Which of the following is paramagnetic?

We recall that a substance is said to be paramagnetic when it possesses at least one unpaired electron. Whenever all electrons are paired the substance is diamagnetic. For homonuclear or simple di-atomic molecules and ions of the second period, the easiest way to check the presence of unpaired electrons is to write the complete Molecular Orbital (MO) electronic configuration and then look for singly-occupied orbitals.

First we note the sequence of MOs that we shall use:

For species having total electrons $$\le 13$$ (for example $$B_2$$) the energetic order is

$$\sigma(2s)\,<\,\sigma^{\*}(2s)\,<\,\pi(2p_x)=\pi(2p_y)\,<\,\sigma(2p_z)\,<\,\pi^{\*}(2p_x)=\pi^{\*}(2p_y)\,<\,\sigma^{\*}(2p_z)$$

For species having $$\ge 14$$ electrons (for example $$CO,\,NO^{+},\,O_2^{2-}$$) the order changes slightly to

$$\sigma(2s)\,<\,\sigma^{\*}(2s)\,<\,\sigma(2p_z)\,<\,\pi(2p_x)=\pi(2p_y)\,<\,\pi^{\*}(2p_x)=\pi^{\*}(2p_y)\,<\,\sigma^{\*}(2p_z)$$

Now we examine every option one by one.

Option A : $$CO$$

Carbon contributes $$6$$ electrons and oxygen contributes $$8$$ electrons, so

$$\text{Total electrons} =6+8 =14$$

Because the total is $$14$$ we use the second ordering. We fill the MOs in the order written:

$$\sigma(2s)^2\,\sigma^{\*}(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2\,\pi(2p_y)^2$$

All orbitals are doubly occupied; there is no unpaired electron. So $$CO$$ is diamagnetic.

Option B : $$NO^{+}$$

Neutral $$NO$$ has $$7+8=15$$ electrons. The positive charge removes one electron, hence

$$\text{Total electrons in }NO^{+}=15-1 =14$$

Again we use the sequence for $$\ge 14$$ electrons and fill exactly as for $$N_2$$ or $$CO$$:

$$\sigma(2s)^2\,\sigma^{\*}(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2\,\pi(2p_y)^2$$

All electrons are paired, therefore $$NO^{+}$$ is diamagnetic.

Option C : $$O_2^{2-}$$

Neutral $$O_2$$ possesses $$8+8=16$$ electrons. The $$2-$$ charge adds two more electrons:

$$\text{Total electrons in }O_2^{2-}=16+2 =18$$

We again employ the $$\ge 14$$ order and fill up to $$18$$ electrons:

$$\sigma(2s)^2\,\sigma^{*}(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2\,\pi(2p_y)^2\,\pi^{*}(2p_x)^2\,\pi^{*}(2p_y)^2$$

The antibonding $$\pi^{\*}$$ orbitals are now completely filled; all electrons remain paired. Therefore $$O_2^{2-}$$ is diamagnetic.

Option D : $$B_2$$

Each boron atom carries $$5$$ electrons, so

$$\text{Total electrons} =5+5 =10$$

Because $$10<13$$ we must adopt the first energy sequence. We now fill ten electrons:

$$\sigma(2s)^2\,\sigma^{\*}(2s)^2\,\pi(2p_x)^1\,\pi(2p_y)^1$$

The $$\pi(2p_x)$$ and $$\pi(2p_y)$$ each contain a single electron. These two electrons are unpaired. As unpaired electrons are present, $$B_2$$ is unequivocally paramagnetic.

Only $$B_2$$ possesses unpaired electrons; the remaining three species have all electrons paired.

Hence, the correct answer is Option D.