Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The Molecular orbital electronic configurations are

${\text{O}}_2^{-}\Rightarrow \text{σ}1{\text{s}}^2<\stackrel{*}{\text{σ}}1{\text{s}}^2<\text{σ}2{\text{s}}^2<\stackrel{*}{\text{σ}}2{\text{s}}^2<\text{σ}2{\text{p}}_{\text{z}}^2<\text{π}2{\text{p}}_{\text{x}}^2=\text{π}2{\text{p}}_{\text{y}}^2<\stackrel{*}{\text{π}}2{\text{p}}_{\text{x}}^2=\stackrel{*}{\text{π}}2{\text{p}}_{\text{y}}^1$(unpaired electron)

${\text{N}}_2^{2+}\Rightarrow \text{σ}1{\text{s}}^2<\stackrel{*}{\text{σ}}1{\text{s}}^2<\text{σ}2{\text{s}}^2<\stackrel{*}{\text{σ}}2{\text{s}}^2<\text{π}2{\text{p}}_{\text{x}}^2=\text{π}2{\text{p}}_{\text{y}}^2<\text{σ}2{\text{p}}_{\text{z}}^0$

${\text{O}}_2^{2-}\Rightarrow \text{σ}1{\text{s}}^2<\stackrel{*}{\text{σ}}1{\text{s}}^2<\text{σ}2{\text{s}}^2<\stackrel{*}{\text{σ}}2{\text{s}}^2<\text{σ}2{\text{p}}_{\text{z}}^2<\text{π}2{\text{p}}_{\text{x}}^2=\text{π}2{\text{p}}_{\text{y}}^2<\stackrel{*}{\text{π}}2{\text{p}}_{\text{x}}^2=\stackrel{*}{\text{π}}2{\text{p}}_{\text{y}}^2$

${\text{N}}_2\Rightarrow \text{σ}1{\text{s}}^2<\stackrel{*}{\text{σ}}1{\text{s}}^2<\text{σ}2{\text{s}}^2<\stackrel{*}{\text{σ}}2{\text{s}}^2<\text{π}2{\text{p}}_{\text{x}}^2=\text{π}2{\text{p}}_{\text{y}}^2<\text{σ}2{\text{p}}_{\text{z}}^2$