, B2 and C2 in their ground state is

(a) NO, B2, and C2 are paramagnetic (b) B2, O2 and NO are paramagnetic

(c) O2, C2 and NO are paramagnetic (d) O2, B2 and C2 are paramagnetic

Ans. b

Sol. Observed magnetic properties depends on electronic arrangement whether electrons are unpaired or paired.

(1) NO there is one unpaired electron

(2) O2 there are 2 unpaired electrons

(3) B2 there are 2 unpaired electrons

(4) C2 all electrons are paired.

So NO, O2, B2 are paramagnetic.

Correct option is (b)

20. The observed magnetic moments of octahedral Mn3+, Fe3+ and Co3+ complexes are 4.95, 6.06 and 0.00 BM, respectively. The correct option for the electronic configuration of Mn3+, Fe3+ and Co3+ metal ions in these complexes, respectively, is

(a) (b)

(c) (d)

Ans. c

Sol. Magnetic moment values suggests no. of unpaired electrons in given metal ions

Correct option is (c)

21. Among the following compounds, the one having the lowest boiling point is

(a) SnCl4 (b) GeCl4 (c) SiCl4 (d) CCl4

Ans. 3

Sol. CCl4 molecules are smaller than SiCl4 molecules, and can get closer to one another which would lead to a larger contact area between molecules and therefore a larger intermolecular force.

Also, C–Cl bond in CCl4 is more polar than SiCl4. Hence, in CCl4 there exist high intermolecular attractions due to polarity.

Therefore, the correct order for the boiling point between CCl4 and SiCl4 is:

CCl4 > SiCl4

SiCl4 has lowest boiling point value.

Correct option is (c)

22. The correct option having one complex from each of the following pairs which is more reactive towards the oxidative addition reaction by hydrogen molecule is

Pair 1: IrCl(PMe3)3 (I) and IrCl(CO)(PMe3)2 (II)

Pair 2: IrCl(CO)(PPh3)2 (III) and IrCl3(PPh3) (IV)

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

Ans. 1

Sol. For oxidative addition reaction, metal center should be electronically rich which can be achieved only when there are relatively poor acceptors ligands in the complex.

Complex I[IrCl(PMe3)3] in pair 1 relatively easily undergo oxidative addition reaction.

In pair 2, complex IV [IrCl3(PPh3)], Ir is in +3 oxidation state.

So, less electron density in case of IV in comparison to III in which metal in +1 oxidation state.

Correct option is (a)

23. Among the following, the correct statement is

(a) The density follows the order, Cs > Rb > Li > Na.

(b) The solubility in water follows the order, Cs2CO3 > K2CO3 > Na2CO3 > Li2CO3.

(c) The first ionization potential follows the order, Li > K > Na > Cs.

(d) The melting point follows the order, MgCl2 > BeCl2 > CaCl2 > SrCl2.

Ans. 2

Sol. Solubility of any ionic compound depends on balance of solvation and lattice enthalpy. If solvation enthalpy is greater than lattice enthalpy then considered ionic compound is soluble.

trend of lattice enthalpy changes as

anion is same in all compounds but metal ion changes from Li+ to Cs+ their ionic size increases and hence charge density decrease, therefore, polarization decrease and lattice enthalpy increases.

Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3.

Correct option is (b)

24.

(a) (b)

(c) (d)

Ans. 2

Sol.

25. In 1H NMR spectrum of the given molecule, the correct order of chemical shifts of the labelled protons (HX, HY, HZ) is

(a) HZ > HX > HY (b) HZ > HY > HX (c) HX > HY > HZ (d) HY > HX > HZ

Ans. 4

Sol.

More deshielded the proton, higher is the value of chemical shift.

Extent of electron deficiency or deshielding around proton depends an electronegativily of side groups.

group is attached to highly electronegative oxygen and sp hybridised C atom. It is most electron deficient and hence proton Y is highly deshielded with maximum value.

Similarly, oxygen attached to sp3 C makes it some what electron deficient and therefore proton X comes second. Whereas proton Z attached sp carbon is least deficient and hence correct order of value is HY > HX > HZ.

