JEE Main 2017 — Chemical Bonding And Molecular Structure Question with Solution

Those species which have unpaired electrons are called paramagnetic species.

(a)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

Here is no unpaired electron so it is diamagnetic.

(b)   $$O_2$$ has 16 electrons.

Moleculer orbital configuration of $$O_2$$ is

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

Here 2 unpaired electron present, so it is paramagnetic.

(c)   B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^1}={\pi }_{2p_y^1}$$

Here two unpaired electrons present. So it is paramagnetic.

(d)   NO has 15 electrons.

Moleculer orbital configuration of NO is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^0}^{\ast }$$

Here is 1 unpaired electron, So it is Paramagnetic.