The bond order and the magnetic characteristics of CN- are :
Solution
CN<sup>-</sup> has 14 electrons.
<br><br>Moleculer orbital configuration of CN<sup>-</sup> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br>N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3
About this question
Subject: Chemistry · Chapter: Chemical Bonding and Molecular Structure · Topic: Ionic and Covalent Bonding
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