Which of the following molecules/ions does not contain unpaired electrons?

Molecular-orbital theory is the most reliable tool for deciding whether a homonuclear diatomic species is paramagnetic (has unpaired electrons) or diamagnetic (all electrons paired).
The procedure is: (i) count the total number of valence electrons present in the species, (ii) fill them into the standard MO energy diagram in the correct order, and (iii) look for unpaired electrons.

The required MO order is different for lighter (up to $$N_2$$) and heavier (from $$O_2$$ onward) species.

Case 1: $$B_2$$

• Atomic number of B = 5 ⇒ valence electrons per B atom = 3.
• Total electrons = $$2 \times 3 = 6$$.

Energy order (same as $$N_2$$-type):
$$\sigma_{2s},\;\sigma_{2s}^*,\;\pi_{2p_x} = \pi_{2p_y},\;\sigma_{2p_z}$$

Filling six electrons gives the configuration
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x}^1 = \pi_{2p_y}^1)$$

The two $$\pi$$ electrons occupy different orbitals with parallel spins, so there are 2 unpaired electrons ⇒ $$B_2$$ is paramagnetic.

Case 2: $$N_2^{+}$$

• Atomic number of N = 7 ⇒ valence electrons per atom = 5.
• Neutral $$N_2$$ has $$10$$ valence electrons; removing one electron (positive charge) leaves $$9$$ electrons.

Energy order (still $$N_2$$-type):
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\sigma_{2p_z}^1$$

The last electron is single in $$\sigma_{2p_z}$$, so there is 1 unpaired electron ⇒ $$N_2^{+}$$ is paramagnetic.

Case 3: $$O_2$$

• Atomic number of O = 8 ⇒ valence electrons per O atom = 6.
• Total electrons = $$12$$.

For oxygen and heavier atoms the MO order changes to
$$\sigma_{2s},\;\sigma_{2s}^*,\;\sigma_{2p_z},\;\pi_{2p_x} = \pi_{2p_y},\;\pi_{2p_x}^* = \pi_{2p_y}^*,\;\sigma_{2p_z}^*$$

Filling twelve electrons:
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,(\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1})$$

The two $$\pi^*$$ orbitals each contain one electron with parallel spins ⇒ 2 unpaired electrons ⇒ $$O_2$$ is paramagnetic.

Case 4: $$O_2^{2-}$$

• Start from $$O_2$$ (12 electrons) and add two more electrons for the $$2-$$ charge ⇒ $$14$$ valence electrons.

Using the same oxygen-type MO order:

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

All orbitals are completely filled; no orbital contains a single electron. Thus every electron is paired ⇒ $$O_2^{2-}$$ is diamagnetic.

Among the given options, the only diamagnetic species is $$O_2^{2-}$$.

Option A which is: $$O_2^{2-}$$