Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Bond order of ${\mathrm{O}}_2{}^{2-}$ : Electronic : $\sigma 1{\mathrm{s}}^2{\sigma }^{\star }({\mathrm{s}}^2\sigma 2{\mathrm{s}}^2{\sigma }^{\ast }2{\mathrm{s}}^2\sigma 2{\mathrm{P}}_{\mathrm{z}}{}^2\left(\pi 2{\mathrm{P}}_{\mathrm{x}}{}^2=\pi 2{\mathrm{P}}_{\mathrm{y}}{}^2\right)$ $\left(\pi 2{\mathrm{P}}_{\mathrm{x}}^2=\pi 2^{\star }{\mathrm{P}}_{\mathrm{y}}^2\right){\sigma }^{\star }2\mathrm{pz}{}^{\circ }$ Bond order $=\frac{\text{ Bonding }-\text{ Antibonding electrons }}{2}=\frac{10-8}{2}=1$ Bond order of $B_2$ : $\sigma 1s^2{\sigma }_{1s}{}^{2}2{\sigma }_{2s}2{\sigma }^{\star }2s^2\sigma 2p_z^2$ Bond order $=\frac{6-4}{2}=1$ $\therefore {\mathrm{O}}_2{}^{2-}$ and ${\mathrm{B}}_2$ has same bond order 1.