The bond order and the magnetic characteristic of CN$$^-$$ are

First, we find the total number of electrons present in the cyanide ion $$\text{CN}^-.$$

Carbon has atomic number 6, so it contributes $$6$$ electrons. Nitrogen has atomic number 7, so it contributes $$7$$ electrons. Because of the extra negative charge, the ion possesses one additional electron.

Hence the total electron count is $$6 + 7 + 1 \;=\; 14.$$

We now recall that a diatomic species with $$14$$ electrons is isoelectronic with the molecule $$\text{N}_2,$$ which also contains $$14$$ electrons. Therefore, $$\text{CN}^-$$ will have exactly the same pattern of molecular-orbital occupation as $$\text{N}_2.$$

For second-period diatomic molecules whose total atomic numbers are $$\le 14$$ (that is, up to nitrogen), the accepted energy order of molecular orbitals is $$\sigma_{1s},\;\sigma_{1s}^{*},\;\sigma_{2s},\;\sigma_{2s}^{*},\;\pi_{2p_x} = \pi_{2p_y},\;\sigma_{2p_z},\;\pi_{2p_x}^{*} = \pi_{2p_y}^{*},\;\sigma_{2p_z}^{*}.$$

We now place the $$14$$ electrons of $$\text{CN}^-$$ into these orbitals, filling from lower to higher energy and pairing spins whenever possible:

$$$\begin{aligned} \sigma_{1s}&^2 \\[-2pt] \sigma_{1s}^{*}&^2 \\[-2pt] \sigma_{2s}&^2 \\[-2pt] \sigma_{2s}^{*}&^2 \\[-2pt] \pi_{2p_x}&^2 \\[-2pt] \pi_{2p_y}&^2 \\[-2pt] \sigma_{2p_z}&^2 \end{aligned}$$$

All electrons are paired, so the species is diamagnetic.

To calculate bond order we use the standard formula

$$\text{Bond order} \;=\; \dfrac{N_b \;-\; N_a}{2},$$

where $$N_b$$ is the number of electrons in bonding molecular orbitals and $$N_a$$ is the number in antibonding molecular orbitals.

From the configuration we have

$$$\begin{aligned} N_b &= 2(\sigma_{1s}) + 2(\sigma_{2s}) + 4(\pi_{2p_x},\pi_{2p_y}) + 2(\sigma_{2p_z}) \;=\; 10,\\[4pt] N_a &= 2(\sigma_{1s}^{*}) + 2(\sigma_{2s}^{*}) \;=\; 4. \end{aligned}$$$

Substituting into the formula, we get

$$\text{Bond order} \;=\; \dfrac{10 - 4}{2} \;=\; \dfrac{6}{2} \;=\; 3.$$

Thus, $$\text{CN}^-$$ has a bond order of $$3$$ and is diamagnetic.

Hence, the correct answer is Option B.