Four diatomic species are given below. Which of the following...

The bond order of a diatomic species can be determined from its total number of electrons.

For $\mathrm{NO}^{+}$, the total number of electrons is $7 + 8 - 1 = 14$. Its bond order is $3$.

For $\mathrm{NO}$, the total number of electrons is $7 + 8 = 15$. Its bond order is $2.5$.

For $\mathrm{O}_2$, the total number of electrons is $8 + 8 = 16$. Its bond order is $2$.

For $\mathrm{O}_2^{-}$, the total number of electrons is $8 + 8 + 1 = 17$. Its bond order is $1.5$.

Bond length is inversely proportional to bond order. Since the bond order decreases in the sequence $\mathrm{NO}^{+} > \mathrm{NO} > \mathrm{O}_2 > \mathrm{O}_2^{-}$, the bond length will increase in the reverse order.

Therefore, the correct sequence of increasing bond length is $\mathrm{NO}^{+} < \mathrm{NO} < \mathrm{O}_2 < \mathrm{O}_2^{-}$.

Answer: $\mathrm{NO}^{+} < \mathrm{NO} < \mathrm{O}_2 < \mathrm{O}_2^{-}$