Students aspiring to join some of the prestigious institutions in the country must get a good score in JEE. The students should definitely solve previous yearsβ question papers to crack the examination. All the questions from JEE Main 2020 Chemistry (January 8, Shift 2) are solved here. Learn these questions to excel in the test.
1. Arrange the following bonds according to their average bond energies in descending order:
C-Cl, C-Br, C-F, C-I
Solution:
In C β F there is 2p-2p overlapping involved, in C β Cl the overlapping involved is 2p-3p whereas for C β Br and C β I the overlapping involved are 2p-4p and 2p-5p, respectively. The bond length for the various type of overlapping can be given as:
2p-2p < 2p-3p < 2p-4p < 2p-5p.
As we know that Bond energy Ξ± (1/Bond length)
The order of bond energy comes out: C-F > C-Cl > C-Br > C-I
Answer: d
2. The radius of second Bohr orbit, in terms of the Bohr radius, π_{0} , in Li^{2+} is:
Solution:
The formula for Bohrβs radius for any one electron species is: r = a_{0}n^{2}/ z
for Li^{2+}
: r = a_{0}2^{2}/3
= 4a_{0}/3
Answer: c
3. A metal (A) on heating in nitrogen gas gives compound B. B on treatment with π»2π gives a colourless gas which when passed through πΆπ’ππ4 solution gives a dark blue-violet coloured solution. A and B respectively, are :
Solution:
As it is provided in the question that nitride is being formed so the option c and d can be eliminated. Amongst Mg and Na we already know that Mg can only form nitride so the correct choice is option a.
Answer: a
4. The correct order of the calculated spin-only magnetic moments of complexes A to D is:
A. π(πΆπ)_{4}
B. [ππ(π»_{2}π)_{6}]^{2+}
C. ππ_{2}[ππ(πΆπ)_{4}]
D. πππΆπ_{2}(ππβ_{3})^{2}
Solution:
[Pd(PPh_{3})_{2}Cl_{2}] : Here Pd is in +2 oxidation state and configuration of Pd ^{2+} is [Kr]4d^{8} . As the CFSE value for Pd is very high so all the electrons will be paired and hence magnetic moment for this complex will be zero.
[Ni(CO)_{4}] : Here Ni is in 0 oxidation state and configuration of Ni is [Ar]3d^{8}4s^{2} . As here the ligand is carbonyl which is a strong field ligand, all the electrons will be paired and hence magnetic moment for this complex will be zero.
[Ni(CN)_{4}]^{2-} : Here Ni is in +2 oxidation state and configuration of Ni^{2+} is [Ar]3d^{8} . As here the ligand is cyanide which is a strong field ligand, all the electrons will be paired and hence magnetic moment for this complex will be zero.
[Ni(H_{2}O)_{6}]^{2+} : Here Ni is in +2 oxidation state and configuration of Ni^{2+} is [Ar]3d^{8} . As here the ligand is water which is a weak field ligand, the electrons will not be paired and there are two unpaired electrons in this complex hence magnetic moment for this complex will be β8 BM.
So the order of magnetic moment is A=B=C<D.
Answer: b
5. Hydrogen has three isotopes (A), (B) and (C). If the number of neutron(s) in (A), (B) and (C) respectively, are (x), (y) and (z), the sum of (x), (y) and (z) is:
Solution:
Number of neutrons in protium = 0
Number of neutrons in deuterium = 1
Number of neutrons in tritium = 2
So, total number of neutrons = 3
Answer: c
6. Consider the following plots of rate constant versus 1/π for four different reactions. Which of the following orders is correct for the activation energies of these reactions?
Solution:
To avoid confusion, in this question weβll be denoting activation energy by E_{x}
K = Ae^{βEx/RT}
log K = log A β (E_{x}/2.303RT) -----(1)
Here, the graph given in the question is of a straight line and we know that the equation of straight
line is
y= mx + c ----(2)
Comparing equation 1 with 2 we get,
Slope = βE_{x}/2.303R
So, from the graph we can conclude that the line with the most negative slope will have the
maximum activation energy value.
E_{c} > E_{a} > E_{d} > E_{b}
Answer: b
7. Which of the following compounds is likely to show both Frenkel and Schottky defects in its
crystalline form?
Solution:
Answer: d
8. White phosphorus on reaction with concentrated NaOH solution in an inert atmosphere of πΆπ_{2} gives phosphine and compound (X). (X) on acidification with HCl gives compound (Y). The basicity of compound (Y) is:
Solution:
Here the product B which is mentioned in the question is H_{3}PO_{2}. The structure of H_{3}PO_{2} can be given as:
Answer: d
9. Among the reactions (a) β (d), the reaction(s) that does/do not occur in the blast furnace during the extraction of iron is/are:
A. CaO + SiO_{2}β CaSiO_{3}
B. 3Fe_{2}O_{3} + CO β 2Fe_{3}O_{4} + CO_{2}
C. FeO +SiO_{2} βFeSiO_{3}
D. FeO β Fe + (1/2)O_{2}
Solution:
In metallurgy of iron, CaO is used as flux which is used to remove the impurities of SiO_{2},
CaO + SiO_{2}β CaSiO_{3} .
Also here Fe_{2}O_{3} is reduced by CO to Fe_{3}O_{4} which is further reduced to FeO which is further reduced
to Fe.
