Among the following species
$$\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}, \mathrm{N}_{2}^{-}, \mathrm{N}_{2}^{2-}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-}$$
the number of species showing diamagnesim is _______________.
Answer (integer)
2
Solution
Those species which have unpaired electrons are called paramagnetic species.
<br><br>And those species which have no unpaired electrons are called diamagnetic species.
<br><br>(1) $N_2$ has 14 electrons.
<br><br>Moleculer orbital configuration of $N_2$
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here no unpaired electron present, so it is diamagnetic.
<br><br>(2) Moleculer orbital configuration of $N_2^{ + }$ (13 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
<br><br>Here in $N_2^{ + }$, 1 unpaired electron present, so it is paramagnetic.
<br><br>(3) $\mathrm{N}_{2}^{2-}$ has 16 electrons.
<br><br>Moleculer orbital configuration of $\mathrm{N}_{2}^{2-}$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present, so it is paramagnetic.
<br><br>(4) $\mathrm{N}_{2}^{-}$ has 15 electrons.
<br><br>Moleculer orbital configuration of $\mathrm{N}_{2}^{-}$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
<br><br>(a) $O_2^{2−}$ has 18 electrons.
<br><br>Moleculer orbital configuration of $O_2^{2−}$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br><br>(b) $O_2^{−}$ has 17 electrons.
<br><br>Moleculer orbital configuration of $O_2^{2−}$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
<br><br>(c) $O_2$ has 16 electrons.
<br><br>Moleculer orbital configuration of $O_2$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present, so it is paramagnetic.
<br><br>(d) $O_2^{+}$ has 15 electrons.
<br><br>Moleculer orbital configuration of $O_2^{+}$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
About this question
Subject: Chemistry · Chapter: Chemical Bonding and Molecular Structure · Topic: Ionic and Covalent Bonding
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