Correct option is (d)

26. In the following reaction of (D)-Glucose, a product P is formed.

Among the following compounds, the one which will give the same product (P) under identical reaction conditions is

1. (a) (b) (c) (d)

Ans. 2

Sol.

Br2/H2O, oxidation does not affect configuration of C2 and in 2nd step C2 carbon atom is eliminated as CO2 without affecting configuration at C3 or any other carbon in given chain. Therefore

Reacts similarly with Br2/H2O and H2O2/Fe+3 reagent to give same D-arabinose as configuration C3 onward in (B) is same as D glucose.

Correct option is (b)

27.

(a) (b)

(c) (d)

Ans. 1

Sol.

28. The correct option for the product(s) of the following reaction is

(a) (b)

(c) (d)

Ans. 3

Sol. both protons are acidic and hence in the presence of sodamide strong base we have

29. The increasing order of acidity of the given molecules in aqueous media is

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

Ans. 3

Sol. pKa values suggest strength of acidity of any chemical species.

Lower is the pKa value, stronger is the acidic behaviour. It also depends on stability of conjugate base. More stable is the conjugate base, stronger is acid.

Due to resonance stablization and 3 resonating structure phenol’s acidity comes second with pKa value nearby 10.

conjugate base is more stable due to elemental electronegative oxygen possess excess –ve change instead carbon in case II.

correct acidity order is

III > I > IV > II with reverse pKa

Sequence

30. The compound formed upon subjecting an aliphatic amine to Lassaigne’s test is

(a) NaNH2 (b) NaNO2 (c) NaCN (d) NaN3

Ans. 3

Sol. if nitrogen is present in the compound, lassaigne’s extract would contain sodium cyanide during fusion. Sodium cyanide is converted to sodium ferrocyanide on treating with ferrons sulphate. On further treating if with ferric chloride, a prussian blue complex, ferriferrocyanide is formed.

31. The eigenvalue(s) of the matrix is/are

(a) –1 (b) 1 (c) 2 (d) 3

Ans. (a, d)

Sol.

32. The unit of the constant ‘a’ in van der Waals equation of state of a real gas can be expressed as

(a) m6 Pa mol–2 (b) m6 J mol–2 (c) m3 Pa mol–2 (d) m3 J mol–2

Ans. (a, d)

Sol. Vander Waal’s equation of state

a = Pa·m6 mol–2

Also m3Pa = J

33. Among the following, microwave active molecule(s) is/are

(a) trans-dichloroethene (b) 1,2-dinitrobenzene

(c) 3-methylphenol (d) para-aminophenol

Ans. (b, c, d)

Sol. For a molecule to be microwave active, it must be a permanent dipole. (Polar in nature)

So B, C and D are IR active molecules.

34. The true statement(s) regarding the brown ring test carried out in the laboratory for the detection of NO3– is/are

(a) Brown ring is due to the formation of the iron nitrosyl complex.

(b) Concentrated nitric acid is used for the test.

(c) The complex formed in the reaction is [Fe(CN)5NO]2–.

(d) The brown colored complex is paramagnetic in nature.

Ans. (a, d)

Sol. A brown ring test is employed for detection of NO3– ions using concentrated sulfuric acid. Presence of nitrate is indicated by appearance of brown ring [Fe(OH2)5NO]2+ at interface of sample solution and concentration H2SO4.

here NO exists as NO– and Fe is in +3 oxidation state which is d5 system and hence paramagnetic.

Correct options are (a and d)

35. The true statement(s) regarding the carbonic anhydrase anzyme is/are

(a) It is involved in peptide bond cleavage.

(b) Redox inactive Zn2+ ion is involved in the catalytic activity of this enzyme.

(c) Activated M–OH2 (M = metal ion) acts as the nucleophile in the enzyme.

(d) The metal ion is coordinated to the side chain of histidine residues.

Ans. (b, c, d)

Sol. Correct options are (b, c and d)

36. The correct statement(s) about NO2, NO2+ and CO2 is/are

(a) Both NO2 and CO2 are paramagnetic.

(b) NO2 is paramagnetic and NO2+ is diamagnetic

(c) Both CO2 and NO2+ have linear geometry.

