To determine the bond order and magnetic characteristics of ${\mathrm{CN}}^{-}$, we first need to find the total number of electrons and then fill them into molecular orbitals.

Number of electrons in Carbon (C) = 6
Number of electrons in Nitrogen (N) = 7
For the ${\mathrm{CN}}^{-}$ ion, there is an additional electron due to the negative charge.
Total electrons = 6 + 7 + 1 = 14 electrons.

The molecular orbital configuration for 14 electrons (similar to ${\mathrm{N}}_2$) is:
$({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\sigma }_{2p_z}{)}^2$

Bond Order (B.O.) = $\frac{1}{2}$ (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals)
Bonding electrons = 2 (from ${\sigma }_{1s}$) + 2 (from ${\sigma }_{2s}$) + 2 (from ${\pi }_{2p_x}$) + 2 (from ${\pi }_{2p_y}$) + 2 (from ${\sigma }_{2p_z}$) = 10
Antibonding electrons = 2 (from ${\sigma }_{1s}^{\ast }$) + 2 (from ${\sigma }_{2s}^{\ast }$) = 4
B.O. = $\frac{1}{2}(10-4)=\frac{1}{2}(6)=3$

Since all electrons in the molecular orbital configuration are paired, ${\mathrm{CN}}^{-}$ is diamagnetic.

• A) $2\frac{1}{2}$, diamagnetic: Incorrect bond order.
• B) 3, paramagnetic: Incorrect magnetic characteristic.
• C) 3, diamagnetic: Correct bond order and magnetic characteristic.
• D) $2\frac{1}{2}$, paramagnetic: Incorrect bond order and magnetic characteristic.

Correct Answer: (C)