JEE MAIN - Chemistry (2007 - No. 35)
Which of the following species exhibits the diamagnetic behaviour?
$$O_2^{2−}$$
NO
$$O_2^+$$
O2
Explanation
Those species which have unpaired electrons are called paramagnetic species.
And those species which have no unpaired electrons are called diamagnetic species.
(a) $$O_2^{2−}$$ 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.
(b) 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.
(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) $$O_2^{+}$$ 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.
And those species which have no unpaired electrons are called diamagnetic species.
(a) $$O_2^{2−}$$ 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.
(b) 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.
(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) $$O_2^{+}$$ 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.
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