The bond order of $\mathrm{CN}^{-}$is

Question

Aicharya · Accepted Answer

Valence electrons in $\mathrm{CN}^{-}=10 \quad$ Electronic configuration $(\sigma 2 \mathrm{~s})^{2}(\sigma * 2 \mathrm{~s})^{2}\left(\pi 2 \mathrm{p}_{x}ight)^{2}\left(\pi 2 \mathrm{p}_{y}ight)^{2}\left(\sigma 2 \mathrm{p}_{z}ight)^{2}$ Bond order $=\frac{\text { (Bonding - antibonding) electrons }}{2}=\frac{8-2}{2}=3$ .

The bond order of \(\mathrm{CN}^{-}\)is

Solution