Question

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

• A. $\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_{x^2}\equiv \pi 2p_{y^2}\left)\right({\pi }^{\ast }2p_{x^1}\equiv {\pi }^{\ast }2p_{y^1})$
• B. $\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_{x^2}\equiv \pi 2p_{y^2}\left)\right({\pi }^{\ast }2p_{x^2})$
• C. $\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_{x^2}\equiv \pi 2p_{y^2}\left)\right({\pi }^{\ast }2p_{y^2})$
• D. $\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_{x^2}\equiv \pi 2p_{y^2}\left)\right({\pi }^{\ast }2p_{y^2})$

Answer 1

The correct option is A

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

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:

$O_2⟶$ Dioxygen$⟶$ Paramagnetic; Coloured

${O_2}^{2+}⟶$ Diamagnetic; Colourless

${O_2}^{2-}⟶$ Peroxide $⟶$ Diamagnetic; Colourless

${O_2}^{+}⟶$ Paramagnetic; Coloured

$O_2^{-}⟶$ Superoxide$⟶$ Paramagnetic; Coloured

Now,

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

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

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

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

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

So, the electronic configuration will be:

$O_2⟶$$\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})$

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