JEE MAIN - Chemistry (2019 - 9th April Evening Slot - No. 11)
Among the following species, the diamagnetic
molecule is
CO
B2
O2
NO
Explanation
Those species which have unpaired electrons are called paramagnetic species.
And those species which have no unpaired electrons are called diamagnetic species.
(a) 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.
(b) 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.
(c) $$O_2$$ has 16 electrons.
Moleculer orbital configuration of $$O_2$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
Here 2 unpaired electron present, so it is paramagnetic.
(d) NO has 15 electrons.
Moleculer orbital configuration of NO is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
Here is 1 unpaired electron, So it is Paramagnetic.
And those species which have no unpaired electrons are called diamagnetic species.
(a) 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.
(b) 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.
(c) $$O_2$$ has 16 electrons.
Moleculer orbital configuration of $$O_2$$ is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
Here 2 unpaired electron present, so it is paramagnetic.
(d) NO has 15 electrons.
Moleculer orbital configuration of NO is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
Here is 1 unpaired electron, So it is Paramagnetic.
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