Outermost electronic configurations of four elements $A,B,C,D$ are given below: (A) $3{s}^{2}$ (B) $3{s}^{2}3{p}^{1}$ (C) $3{s}^{2}3{p}^{3}$ (D) $3{s}^{2}3{p}^{4}$ The correct order of first ionization enthalpy for them is

JEE Main 2022 — Chemistry Inorganic Chemistry

2022mcqmedium

Outermost electronic configurations of four elements $A,B,C,D$ are given below:

(A) $3{s}^{2}$

(B) $3{s}^{2}3{p}^{1}$

(D) $3{s}^{2}3{p}^{4}$

The correct order of first ionization enthalpy for them is

Official previous-year question

Held on 27 Jul 2022 · Verified 6 Jul 2026.

Options

A
$A<B<C<D$
B
$B<A<D<C$
C
$B<D<A<C$
D
$B<A<C<D$

Did you get this right?

Solution

(A) $3{s}^{2}\rightarrow \mathrm{Mg}$

(B) $3{s}^{2}3{p}^{1}\rightarrow \mathrm{Al}$

(D) $3{s}^{2}3{p}^{4}\rightarrow S$

$\mathrm{Al}$ has one unpaired electron in its highest energy orbital $(3p)$ , and $\mathrm{Mg}'s$ highest energy orbital $(3s)$ has the paired electrons.

It is energetically favourable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.

Compared to $16\mathrm{th}$ group, $15$ group have high ionisation energy. It is because of the stable half filled configuration of he $15\mathrm{th}$ group elements.

$P-3{s}^{2}3{p}^{3}$

$S-3{s}^{2}3{p}^{4}$

JEE Main 2022 Chemistry, Classification of Elements and Periodicity in Properties — solution figure

Therefore, P has more ionisation energy than S.

Hence, the correct order of first ionization enthalpy is $B<A<D<C$

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JEE Main 2022 — Chemistry Inorganic Chemistry

2022mcqmedium

Outermost electronic configurations of four elements $A,B,C,D$ are given below:

(A) $3{s}^{2}$

(B) $3{s}^{2}3{p}^{1}$

(D) $3{s}^{2}3{p}^{4}$

The correct order of first ionization enthalpy for them is

Official previous-year question

Held on 27 Jul 2022 · Verified 6 Jul 2026.

Options

A
$A<B<C<D$
B
$B<A<D<C$
C
$B<D<A<C$
D
$B<A<C<D$

Did you get this right?

Solution

(A) $3{s}^{2}\rightarrow \mathrm{Mg}$

(B) $3{s}^{2}3{p}^{1}\rightarrow \mathrm{Al}$

(D) $3{s}^{2}3{p}^{4}\rightarrow S$

$\mathrm{Al}$ has one unpaired electron in its highest energy orbital $(3p)$ , and $\mathrm{Mg}'s$ highest energy orbital $(3s)$ has the paired electrons.

It is energetically favourable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.

Compared to $16\mathrm{th}$ group, $15$ group have high ionisation energy. It is because of the stable half filled configuration of he $15\mathrm{th}$ group elements.

$P-3{s}^{2}3{p}^{3}$

$S-3{s}^{2}3{p}^{4}$

JEE Main 2022 Chemistry, Classification of Elements and Periodicity in Properties — solution figure

Therefore, P has more ionisation energy than S.

Hence, the correct order of first ionization enthalpy is $B<A<D<C$

Your note

JEE Main Inorganic Chemistry in other years

Keep practicing Inorganic Chemistry

Work through every JEE Main Inorganic Chemistry PYQ, year by year.

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JEE Main 2022 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main 2022 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

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Other Chemistry topics for JEE Main

Physical Chemistry

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