Match List- I with List- II . List- I Electronic configuration of elements List- II Δ _ i H in kJ mol ^ - 1 (a) 1 s ^ 2 2 s ^ 2 (p) 801 (b) 1 s ^ 2 2 s ^ 2 2 p ^ 4 (q) 899 (c) 1 s ^ 2 2 s ^ 2 2 p ^ 3 (r) 1314 (d) 1 s ^

Correct answer: A. (a) → (q), (b) → (r), (c) → (s), (d) → (p) (a) 1 s ^ 2 2 s ^ 2 → Be (b) 1 s ^ 2 2 s ^ 2 2 p ^ 4 → O (c) 1 s ^ 2 2 s ^ 2 2 p ^ 3 → N (d) 1 s ^ 2 2 s ^ 2 2 p ^ 1 → B The ionization enthalpy order is B Be has more IE compared to B due to extra stability and N has more IE compared to oxygen due to extra stability Hence, N → 1402kJ mol ^ - 1 O → 1314kJ mol ^ - 1 B → 801kJ mol ^ - 1 Be → 899kJ mol ^ - 1

JEE Main 202126 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Morning Shift
Subject: Chemistry
Chapter: Classification of Elements and Periodicity in Properties

JEE Main 2021 · 26 February · Morning Shift · Chemistry · Classification of Elements and Periodicity in Properties

Match List- I with List- II . List- I Electronic configuration of...

Match List- $I$ with List- $II$ .

	List- $I$ Electronic configuration of elements	List- $II$ ${\Delta}_{i}H$ in kJ{mol}^{- 1}
\left(a\right) $</td> <td>$ 1{s}^{2}2{s}^{2} $</td> <td style="text-align: center;">\left(p\right)$	$801$
\left(b\right) $</td> <td>$ 1{s}^{2}2{s}^{2}2{p}^{4} $</td> <td style="text-align: center;">\left(q\right)$	$899$
\left(c\right) $</td> <td>$ 1{s}^{2}2{s}^{2}2{p}^{3} $</td> <td style="text-align: center;">\left(r\right)$	$1314$
\left(d\right) $</td> <td>$ 1{s}^{2}2{s}^{2}2{p}^{1} $</td> <td style="text-align: center;">\left(s\right)$	$1402$

Choose the most appropriate answer from the options given below:

Options

A $\left(a\right) \to \left(q\right), \left(b\right) \to \left(r\right), \left(c\right) \to \left(s\right), \left(d\right) \to \left(p\right)$
B $\left(a\right) \to \left(p\right), \left(b\right) \to \left(r\right), \left(c\right) \to \left(s\right), \left(d\right) \to \left(q\right)$
C $\left(a\right) \to \left(p\right), \left(b\right) \to \left(s\right), \left(c\right) \to \left(r\right), \left(d\right) \to \left(q\right)$
D $\left(a\right) \to \left(s\right), \left(b\right) \to \left(p\right), \left(c\right) \to \left(q\right), \left(d\right) \to \left(r\right)$

Correct answer

A. $\left(a\right) \to \left(q\right), \left(b\right) \to \left(r\right), \left(c\right) \to \left(s\right), \left(d\right) \to \left(p\right)$

Step-by-step solution

(a) $1{s}^{2}2{s}^{2} \to Be$

(b) $1{s}^{2}2{s}^{2}2{p}^{4} \to O$

(d) $1{s}^{2}2{s}^{2}2{p}^{1} \to B$

The ionization enthalpy order is $B < Be < O < N$

$Be$ has more $IE$ compared to $B$ due to extra stability $\text{ and } N$ has more $IE$ compared to oxygen due to extra stability Hence,

$N \to 1402kJ{mol}^{- 1}$

$O \to 1314kJ{mol}^{- 1}$

$B \to 801kJ{mol}^{- 1}$

$Be \to 899kJ{mol}^{- 1}$

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