3Fe_{2}O_{3} + CO β 2Fe_{3}O_{4} + CO_{2}
Fe_{3}O_{4} + CO β 3FeO + CO_{2}
FeO + CO β Fe + CO_{2}
Answer: c
10. The increasing order of the atomic radii of the following elements is:
A. C
B. O
C. F
D. Cl
E. Br
Solution:
Across the period size decreases, so the order that follows is: C > O >N> F
Down the group size increases and the order is: Br > Cl > F
Change in size down the group is much more significant as compared to across the period.
So, the overall order of radius of elements is: Br > Cl > C > O > F.
Answer: a
11. Among (a) β (d), the complexes that can display geometrical isomerism are:
A. [Pt(NH_{3})_{3}Cl]^{+}
B. [Pt(NH_{3})Cl_{5}]
C. [Pt(NH_{3})_{2}Cl(NO_{2})]
D. [Pt(NH_{3})_{4}ClBr^{] 2+}
Solution:
Answer: a
12. For the following Assertion and Reason, the correct option is:
Assertion: The pH of water increases with increase in temperature.
Reason: The dissociation of water into H+ and OH- an exothermic reaction.
Solution:
H2O β H^{+} + OH^{-} is an endothermic process.
On increasing the temperature the value of K_{w}
increases which will result in decrease in pK_{w} . So we can say that pH of water will decrease on
increasing temperature because pH for water =1/2pK_{w}.
Answer: d
13. For the following Assertion and Reason, the correct option is:
Assertion: For hydrogenation reactions, the catalytic activity increases from group-5 to group11 metals with maximum activity shown by group 7-9 elements
Reason: The reactants are most strongly adsorbed on group 7-9 elements
Solution:
Group 7-9 elements of the periodic table show variable valencies so they have maximum activity because of the increase in adsorption rate.
Answer: a
14. The major product of the following reactions is:
Solution:
Answer: d
15. Find The major product [B] of the following sequence of reactions is:
Solution:
Answer: c
16. Among the following compounds A and B with molecular formula C_{9}H_{18}O_{3}, A is having higher boiling point than B. The possible structures of A and B are
Solution:
Answer: b
17. Kjeldahlβs method cannot be used to estimate nitrogen for which of the following compounds?
Solution:
Answer: c
18. An unsaturated hydrocarbon absorbs two hydrogen molecules on catalytic hydrogenation, and also gives following reaction;
π will be:
Solution:
Answer : b
19: Preparation of Bakelite proceeds via reactions:
Solution:
Bakelite is a condensation polymer of phenol and formaldehyde.
Answer: a
20. Two monomers of maltose are:
Solution:
Maltose is formed by the glycosidic linkage between C-1 of one alpha -D-Glucose unit to the C-4 of
another alpha -D-Glucose.
Answer: b
21: At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is _____.
Solution:
βU = nCvβT
5000 = 4 Γ Cv (500 β 300)
C_{v} = 6.25 JK^{β1} mol^{β1}
Answer: 6.25
22: For an electrochemical cell he ratio [Sn2+]/[Pb2+] when this cell attains equilibrium is ______.
Solution:
Anodic half: Sn β Sn ^{2+} + 2e^{β}
Cathodic half: Pb ^{2+} + 2e^{β} β Pb
Net reaction: Sn + Pb ^{2+} β Pb + Sn ^{2+}
E^{0}_{cell} = E^{0}_{cathode} β E^{0}_{anode}
E^{0}_{cell} = 0.01 V
E_{cell} = E^{0}_{cell}β (0.06/2) logQ
At equilibrium state E_{cell} = 0
So,
0 = 0.01 β (0.06/2)log [Sn ^{2+}]/[Pb ^{2+}]
0.01 = (0.06/2)log [Sn ^{2+}]/[Pb ^{2+}]
log [Sn ^{2+}]/[Pb ^{2+}] =1/3
[Sn ^{2+}]/[Pb ^{2+}] = 2.154
Answer: 2.15
23. NaClO_{3} is used, even in spacecrafts, to produce O_{2}. The daily consumption of pure O_{2} by a person is 492 L at 1 atm, 300 K. How much amount of NaClO_{3}, in grams, is required to produce O_{2} for the daily consumption of a person at 1 atm, 300 K ?
NaClO_{3}(s) + Fe(s) β O_{2}(g) + FeO(s) + NaCl(s)
R = 0.082 L atm molβ1 Kβ1
Solution:
Mol of NaClO_{3} = mol of O_{2}
Mol of O_{2} =PV/RT =(1x492)/(0.082x300)= 20 mol
Molar mass of NaClO_{3} is 106.5
So, mass = 20 x106.5 = 2130 g = 2.13 kg
Answer: 2.13
24. Complexes [ML_{5}] of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal and geometries, respectively. The sum of the 90^{0}, 120^{0} and 180^{0} L-M-L angles in the two complexes is ______.
Solution:
For trigonal bipyramidal geometry
Total number of 180^{0} L-M-L bond angles = 1
Total number of 90^{0} L-M-L bond angles = 6
Total number of 120^{0} L-M-L bond angles = 3
Total = 10
For square pyramidal geometry
Total number of 180^{0} L-M-L bond angles = 2
Total number of 90^{0} L-M-L bond angles = 8
Total number of 120^{0} L-M-L bond angles = 0
Total = 10
Total for both the structures = 20
Answer: 20
25: In the following sequence of reactions, the maximum number of atoms present in molecule βCβ in
one plane is _____.
(Where A is a lowest molecular weight alkyne).
Solution:
Answer: 13