Which of the following species is not paramagnetic?

First, we recall the basic rule for magnetic behaviour: a species that possesses at least one unpaired electron is called paramagnetic, while a species in which all electrons are paired is termed diamagnetic. So we must examine every option and decide whether unpaired electrons are present.

To do this rigorously we use the Molecular Orbital (MO) theory for homonuclear and heteronuclear diatomic molecules of the second period. The standard energy order of the MOs that we shall employ is stated below.

For species with a total of $$\le 14$$ electrons (that is, up to $$N_2$$) the order is

$$\sigma_{1s}\,<\,\sigma^*_{1s}\,<\,\sigma_{2s}\,<\,\sigma^*_{2s}\,<\,\pi_{2p_x}=\pi_{2p_y}\,<\,\sigma_{2p_z}\,<\,\pi^*_{2p_x}=\pi^*_{2p_y}\,<\,\sigma^*_{2p_z}$$

For species with $$>14$$ electrons (starting from $$O_2$$) the order changes slightly:

$$\sigma_{1s}\,<\,\sigma^*_{1s}\,<\,\sigma_{2s}\,<\,\sigma^*_{2s}\,<\,\sigma_{2p_z}\,<\,\pi_{2p_x}=\pi_{2p_y}\,<\,\pi^*_{2p_x}=\pi^*_{2p_y}\,<\,\sigma^*_{2p_z}$$

Now we treat each molecule one by one, always writing the electron count, filling the MOs in the proper sequence, and looking for unpaired electrons.

1. $$CO$$

The total number of valence electrons is obtained by simple addition:

$$\text{Electrons} = Z_{\text{C}} + Z_{\text{O}} = 6 + 8 = 14$$

Because the total equals $$14$$, we must use the first sequence of MO energies. We now fill the orbitals two electrons at a time (opposite spins) until all $$14$$ are placed:

$$\sigma_{1s}^2\,\sigma_{1s}^{*2}\; \sigma_{2s}^2\,\sigma_{2s}^{*2}\; \pi_{2p_x}^2\,\pi_{2p_y}^2\; \sigma_{2p_z}^2$$

We reach $$14$$ electrons exactly at $$\sigma_{2p_z}^2$$, and no higher orbital contains any electrons. Because every occupied orbital carries an even superscript, all electrons are paired. Therefore $$CO$$ is diamagnetic.

2. $$O_2$$

Here

$$\text{Electrons} = Z_{\text{O}} + Z_{\text{O}} = 8 + 8 = 16$$

Since the total exceeds $$14$$, we choose the second order of energy. Filling step by step:

$$\sigma_{1s}^2\,\sigma_{1s}^{*2}\; \sigma_{2s}^2\,\sigma_{2s}^{*2}\; \sigma_{2p_z}^2\; \pi_{2p_x}^2\,\pi_{2p_y}^2\; \pi_{2p_x}^{*1}\,\pi_{2p_y}^{*1}$$

We have now placed all $$16$$ electrons. Observe the last line: each $$\pi^*$$ orbital contains a single electron, so two electrons remain unpaired. Hence $$O_2$$ is paramagnetic.

3. $$B_2$$

For $$B_2$$ we count

$$\text{Electrons} = Z_{\text{B}} + Z_{\text{B}} = 5 + 5 = 10$$

Because $$10 \le 14$$, the first sequence applies. Filling the orbitals gives

$$\sigma_{1s}^2\,\sigma_{1s}^{*2}\; \sigma_{2s}^2\,\sigma_{2s}^{*2}\; \pi_{2p_x}^1\,\pi_{2p_y}^1$$

Only eight electrons are required up to $$\sigma_{2s}^{*2}$$, so the remaining two fill the degenerate $$\pi_{2p_x}$$ and $$\pi_{2p_y}$$ orbitals singly according to Hund’s rule. Both of these electrons are unpaired, and consequently $$B_2$$ is paramagnetic.

4. $$NO$$

This heteronuclear molecule has

$$\text{Electrons} = Z_{\text{N}} + Z_{\text{O}} = 7 + 8 = 15$$

The total exceeds $$14$$, so we again use the second energy order. Adding electrons stepwise yields

$$\sigma_{1s}^2\,\sigma_{1s}^{*2}\; \sigma_{2s}^2\,\sigma_{2s}^{*2}\; \sigma_{2p_z}^2\; \pi_{2p_x}^2\,\pi_{2p_y}^2\; \pi_{2p_x}^{*1}$$

The final electron enters a $$\pi^*$$ orbital alone, leaving it unpaired. This single unpaired electron renders $$NO$$ paramagnetic.

Putting all the conclusions together:

$$CO$$ diamagnetic; $$O_2$$ paramagnetic; $$B_2$$ paramagnetic; $$NO$$ paramagnetic

Since the question asks for the species that is not paramagnetic, we identify $$CO$$.

Hence, the correct answer is Option A.