JEE MAIN - Chemistry (2022 - 28th July Morning Shift - No. 15)

The number of paramagnetic species among the following is ___________.

$$\mathrm{B}_{2}, \mathrm{Li}_{2}, \mathrm{C}_{2}, \mathrm{C}_{2}^{-}, \mathrm{O}_{2}^{2-}, \mathrm{O}_{2}^{+}$$ and $$\mathrm{He}_{2}^{+}$$

Answer
4

Explanation

Those species which have unpaired electrons are called paramagnetic species.

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

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.

$$O_2^{2−}$$ 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 is no unpaired electron so it is diamagnetic.

$$O_2^{+}$$ has 15 electrons.

Moleculer orbital configuration of $$O_2^{+}$$ is

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

Here 1 unpaired electron present, so it is paramagnetic.

$$C_2$$ has 12 electrons.

Moleculer orbital configuration of $$C_2$$

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

Here no unpaired electron present, so it is diamagnetic.

$$C_2^{ - }$$ has 13 electrons.

Moleculer orbital configuration of $$C_2^{ - }$$ is

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

Here 1 unpaired electron present, so it is paramagnetic.

Li2 has 6 electrons.

Li2 = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}} \,$$

Here no unpaired electron present, so it is diamagnetic.

Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$

Here 1 unpaired electron present, so it is paramagnetic.

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