Question

Which of the following represent the correct electronic configuration of $O_2$?

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

The correct option is A

(a)

To write the electronic configuration of $O_2$,

$\to$ First we have to draw the energy level diagram (MOT) for $O_2$

$\to$ While filling the electrons in the energy levels, we have to follow the rules for filling the electrons.

The energy level diagram of $O_2$ will be:

Now,

As we can see from the diagram, the electrons present in different energy levels are:

Let us start filling the electrons:

$\sigma 1s=2electrons,{\sigma }^{\ast }1s=2electrons$

$\sigma 2s=2electrons,{\sigma }^{\ast }2s=2electrons$

$\sigma 2p_z=2electrons$

$\pi 2p_x=2electrons,\pi 2p_y=2electrons$

${\pi }^{\ast }2p_x=1electrons,{\pi }^{\ast }2p_y=1electrons$

So, the electronic configuration will be:

$\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_{x^2}\equiv \pi 2p_{y^2}\left)\right({\pi }^{\ast }2p_{x^1}\equiv {\pi }^{\ast }2p_{y^1})$

We know, that in p-orbitals, $\sigma$ bond will be formed by Z-axis (i.e., $p_z$).
Whereas, $\pi$ bond can be formed either by x or y, that's why they are shown in one single bracket

$O_2\to$ Dioxygen$\to$ Paramagnetic; Coloured

${O_2}^{2+}\to$ Diamagnetic; Colourless

${O_2}^{2-}\to$ Peroxide $\to$ Diamagnetic; Colourless

${O_2}^{+}\to$ Paramagnetic; Coloured

$O_2^{-}\to$ Superoxide$\to$ Paramagnetic; Coloured