Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{O}}_2^{2-}\left(18e^{-}\right):KK\sigma (2s{)}^2{\sigma }^{\ast }(2s{)}^2\sigma {\left(2p_z\right)}^2\pi \\ & {\left(2p_x\right)}^2{\pi }^{\ast }{\left(2p_y\right)}^2{\pi }^{\ast }{\left(2p_x\right)}^2{\pi }^{\ast }{\left(2p_y\right)}^2\end{array}$ $\begin{array}{c}\text{ Bond order }(\mathrm{B}.\mathrm{O})=\frac{8-6}{2}=1\\ {\mathrm{B}}_2\left(10e^{-}\right):KK\sigma (2s{)}^2{\sigma }^{\ast }(2s{)}^2\pi {\left(2p_x\right)}^1\pi {\left(2p_y\right)}^1\\ \text{ B.O }=\frac{4-2}{2}=1\end{array}$ Therefore, ${\mathrm{O}}_2^{2-}$ and ${\mathrm{B}}_2$ have identical bond order.