JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Chemical Bonding and Molecular Structure

Calculate the bond order for each species using MO theory:

Bond order = $\frac{(N_b-N_a)}{2}$ where $N_b$ = electrons in bonding MOs, $N_a$ = electrons in antibonding MOs

• $\mathrm{CN}{}^{-}$: 14 electrons → Bond order = $\frac{10-4}{2}=3$
• $\mathrm{NO}{}^{+}$: 14 electrons → Bond order = $\frac{10-4}{2}=3$
• $\mathrm{O}{}_2$: 16 electrons → Bond order = $\frac{10-6}{2}=2$
• $\mathrm{O}{}_2{}^{+}$: 15 electrons → Bond order = $\frac{10-5}{2}=2.5$
• $\mathrm{O}{}_2{}^{2+}$: 14 electrons → Bond order = $\frac{10-4}{2}=3$

Species with bond order 3: $\mathrm{CN}{}^{-}$, $\mathrm{NO}{}^{+}$, $\mathrm{O}{}_2{}^{2+}$

Therefore, 3 species have identical bond order.