(d) CO2 and NO2+ are isoelectronic.

Ans. (b, c, d)

Sol. In NO2 wich is a 17e– species has last e– in i/5 antibonding orbital which imparts paramagnetism to this species whereas in NO2+ and CO2 there are all e–s paired with linear geometry and hence there both are dimagnetic and isoelectronic 16e– species.

Correct options are (b, c and d)

37. The compound(s) formed as intermediate(s) in the following reaction sequence is/are

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

Ans. (b)

Sol.

Correct option is (b)

38. The correct statement(s) among the following is/are

(a) Secondary structure of a polypeptide describes the number and type of amino acid residues.

(b) Uracil is a pyrimidine nucleobase.

(c) Natural fatty acids have odd number of carbon atoms.

(d) Reaction of (D)-glucose with Ca(OH)2 gives a product mixture containing (D)-fructose, (D)-mannose, and (D)-glucose.

Ans. (b, d)

Sol. Statements (b and d) are correct

39. The diastereomeric pair(s) among the following option(s) is/are

(a) (b)

(c) (d)

Ans. (a, b, d)

Sol.

Configuration at C2 and C3 are same in both isomers but there is a difference in configuration at C1 of both isomers and hence these are not mirror image therefore diasteromers

Configuration at C2 and C5 are same whereas at C4 there is difference in configuration therefore both are diasteromers.

Correct options are (a, b and d)

40. The reaction(s) that result(s) in the formation of aromatic species is/are

(a) (b)

(c) (d)

Ans. (a, d)

Sol.

Option (a) and (d) are aromatic.

Correct options are (a and d)

41. The bond order of N2+ ion is __________. (Round off to one decimal place)

Ans. (2.5)

Sol. Bond order of N2+ is calculate through its molecular electronic configuration.

42. One liter of a buffer solution contains 0.004 mole of acetic acid (pKa = 4.76) and 0.4 mole of sodium acetate. The pH of the solution is ________. (Round off to two decimal places)

Ans. (6.76)

Sol. This is a acidic buffer composed of weak acid and its conjugate bane.

pH = 4.76 + 2

pH = 6.76

43. The limiting molar conductivity of La3+ and Cl ions in aqueous medium at 298 K are 209.10 × 10–4 and 76.35 × 10–4 S m2 mol–1, respectively. The transport number of Cl in an infinitely dilute aqueous solution of LaCl3 at 298 K is ________. (Round off to two decimal places)

Ans. (0.52)

Sol.

44. The magnetic field strength required to excite an isolated proton to its higher spin state with an electromagnetic radiation of 300 MHz is ________ Tesla (T). (Round off to two decimal places)

[Magnetogyric ratio of proton is 26.75 × 107 rad T–1 s–1]

Ans. (7.0)

Sol.

45. The value of n for the complex [Fe(CO)4(SiMe3)]n satisfying the 18-electron rule is ________.

Ans. (–1)

Sol.

8 + 8 + 1 + n = 18

17 + n = 18

n = 18 – 17 = 1

As n = 1 must be added to achieve 18e

n must be an extra e and hence n = –1

So correct formula of complex is

46. In the structure of P4O10, the number of P-O-P bond(s) is _______.

Ans. (6)

Sol. P4O10

Number of P–O–P bonds = 6

Number of P = O bonds = 4.

47. Number of vertices in an icosahedral closo-borane is _________.

Ans. (12)

Sol. Number of vertices in iscosahedral closo borane is 12.

48. Based on the information given below, the isoelectric point (pI) of lysine is _________. (Round off to one decimal place)

Ans. (9.8)

Sol. Lysine is a basic amino acid. To determine value we need to add two higher pKa values out of given three and then find the average of this value.

49. (R)-2-methyl-1-butanol has a specific rotation of +13.5º. The specific rotation of 2-methyl-1-butanol containing 40% of the (S)-enantiomer is _______º. (Round off to one decimal place)

Ans. (2.7)

Sol. If R isomer has specific rotation = 13.5º then S isomer must have rotation value –13.5º

Also gives

Specific rotation of mixture containing 40% S and 60% R isomer

= 8.1º – 5.4º = 2.7º.

