IIT JAM Chemistry - Previous Year Paper 019

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 n^th 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

(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)

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

O₂^2– 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

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 e_g^* 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)

Ans. 3

Sol.

Correct option is (c).

9. The major product formed in the following reaction is

(a)
(b)

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

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)

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.

(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)

Ans. 2

Sol.

25. In ¹H NMR spectrum of the given molecule, the correct
order of chemical shifts of the labelled protons (H^X,
H^Y, H^Z) is

(a) H^Z > H^X > H^Y (b) H^Z > H^Y > H^X (c) H^X > H^Y > H^Z (d) H^Y > H^X > H^Z

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 H^Y > H^X > H^Z.

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

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

Ans. 2

Sol.

Br₂/H₂O, oxidation does not affect configuration of C₂ and in 2^nd 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)

Ans. 1

Sol.

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

(a) (b)

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 pK_a 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 pK_a 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 pK_a

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

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.

(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.

(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

(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.

(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)

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)

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 (pK_a = 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 pK_a 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.

Correct answer is (3)

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 3^rd 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

IIT JAM Chemistry – Previous Year Paper 019

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