Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

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)&nbsp;&nbsp; $N_2$ 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)&nbsp;&nbsp; $\mathrm{N}_{2}^{2-}$ 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)&nbsp;&nbsp; $\mathrm{N}_{2}^{-}$ 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)&nbsp;&nbsp;&nbsp; $O_2^{2−}$ 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)&nbsp;&nbsp;&nbsp; $O_2^{−}$ 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)&nbsp;&nbsp; $O_2$ 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)&nbsp;&nbsp; $O_2^{+}$ 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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