50. The number of gauche-butane interaction(s) in the following compound is __________.

Ans. (3)

Sol.

There are 3 butane gauche interactions present in given system.

51. The ionization energy of hydrogen atom is 13.6 eV and the first ionization energy of sodium atom is 5.1 eV. The effective nuclear charge experienced by the valance electron of sodium atom is ________. (Round off to one decimal place)

Ans. (1.8)

Sol.

Where –ve sign suggests binding energy.

n value is orbit from which electron is ejected.

For Na last e– stays in 3s orbital means 3rd shere

52. One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is ________ kJ mol–1. (Round off to one decimal place)

[Given: Gas constant (R) = 8.3 J K–1 mol–1]

Ans. 5.73 kJ mol–1.

Sol.

53. One liter of an aqueous urea solution contains 6 g of urea. The osmotic presure of the solution at 300 K (assuming an ideal behavior) is ____________ kPa. (Round off to one decimal place)

[Given: Molecular weight of urea is 60, gas constant (R) is 8.3 J K–1 mol–1]

Ans. 249.56 kPa.

Sol.

54. A first order reflection of X-ray from {220} plane of copper crystal is observed at a glancing angle of 22º. The wavelength of the X-ray used is _________ pm. (Round off to one decimal place)

[Given: Copper forms fcc crystal with unit cell edge length of 361 pm.]

Ans. 95.6 pm

Sol.

55. The collision flux of a monoatomic gas on copper surface is 3.0 × 1018 m–2 s–1. Note that copper surface forms a square lattice with lattice constant of 210 pm. If the sticking coefficient of the atom with copper is 1.0, the time taken by the gas to form a complete monolayer on the surface is _______ s. (Round off to one decimal place)

Ans. 7.6 s

Sol. At copper surface, monoatomic gas adsorbs with sticking cofficient equal to 1 means all that atoms of monoatomic gas which collide with copper get adsorb upon it.

Copper form square lattice with lattice constant = 210 pm

Area of unit cell = 210 × 210 × 10–24 m2

= 441 × 10–22 m2

No. of copper unit cells in 1m2.

441 × 10–22 × no. of unit cells = 1

As each unit cell contains only 1 copper atom, therefore for monolayer formation all copper atom must have monoatomic gases atom adsorbed upon them. Therefore to adsorb 2.3 × 1019 gases atom time taken would be

56. The turnover frequency (TOF) for the catalytic reaction,

with 90% yield of the product is ________ hour–1. (Round off to the nearest integer)

Ans. 18 hr–1.

Sol.

turnover frequency means amount of reactant converted to product per mole of catalyst per hour. As 90% of product is formed and 1 mole of A gives 1 mole B

57. A radioactive sample decays to 10% of its initial amount in 4600 minutes. The rate constant of this process is ________ hour–1. (Round off to two decimal places)

Ans. 0.03 hr–1.

Sol. Radioactive decays occur via 1 order kinetics

58. Given that the radius of the first Bohr orbit of hydrogen atom is 53 pm, the radius of its third Bohr orbit is ________ pm. (Round off to the nearest integer)

Ans. 476.1 pm

Sol.

59. 5.3 g of benzaldehyde was reacted with an excess of acetophenone to produce 5.2 g of the enone product as per the reaction shown below. The yield of the reaction is __________%. (Round off to the nearest integer)

Ans. 50%

Sol.

60. Assume that the reaction of MeMgBr with ethylacetate proceeds with 100% conversion to give tert-butanol. The volume of 0.2 M solution of MeMgBr required to convert 10 mL of a 0.025 M solution of ethylacetate to tert-butanol is ________ mL. (Round off to one decimal place)

Ans. 2.5 ml

Sol.

1 mol of ethylacetate reacts with 2 mol of MeMgBr.

Millimoles of ethylacetate = 10 × 0.025 = 0.25 millimoles.

So for complete reaction MeMgBr must be 0.50 millimoles

Volume of 0.2M MeMgBr containing 0.50 millimoles would be

0.2 × V = 0.5