JEE Main 2013 — Chemical Bonding And Molecular Structure Question with Solution

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(A)   $$N_2$$ has 14 electrons.

Moleculer orbital configuration of $$N_2$$

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

Here no unpaired electron present, 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 electrons present, so it is paramagnetic.

(C)    $$S_2$$ has 32 electrons.

It also has 2 unpaired electrons like $$O_2$$, so it is paramagnetic.

(D)   $$C_2$$ has 12 electrons.

Moleculer orbital configuration of $$C_2$$

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

Here no unpaired electron present, so it is diamagnetic.