CHEM_REMOVED - CHEMICAL BONDING Question with Solution | TestHub

The orbital in which last electron is present is called the highest occupied molecular orbital(HOMO).

${\mathrm{N}}_2$ : $\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\frac{\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2}{\mathrm{HOMO}}$

${\mathrm{N}}_2^{+}: \mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\frac{\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^1}{\mathrm{HOMO}}$

${\mathrm{O}}_2: \mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1={\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\left(\mathrm{HOMO}\right)$

${\mathrm{O}}_2^{+}: \mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2 {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1={\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^0\left(\mathrm{HOMO}\right)$

${\mathrm{N}}_2\Rightarrow 0$ unpaired ${\mathrm{e}}^{–}$ in $\mathrm{HOMO}$

${\mathrm{N}}_2^{+}\Rightarrow 1$ unpaired ${\mathrm{e}}^{-}$ in $\mathrm{HOMO}$

${\mathrm{O}}_2\Rightarrow 2$ unpaired ${\mathrm{e}}^{-}$ in $\mathrm{HOMO}$

${\mathrm{O}}_2^{+}\Rightarrow 1$ unpaired ${\mathrm{e}}^{-}$ in $\mathrm{HOMO}$