JEE · Chemistry
JEE Main 2021 (24 Feb, Shift 1)
2021
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18 questions
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Q31
mcq
Which of the following are isostructural pairs ?
A. $\text{SO}_4^{2-}$ and $\text{CrO}_4^{2-}$
B. $\text{SiCl}_4$ and $\text{TiCl}_4$
C. $\text{NH}_3$ and $\text{NO}_3^-$
D. $\text{BCl}_3$ and $\text{BrCl}_3$
a.
A and B only
Correct Answer
b.
A and C only
c.
B and C only
d.
C and D only
Step-by-Step AI Explanation
Verified Method
- $\text{SO}_4^{2-}$ and $\text{CrO}_4^{2-}$ both have tetrahedral geometry (isostructural).
- $\text{SiCl}_4$ and $\text{TiCl}_4$ both have tetrahedral geometry (isostructural).
- $\text{NH}_3$ is trigonal pyramidal while $\text{NO}_3^-$ is trigonal planar (not isostructural).
- $\text{BCl}_3$ is trigonal planar while $\text{BrCl}_3$ is T-shaped (not isostructural).
Hence, only pairs A and B are isostructural.
Q33
mcq
Consider the elements Mg, Al, S, P and Si, the correct increasing order of their first ionization enthalpy is :
a.
$\text{Al} < \text{Mg} < \text{Si} < \text{S} < \text{P}$
Correct Answer
b.
$\text{Mg} < \text{Al} < \text{Si} < \text{P} < \text{S}$
c.
$\text{Mg} < \text{Al} < \text{Si} < \text{S} < \text{P}$
d.
$\text{Al} < \text{Mg} < \text{S} < \text{Si} < \text{P}$
Step-by-Step AI Explanation
Verified Method
Across the 3rd period, first ionization enthalpy generally increases with increasing nuclear charge, with two anomalies:
1. $\text{Mg}$ ($[\text{Ne}] 3s^2$, fully-filled $s$-orbital) has a higher first ionization enthalpy than $\text{Al}$ ($[\text{Ne}] 3s^2 3p^1$).
2. $\text{P}$ ($[\text{Ne}] 3s^2 3p^3$, half-filled $p$-subshell) has a higher first ionization enthalpy than $\text{S}$ ($[\text{Ne}] 3s^2 3p^4$).
Thus, the correct increasing order of first ionization enthalpy is:
$$\text{Al} < \text{Mg} < \text{Si} < \text{S} < \text{P}$$
Q35
mcq
(A) $\text{HOCl} + \text{H}_2\text{O}_2 \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^- + \text{O}_2$
(B) $\text{I}_2 + \text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow 2\text{I}^- + 2\text{H}_2\text{O} + \text{O}_2$
Choose the correct option.
a.
$\text{H}_2\text{O}_2$ acts as oxidising agent in equations (A) and (B).
b.
$\text{H}_2\text{O}_2$ acts as reducing agent in equations (A) and (B).
Correct Answer
c.
$\text{H}_2\text{O}_2$ act as oxidizing and reducing agent respectively in equations (A) and (B).
d.
$\text{H}_2\text{O}_2$ acts as reducing and oxidising agent respectively in equations (A) and (B).
Step-by-Step AI Explanation
Verified Method
In reaction (A), chlorine in $\text{HOCl}$ is reduced from oxidation state $+1$ to $-1$ in $\text{Cl}^-$. In reaction (B), iodine is reduced from oxidation state $0$ in $\text{I}_2$ to $-1$ in $\text{I}^-$. In both reactions, oxygen in $\text{H}_2\text{O}_2$ is oxidised from $-1$ to $0$ (in $\text{O}_2$), which means $\text{H}_2\text{O}_2$ acts as a reducing agent in both equations.
Q36
mcq
$\text{Al}_2\text{O}_3$ was leached with alkali to get X. The solution of X on passing of gas Y, forms Z.
X, Y and Z respectively are :
a.
$\text{X} = \text{Na}[\text{Al}(\text{OH})_4],\; \text{Y} = \text{SO}_2,\; \text{Z} = \text{Al}_2\text{O}_3$
b.
$\text{X} = \text{Al}(\text{OH})_3,\; \text{Y} = \text{SO}_2,\; \text{Z} = \text{Al}_2\text{O}_3\cdot x\text{H}_2\text{O}$
c.
$\text{X} = \text{Al}(\text{OH})_3,\; \text{Y} = \text{CO}_2,\; \text{Z} = \text{Al}_2\text{O}_3$
d.
$\text{X} = \text{Na}[\text{Al}(\text{OH})_4],\; \text{Y} = \text{CO}_2,\; \text{Z} = \text{Al}_2\text{O}_3\cdot x\text{H}_2\text{O}$
Correct Answer
Step-by-Step AI Explanation
Verified Method
During the Bayer process for the extraction of aluminium:
1. $\text{Al}_2\text{O}_3(\text{s}) + 2\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4](\text{aq})$ (X)
2. The aluminate solution is neutralised by passing $\text{CO}_2$ (Y) gas:
$2\text{Na}[\text{Al}(\text{OH})_4](\text{aq}) + \text{CO}_2(\text{g}) \rightarrow \text{Al}_2\text{O}_3\cdot x\text{H}_2\text{O}(\text{s})\text{ (Z)} + 2\text{NaHCO}_3(\text{aq})$.
