During the change of ${\mathrm{O}}_2\text{ to }{\mathrm{O}}_2^{-}$, the incoming electron goes to the orbital.

(a) $\pi 2p_x$             (b) ${\pi }^{\ast }2p_x$       (c) $\pi 2p_y$           (d) ${\sigma }^{\ast }2p_z$

or

Which of the following options represents the correct bond order ?

E.C. of O₂ (16 electrons)

$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\pi 2p_x^1\approx \pi 2p_y^1$

$\text{ Bond order }=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-6)=2$

$\text{ E.C. of }{\mathrm{O}}_2^{+}\text{(15 electrons) }$

$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\pi 2p_x^1\approx \pi 2p_y$

$\text{ Bond order }=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-5)=2.5$

$\text{ E.C. of }{\mathrm{O}}_2^{-}\text{(17 electrons) }$

$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\pi 2p_x^2\approx \pi 2p_y^{\ast }$

$\text{ Bond order }=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-7)=1.5$

Thus, the order of bond order is ${\mathrm{O}}_2^{-}<{\mathrm{O}}_2<{\mathrm{O}}_2^{+}$ and the incoming $\text{ electron goes to }\pi \ast 2p_x-\text{ orbital }$