Question

Among the following species, identify the pair having same bond order $C{N}^{-},O_2^{-},N{O}^{+},C{N}^{+}$

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

Answer 1

Answer: (C)

Species having the same number of electrons, have same bond order.

Since, $C{N}^{-}$ and $N{O}^{-}$ have same number of electrons, they have same bond order, ie, 3.
$C{N}^{-}$ or $N{O}^{+}=14=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,$
$\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2$
Bond order $=\frac{N_b-N_a}{2}=\frac{10-4}{2}=3$