Q37
mcq
The electrode potential of $\text{M}^{2+}/\text{M}$ of 3d-series elements shows positive value for :
a.
Fe
b.
Co
c.
Zn
d.
Cu
Correct Answer
Step-by-Step AI Explanation
Verified Method
Among the 3d transition series elements, copper has a positive standard electrode potential ($E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$) because the sum of its high enthalpy of atomisation and enthalpy of ionisation is not balanced by its enthalpy of hydration.
Q38
mcq
The major components in "Gun Metal" are :
a.
Cu, Sn and Zn
Correct Answer
b.
Cu, Zn and Ni
c.
Cu, Ni and Fe
d.
Al, Cu, Mg and Mn
Step-by-Step AI Explanation
Verified Method
Gunmetal (also known as red brass) is an alloy primarily composed of copper ($\sim 88\%$), tin ($\sim 10\%$), and zinc ($\sim 2\%$).
Q39
mcq
The gas released during anaerobic degradation of vegetation may lead to :
a.
Acid rain
b.
Global warming and cancer
Correct Answer
c.
Corrosion of metals
d.
Ozone hole
Step-by-Step AI Explanation
Verified Method
Anaerobic degradation of vegetation produces methane ($\text{CH}_4$). Methane is a potent greenhouse gas that causes global warming and can contribute to toxic photochemical reactions and health hazards.
Q40
mcq
Which of the following compound gives pink colour on reaction with phthalic anhydride in conc. $\text{H}_2\text{SO}_4$ followed by treatment with $\text{NaOH}$ ?
a.
4-propylphenol
b.
2-propylphenol
Correct Answer
c.
4-methyl-6-propylbenzene-1,3-diol
d.
4-propyl-5-methylbenzene-1,2-diol
Step-by-Step AI Explanation
Verified Method
The phthalein dye test requires the para-position with respect to the phenolic $-\text{OH}$ group to be unoccupied, as the condensation of phenol with phthalic anhydride occurs at the para positions to form a phthalein indicator (which turns pink in alkaline solution). In 2-propylphenol (ortho-propylphenol), the para-position is free, whereas in 4-propylphenol it is blocked.
Q41
mcq
What is the major product formed by HI on reaction with $(\text{CH}_3)_3\text{C}-\text{CH}=\text{CH}_2$ ?
a.
$(\text{CH}_3)_3\text{C}-\text{CH}_2-\text{CH}_2\text{I}$
b.
$(\text{CH}_3)_2\text{C}(\text{I})-\text{CH}(\text{CH}_3)_2$
Correct Answer
c.
$(\text{CH}_3)_3\text{C}-\text{CH}(\text{I})-\text{CH}_3$
d.
$\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{I})-\text{CH}_2-\text{CH}_3$
Step-by-Step AI Explanation
Verified Method
Electrophilic addition of $\text{H}^+$ to 3,3-dimethylbut-1-ene initially forms the secondary carbocation $(\text{CH}_3)_3\text{C}-\text{CH}^+-\text{CH}_3$. A 1,2-methyl shift occurs to generate the more stable tertiary carbocation $(\text{CH}_3)_2\text{C}^+-\text{CH}(\text{CH}_3)_2$. Attack of iodide ion ($\text{I}^-$) on this tertiary carbocation gives 2-iodo-2,3-dimethylbutane as the major product.
Q42
mcq
Which of the following reagent is used for the following reaction ?
$$\text{CH}_3\text{CH}_2\text{CH}_3 \xrightarrow{?} \text{CH}_3\text{CH}_2\text{CHO}$$
a.
Copper at high temperature and pressure
b.
Molybdenum oxide
Correct Answer
c.
Manganese acetate
d.
Potassium permanganate
Step-by-Step AI Explanation
Verified Method
Controlled oxidation of alkanes with molybdenum oxide ($\text{Mo}_2\text{O}_3$) gives aldehydes, whereas copper under heat and pressure gives alcohols, and manganese acetate gives carboxylic acids.
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Q43
mcq
The product formed in the first step of the reaction of
$$\text{CH}_3-\text{CH}_2-\overset{\begin{matrix}\text{Br}\\|\end{matrix}}{\text{CH}}-\text{CH}_2-\underset{\begin{matrix}|\\\text{Br}\end{matrix}}{\text{CH}}-\text{CH}_3$$
with excess $\text{Mg}/\text{Et}_2\text{O} (\text{Et} = \text{C}_2\text{H}_5)$ is :
a.
$$\begin{matrix} \text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2-\text{CH}-\text{CH}_3 \\ \quad\quad\quad\quad\quad|\quad\quad\quad\quad| \\ \text{CH}_3-\text{CH}-\text{CH}_2-\text{CH}-\text{CH}_2-\text{CH}_3 \end{matrix}$$
b.
