CHEM_REMOVED - CHEMICAL BONDING Question with Solution | TestHub

(i) Molecular orbital configuration of ${\mathrm{O}}_2^{2-}$ is $\mathrm{KK} (\mathrm{\sigma }2\mathrm{s}{)}^2 (\mathrm{\sigma }*2\mathrm{s}{)}^2 𝜋(2{\mathrm{p}}_{\mathrm{x}}{)}^2 𝜋(2{\mathrm{p}}_{\mathrm{y}}{)}^2 (\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}{)}^2 𝜋*(2{\mathrm{p}}_{\mathrm{x}}{)}^2 𝜋*(2{\mathrm{p}}_{\mathrm{y}}{)}^2$

Bond order = $\frac{1}{2} ({\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}})$ = $\frac{1}{2} \left(8 - 6\right) = 1$

(ii) Molecular orbital configuration of $\mathrm{CO}$ is $\mathrm{KK} (\mathrm{\sigma }2\mathrm{s}{)}^2 (\mathrm{\sigma }*2\mathrm{s}{)}^2 𝜋(2{\mathrm{p}}_{\mathrm{x}}{)}^2 𝜋(2{\mathrm{p}}_{\mathrm{y}}{)}^2 (\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}{)}^{2 }$

Bond order = $\frac{1}{2} ({\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}})$ = $\frac{1}{2} \left(8 - 2\right) = 3$

(iii) Molecular orbital configuration of ${\mathrm{NO}}^{+}$ is $\mathrm{KK} (\mathrm{\sigma }2\mathrm{s}{)}^2 (\mathrm{\sigma }*2\mathrm{s}{)}^2 𝜋(2{\mathrm{p}}_{\mathrm{x}}{)}^2 𝜋(2{\mathrm{p}}_{\mathrm{y}}{)}^2 (\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}{)}^{2 }$

Bond order = $\frac{1}{2} ({\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}})$ = $\frac{1}{2} \left(8 - 2\right) = 3$

So, according to molecular orbital theory, the bond orders respectively are 1, 3 and 3.