Chemistry - CHEMICAL BONDING Question with Solution | TestHub

${\mathrm{O}}_2(16$ electrons)

${\sigma }_{1s}^2,{\sigma }_{1s}^{*2},{\sigma }_{2s}^2,{\sigma }_{2s}^{*2},{\sigma }_{2p_2}^2$

${\pi }_{2{\mathrm{p}}_{\mathrm{x}}}^2={\pi }_{2{\mathrm{p}}_{\mathrm{y}}}^2,{\pi }_{2{\mathrm{p}}_{\mathrm{x}}}^{{*}_1}={\pi }_{2{\mathrm{p}}_{\mathrm{y}}}^{{*}_1},{\sigma }_{2{\mathrm{p}}_{\mathrm{z}}}^{*}$

Bond order of ${\mathrm{O}}_2\Rightarrow 2$
Bond order of ${\mathrm{O}}_2^{-}\Rightarrow 1.5$
Bond order of ${\mathrm{O}}_2^{2-}\Rightarrow 1$
Bond order of ${\mathrm{O}}_2^{+}\Rightarrow 2.5$