JEE Main 2020 — Chemistry Inorganic Chemistry

Given, magnetic moment of di-oxygen species$=1.73B.M.$=1.73B.M.

We know, magnetic moment $=\sqrt{ \text{n}( \text{n}+2) }$=n(n+2)​, where $n=$n=number of unpaired electrons

$$1.73=\sqrt{ \text{n}( \text{n}+2) }$$1.73=n(n+2)​

$$\text{n}=1$$n=1

So, species should contain one unpaired electron.

Let us check by electronic configuration of each species

$${O}_{2}=\sigma 1{s}^{2}{\sigma }^{*}1{s}^{2}\sigma 2{s}^{2}{\sigma }^{*}2{s}^{2}\sigma 2{{p}^{2}}_{z}\pi 2{{p}^{2}}_{x}=\pi 2{{p}^{2}}_{y}{\pi }^{*}2{{p}^{1}}_{x}={\pi }^{*}2{{p}^{1}}_{y} {{O}_{2}}^{+}=\sigma 1{s}^{2}{\sigma }^{*}1{s}^{2}\sigma 2{s}^{2}{\sigma }^{*}2{s}^{2}\sigma 2{{p}^{2}}_{z}\pi 2{{p}^{2}}_{x}=\pi 2{{p}^{2}}_{y}{\pi }^{*}2{{p}^{1}}_{x}={\pi }^{*}2{{p}^{0}}_{y} {{O}_{2}}^{-}=\sigma 1{s}^{2}{\sigma }^{*}1{s}^{2}\sigma 2{s}^{2}{\sigma }^{*}2{s}^{2}\sigma 2{{p}^{2}}_{z}\pi 2{{p}^{2}}_{x}=\pi 2{{p}^{2}}_{y}{\pi }^{*}2{{p}^{2}}_{x}={\pi }^{*}2{{p}^{1}}_{y}$$O2​=σ1s2σ∗1s2σ2s2σ∗2s2σ2p2z​π2p2x​=π2p2y​π∗2p1x​=π∗2p1y​O2​+=σ1s2σ∗1s2σ2s2σ∗2s2σ2p2z​π2p2x​=π2p2y​π∗2p1x​=π∗2p0y​O2​−=σ1s2σ∗1s2σ2s2σ∗2s2σ2p2z​π2p2x​=π2p2y​π∗2p2x​=π∗2p1y​

As clear by above electronic configurations that, ${O}_{2}$O2​ has two unpaired electrons and {{O}_{2}}^{-}&{{O}_{2}}^{+} Each has only one unpaired electron; hence magnetic moment of {{O}_{2}}^{+}&{{O}_{2}}^{-} is $1.73B.M.$1.73B.M.