JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqmedium

According to $\mathrm{MO}$ theory the bond orders for ${{O}_{2}}^{2-}$ , $\mathrm{CO}$ and ${\mathrm{NO}}^{+}$ respectively, are

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

Options

A
$1,3$ and $3$
B
$1,3$ and $2$
C
$1,2$ and $3$
D
$2,3$ and $3$

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Solution

(i) Molecular orbital configuration of ${O}_{2}^{2-}$ is $\mathrm{KK}(\sigma 2s{)}^{2}(\sigma *2s{)}^{2}𝜋(2{p}_{x}{)}^{2}𝜋(2{p}_{y}{)}^{2}(\sigma 2{p}_{z}{)}^{2}𝜋*(2{p}_{x}{)}^{2}𝜋*(2{p}_{y}{)}^{2}$

Bond order = $\frac{1}{2}({N}_{b}-{N}_{a})$ = $\frac{1}{2}(8-6)=1$

(ii) Molecular orbital configuration of $\mathrm{CO}$ is $\mathrm{KK}(\sigma 2s{)}^{2}(\sigma *2s{)}^{2}𝜋(2{p}_{x}{)}^{2}𝜋(2{p}_{y}{)}^{2}(\sigma 2{p}_{z}{)}^{2}$

Bond order = $\frac{1}{2}({N}_{b}-{N}_{a})$ = $\frac{1}{2}(8-2)=3$

(iii) Molecular orbital configuration of ${\mathrm{NO}}^{+}$ is $\mathrm{KK}(\sigma 2s{)}^{2}(\sigma *2s{)}^{2}𝜋(2{p}_{x}{)}^{2}𝜋(2{p}_{y}{)}^{2}(\sigma 2{p}_{z}{)}^{2}$

Bond order = $\frac{1}{2}({N}_{b}-{N}_{a})$ = $\frac{1}{2}(8-2)=3$

So, according to molecular orbital theory, the bond orders respectively are 1, 3 and 3.

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Solution

Bond order = $\frac{1}{2}({N}_{b}-{N}_{a})$ = $\frac{1}{2}(8-6)=1$

(ii) Molecular orbital configuration of $\mathrm{CO}$ is $\mathrm{KK}(\sigma 2s{)}^{2}(\sigma *2s{)}^{2}𝜋(2{p}_{x}{)}^{2}𝜋(2{p}_{y}{)}^{2}(\sigma 2{p}_{z}{)}^{2}$

Bond order = $\frac{1}{2}({N}_{b}-{N}_{a})$ = $\frac{1}{2}(8-2)=3$

(iii) Molecular orbital configuration of ${\mathrm{NO}}^{+}$ is $\mathrm{KK}(\sigma 2s{)}^{2}(\sigma *2s{)}^{2}𝜋(2{p}_{x}{)}^{2}𝜋(2{p}_{y}{)}^{2}(\sigma 2{p}_{z}{)}^{2}$

Bond order = $\frac{1}{2}({N}_{b}-{N}_{a})$ = $\frac{1}{2}(8-2)=3$

So, according to molecular orbital theory, the bond orders respectively are 1, 3 and 3.

JEE Main 2023 — Chemistry Inorganic Chemistry

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JEE Main 2023 — Chemistry Inorganic Chemistry

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Solution

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