To determine which pair does not have the same bond order, we need to calculate the bond order for each species using the Molecular Orbital Theory (MOT) formula:

$\text{Bond Order}=\frac{1}{2}(\text{Number of electrons in bonding MOs}-\text{Number of electrons in antibonding MOs})$

Let's calculate the total number of electrons and then the bond order for each species:

A) N${}_2$ and CN${}^{-}$

• N${}_2$: Total electrons = 7 + 7 = 14. Electron configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\sigma }_{2p_z}{)}^2$.
  Bond Order = $\frac{1}{2}(10-4)=3$
• CN${}^{-}$: Total electrons = 6 + 7 + 1 (for negative charge) = 14. Electron configuration is similar to N${}_2$.
  Bond Order = $\frac{1}{2}(10-4)=3$
• Bond orders are the same (3).

B) O${}_2^{+}$ and NO

• O${}_2^{+}$: Total electrons = 8 + 8 - 1 = 15. Electron configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\pi }_{2p_x}^{\ast }{)}^1$.
  Bond Order = $\frac{1}{2}(10-5)=2.5$
• NO: Total electrons = 7 + 8 = 15. Electron configuration is similar to O${}_2^{+}$.
  Bond Order = $\frac{1}{2}(10-5)=2.5$
• Bond orders are the same (2.5).

C) F${}_2^{-}$ and O${}_2^{2-}$

• F${}_2^{-}$: Total electrons = 9 + 9 + 1 = 19. Electron configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\pi }_{2p_x}^{\ast }{)}^2({\pi }_{2p_y}^{\ast }{)}^2({\sigma }_{2p_z}^{\ast }{)}^1$.
  Bond Order = $\frac{1}{2}(10-9)=0.5$
• O${}_2^{2-}$: Total electrons = 8 + 8 + 2 = 18. Electron configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\pi }_{2p_x}^{\ast }{)}^2({\pi }_{2p_y}^{\ast }{)}^2$.
  Bond Order = $\frac{1}{2}(10-8)=1$
• Bond orders are different (0.5 and 1).

D) B${}_2^{2-}$ and CN${}^{+}$

• B${}_2^{2-}$: Total electrons = 5 + 5 + 2 = 12. Electron configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2$.
  Bond Order = $\frac{1}{2}(8-4)=2$
• CN${}^{+}$: Total electrons = 6 + 7 - 1 = 12. Electron configuration is similar to B${}_2^{2-}$.
  Bond Order = $\frac{1}{2}(8-4)=2$
• Bond orders are the same (2).

Therefore, the pair F${}_2^{-}$ and O${}_2^{2-}$ does not have the same bond order.

Correct Answer: (C)