Chemistry - CHEMICAL BONDING Question with Solution | TestHub

As per Molecular orbital theory,

${\mathrm{N}}_2 \left(14 \mathrm{electrons}\right)$all paired electrons, so diamagnetic.

${\mathrm{N}}_2\Rightarrow \mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2<\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2<\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

${\mathrm{O}}_3-24$electrons and diamagnetic.​

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

$\mathrm{CO}\Rightarrow \mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2<\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2<\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

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

$\mathrm{NO}\Rightarrow \mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2<\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2<\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2<\stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{x}}^1=\stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{y}}^0$