Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{O}}_2(16\mathrm{e}):{\left({\sigma }_{1\mathrm{s}}\right)}^2{\left({\sigma }_{1\mathrm{s}}^{\ast }\right)}^2{\left({\sigma }_{2\mathrm{s}}\right)}^2{\left({\sigma }_{2\mathrm{s}}^{\ast }\right)}^2\\ & {\left({\sigma }_{2\mathrm{p}}\right)}^2\left[{\left({\pi }_{2\mathrm{p}}\right)}^2={\left({\pi }_{2\mathrm{p}}\right)}^2\right],\left[{\left({\pi }_{2\mathrm{p}}^{\ast }\right)}^1={\left({\mathrm{m}}_{2\mathrm{p}}^{\ast }\right)}^1\right]\end{array}$
Number of ${\mathrm{e}}^{-}$present in $\left({\pi }^{\ast }\right)$ of ${\mathrm{O}}_2=2$ Number of ${\mathrm{e}}^{-}$present in $\left({\pi }^{\ast }\right)$ of ${\mathrm{O}}_2^{+}=1$ Number of ${\mathrm{e}}^{-}$present in $\left({\pi }^{\ast }\right)$ of ${\mathrm{O}}_2^{-}=3$ So total ${\mathrm{e}}^{-}$in $\left({\pi }^{\ast }\right)=2+1+3=6$