Chemistry - CHEMICAL BONDING Question with Solution | TestHub

CO has no unpaired electron while  other have unpaired electrons so, paramagnetic in nature.(Apply M.O.T)

$\text{Molecular orbital configuration}\Rightarrow$

$\text{CO }:{\text{like N}}_2;\text{ σ}1{\mathrm{s}}^2<{\text{σ}}^{*}{\text{1S}}^2<\text{σ}2{\text{S}}^2<{\sigma }^{*}2{\text{S}}^2<\text{π}2{\text{P}}_{\text{x}}^2\text{= π}2{\text{P}}_{\text{y}}^2<\text{σ}2{\text{P}}_{\text{z}}^2(\text{No unpaired electrons)}$

${\text{O}}_2:\text{σ}1{\text{S}}^2<{\text{σ}}^{*}{\text{1S}}^2<\text{σ}2{\text{S}}^2<{\text{σ}}^{*}2{\text{S}}^2<\text{σ}2{\text{P}}_{\text{z}}^2<\text{π}2{\text{P}}_{\text{x}}^2=\text{π}2{\text{P}}_{\text{y}}^2<{\text{π}}^{*}2{\text{P}}_{\text{x}}^{\text{1}}\text{= }{\mathrm{\pi }}^{*}2{\text{P}}_{\text{y}}^{\text{1}}\left(2\text{unpaired electrons}\right)$

${\text{B}}_2:\text{σ}1{\text{S}}^2<{\text{σ}}^{*}1{\text{S}}^2<\text{σ}2{\text{S}}^2<{\text{σ}}^{*}2{\text{S}}^2<\text{π}2{\text{P}}_{\text{x}}^1=\text{π}2{\text{P}}_{\text{y}}^{\text{1}}(2\text{unpaired electrons})$

$\text{NO }:\text{σ}1{\text{S}}^2<{\text{σ}}^{*}1{\text{S}}^2<\text{σ}2{\text{S}}^2<{\text{σ}}^{*}2{\text{S}}^2<\text{σ}2{\text{P}}_{\text{z}}^2<\text{π}2{\text{P}}_{\text{x}}^2=\text{π}2{\text{P}}_{\text{y}}^2<{\text{π}}^{*}2{\text{P}}_{\text{y}}^{\text{1}}={\text{ π}}^{*}2{\text{P}}_{\text{x}}^0\left(1\text{unpaired electrons}\right)$