The correct order of bond orders of $\mathrm{C}_{2}^{2-}$, $\mathrm{N}_{2}^{2-}$ and $\mathrm{O}_{2}^{2-}$ is, respectively.

(B) $\mathrm{O}_{2}^{2-}<\mathrm{N}_{2}^{2-}<\mathrm{C}_{2}^{2-}$

(A) $\mathrm{C}_{2}^{2-}<\mathrm{N}_{2}^{2-}<\mathrm{O}_{2}^{2-}$

(C) $\mathrm{C}_{2}^{2-}<\mathrm{O}_{2}^{2-}<\mathrm{N}_{2}^{2-}$

(D) $\mathrm{N}_{2}^{2-}<\mathrm{C}_{2}^{2-}<\mathrm{O}_{2}^{2-}$

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Chemical Bonding and Molecular Structure

The correct order of bond orders of

C_{2}^{2 -}

N_{2}^{2 -}

and

O_{2}^{2 -}

is, respectively.

Answer: (B)

Using molecular orbital theory, calculate the bond order for each species:

C_{2}^{2 -}

: Total electrons = 14. Bond order =

\frac{10 - 4}{2} = 3

N_{2}^{2 -}

: Total electrons = 16. Bond order =

\frac{10 - 6}{2} = 2

O_{2}^{2 -}

: Total electrons = 18. Bond order =

\frac{10 - 8}{2} = 1

Therefore, the correct order is

O_{2}^{2 -} < N_{2}^{2 -} < C_{2}^{2 -}

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option