Count total electrons in $CN^-$

• Carbon has $6$ electrons.
• Nitrogen has $7$ electrons.
• The negative charge adds $1$ extra electron.

So, total electrons in $CN^-$ are: $6+7+1=14$

Recognize similarity with $N_2$

The species $CN^-$ has $14$ electrons, same as $N_2$. Hence, its molecular orbital filling is analogous to $N_2$.

Molecular orbital configuration

For $CN^-$ (like $N_2$), the MO order up to valence shell is: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,(\pi 2p_x)^2(\pi 2p_y)^2\,\sigma(2p_z)^2$

Calculate bond order

Bond order is: $\text{Bond order} = \frac{N_b - N_a}{2}$ where $N_b$ = number of bonding electrons, $N_a$ = number of antibonding electrons.

• Bonding electrons = $2$ in $\sigma(1s)$, $2$ in $\sigma(2s)$, $4$ in $\pi(2p)$, $2$ in $\sigma(2p)$
• Antibonding electrons = $2$ in $\sigma^*(1s)$, $2$ in $\sigma^*(2s)$

So, $N_b=10, \quad N_a=4$ $\text{Bond order} = \frac{10-4}{2}=3$

(Equivalently, considering only valence electrons also gives the same result.)

Check magnetic character

All molecular orbitals are completely filled with paired electrons. Therefore, $CN^-$ has no unpaired electron.

So, it is diamagnetic.

Evaluate options

• A: $3$, paramagnetic $\rightarrow$ incorrect
• B: $2\dfrac{1}{2}$, paramagnetic $\rightarrow$ incorrect
• C: $3$, diamagnetic $\rightarrow$ correct
• D: $2\dfrac{1}{2}$, diamagnetic $\rightarrow$ incorrect

Final answer

The bond order and magnetic nature of $CN^-$ are: $\boxed{3,\ \text{diamagnetic}}$