Question

Which one of the following pairs of species have the same bond order?

• A. $O_2^{-}$ and $C{N}^{-}$
• B. $N{O}^{+}$ and $C{N}^{+}$
• C. $C{N}^{-}$ and $N{O}^{+}$
• D. $C{N}^{-}$ and $C{N}^{+}$

Answer 1

Answer: (C)

$C{N}^{-}$ and $N{O}^{+}$

Both $C{N}^{-}$ and $N{O}^{+}$ contain same number of electrons that is 14.

$Bondorder=\frac{(N_b-Na)}{2}$

For $O_2^{-}:$

Bond order $=\frac{10-7}{2}=1.5$

For $C{N}^{-}:$

Bond order $=\frac{10-4}{2}=3$

For $N{O}^{+}:$

Bond order $=\frac{10-4}{2}=3$

So, option (C) is right.