JEE Main 2014 — Chemistry Inorganic Chemistry

As per Molecular orbital theory,

${N}_{2}(14\mathrm{electrons})$N2​(14electrons) all paired electrons, so diamagnetic.

$${N}_{2}\Rightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}<\sigma 2{p}_{z}^{2}$$N2​⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px2​=π2py2​<σ2pz2​

${O}_{3}-24$O3​−24 electrons and diamagnetic.

$\mathrm{CO}(14\mathrm{electrons})$CO(14electrons) all paired so, diamagnetic.

$$\mathrm{CO}\Rightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}<\sigma 2{p}_{z}^{2}$$CO⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px2​=π2py2​<σ2pz2​

$\mathrm{NO}(15\mathrm{electrons})$NO(15electrons) one unpaired electron so, paramagnetic.

$$\mathrm{NO}\Rightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}<\sigma 2{p}_{z}^{2}<\overset{*}{\pi }2{p}_{x}^{1}=\overset{*}{\pi }2{p}_{y}^{0}$$NO⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px2​=π2py2​<σ2pz2​<π∗2px1​=π∗2py0​