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 (integer)
4
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>B<sub>2</sub> has 10 electrons.
<br><br>Molecular orbital configuration of B<sub>2</sub> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$
<br><br>Here two unpaired electrons present. So it is paramagnetic.
<br><br>$O_2^{2−}$ 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>$O_2^{+}$ 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.
<br/><br/>$C_2$ has 12 electrons.
<br><br>Moleculer orbital configuration of $C_2$
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$$
<br><br>Here no unpaired electron present, so it is diamagnetic.
<br/><br/>$C_2^{ - }$ has 13 electrons.
<br><br>Moleculer orbital configuration of $C_2^{ - }$ is
<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 1 unpaired electron present, so it is paramagnetic.
<br/><br/>Li<sub>2</sub> has 6 electrons.
<br/><br/>Li<sub>2</sub> = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$
<br><br>Here no unpaired electron present, so it is diamagnetic.
<br/><br/>Configuration of $He_2^ +$ (3 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^1}}^ *$
<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
This question is part of PrepWiser's free JEE Main question bank. 97 more solved questions on Chemical Bonding and Molecular Structure are available — start with the harder ones if your accuracy is >70%.