JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 17)

Which of the following is paramagnetic ?
NO+
CO
$$O_2^{2 - }$$
B2

Explanation

Those species which have unpaired electrons are called paramagnetic species.

(a)   NO+ has 14 electrons.

Moleculer orbital configuration of NO+ is

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

Here is no unpaired electron, So it is Diamagnetic.

(b)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

Here is no unpaired electron so it is diamagnetic.

(c)   $$O_2^{2 - }$$ has 18 electrons.

Moleculer orbital configuration of $$O_2^{2 - }$$ is

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

Here also no unpaired electron present, so it is diamagnetic.

(d)   B2 has 10 electrons.

Molecular orbital configuration of B2 is

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

Here two unpaired electrons present.

So it is paramagnetic.

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