(A) $${B_2} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\pi 2p_x^1 = \pi 2p_y^1$$

It is paramagnetic due to two unpaired electrons. The bond order is

$${{{N_b} - {N_a}} \over 2} = {{6 - 4} \over 2} = 1$$

The gain of electron increases bond order, so reduction is possible.

(B) $${N_2} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\pi 2p_x^2 = \pi 2p_y^2\sigma 2p_z^2$$

There are no unpaired electrons. Bond order is

$${{{N_b} - {N_a}} \over 2} = {{10 - 4} \over 2} = {6 \over 2} = 3$$

Mixing of 2s and 2p orbitals is possible because of similar energies.

(C) $$O_2^ - \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\sigma 2p_z^2\pi 2p_x^2 = \pi 2p_y^2\,\pi * 2p_x^2\,\pi * 2p_y^1$$

The molecule is paramagnetic due to presence of unpaired electron. Bond order is less than 2.

$${{{N_b} - {N_a}} \over 2} = {{10 - 7} \over 2} = {3 \over 2}$$

Loss of electron increases the bond order so oxidation is possible.

(D) $$O_2^{} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\sigma 2p_z^2\pi 2p_x^2 = \pi 2p_y^2\,\pi * 2p_x^1\,\pi * 2p_y^1$$

The molecule is paramagnetic due to presence of unpaired electrons. Bond order is

$${{{N_b} - {N_a}} \over 2} = {{10 - 6} \over 2} = {4 \over 2} = 2$$

Loss of electron causes increase in bond order, so it undergoes oxidation.