1-ethyl-2-methylcyclopropane (shown as a 3-membered ring with ethyl and methyl substituents)
c.
$$\begin{matrix} \text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2-\text{CH}-\text{CH}_3 \\ \quad\quad\quad\quad\quad|\quad\quad\quad\quad| \\ \text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2-\text{CH}-\text{CH}_3 \end{matrix}$$
d.
$$\text{CH}_3\text{CH}_2-\overset{\begin{matrix}\text{MgBr}\\|\end{matrix}}{\text{CH}}-\text{CH}_2-\underset{\begin{matrix}|\\\text{MgBr}\end{matrix}}{\text{CH}}-\text{CH}_3$$
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Q44
mcq
What is the final product (major) 'A' in the given reaction ?
1-(2-methylcyclohexyl)ethan-1-ol $\xrightarrow{\text{HCl}} \text{'A' (major product)}$
a.
1-(1-chloroethyl)-2-methylcyclohexane
b.
1-ethenyl-2-methylcyclohexane
c.
1-ethylidene-2-methylcyclohexane
d.
1-chloro-1-ethyl-2-methylcyclohexane
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Q45
mcq
Identify products A and B.
1-methylcyclopentene $\xrightarrow[273\text{ K}]{\text{dil. KMnO}_4} \text{A} \xrightarrow{\text{CrO}_3} \text{B}$
a.
A : 1-methylcyclopentane-1,2-diol, B : 2-methylcyclopent-2-en-1-one
b.
A : 1-methylcyclopentane-1,2-diol, B : 2-hydroxy-2-methylcyclopentan-1-one
c.
A : 2-methylcyclopentan-1-ol, B : 2-methylcyclopentan-1-one
d.
A : $\text{OHC}-\text{CH}_2\text{CH}_2\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3$, B : $\text{HOOC}-\text{CH}_2\text{CH}_2\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3$
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Q46
mcq
In the following reaction the reason why *meta*-nitro product also formed is :
Aniline $\xrightarrow[\text{Conc. H}_2\text{SO}_4,\, 288\text{ K}]{\text{Conc. HNO}_3} \text{[ A ] (51\%)} + \text{[ B ] (47\%)} + \text{[ C ] (2\%)}$
a.
$-\text{NH}_2$ group is highly *meta*-directive
b.
$-\text{NO}_2$ substitution always takes place at *meta*-position
c.
Formation of anilinium ion
d.
low temperature
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Q47
mcq
'A' and 'B' in the following reactions are :
Aniline $\xrightarrow[\text{KCN}]{\text{NaNO}_2/\text{HCl}} \text{(A)} \xrightarrow{\text{SnCl}_2/\text{HCl}/\text{H}_3\text{O}^+} \text{(B)}$
a.
(A) : $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$, (B) : $\text{C}_6\text{H}_5\text{Cl}$
b.
(A) : $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$, (B) : $\text{C}_6\text{H}_5\text{CHO}$
c.
(A) : $\text{C}_6\text{H}_5\text{CN}$, (B) : $\text{C}_6\text{H}_5\text{Cl}$
d.
(A) : $\text{C}_6\text{H}_5\text{CN}$, (B) : $\text{C}_6\text{H}_5\text{CHO}$
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Q48
mcq
Match List I with List II.
List I (Monomer Unit)
(a) Caprolactam
(b) 2-Chloro-1,3-butadiene
(c) Isoprene
(d) Acrylonitrile
List II (Polymer)
(i) Natural rubber
(ii) Buna-N
(iii) Nylon 6
(iv) Neoprene
Choose the correct answer from the options given below :
a.
(a) $\rightarrow$ (i), (b) $\rightarrow$ (ii), (c) $\rightarrow$ (iii), (d) $\rightarrow$ (iv)
b.
(a) $\rightarrow$ (iv), (b) $\rightarrow$ (iii), (c) $\rightarrow$ (ii), (d) $\rightarrow$ (i)
c.
(a) $\rightarrow$ (ii), (b) $\rightarrow$ (i), (c) $\rightarrow$ (iv), (d) $\rightarrow$ (iii)
d.
(a) $\rightarrow$ (iii), (b) $\rightarrow$ (iv), (c) $\rightarrow$ (i), (d) $\rightarrow$ (ii)
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Q49
mcq
Out of the following, which type of interaction is responsible for the stabilisation of $\alpha$-helix structure of proteins ?
a.
vander Waals forces
b.
Covalent bonding
c.
Ionic bonding
d.
Hydrogen bonding
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Q50
mcq
Given below are two statements :
Statement I : Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame.
Statement II : Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame.
In the light of the above statements, choose the most appropriate answer from the options given below.
a.
Both Statement I and Statement II are true
b.
Both Statement I and Statement II are false
c.
Statement I is true but Statement II is false
d.
Statement I is false but Statement II is true
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