Which one of the following species doesn't have a magnetic moment of 1.73 BM, (spin only value)?
Solution
Species does not have a magnetic moment of 1.7 BM means species must not contain single unpaired electron.<br/><br/>(a) Molecular orbital configuration of O$_2^ +$ (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$<br/><br/>Unpaired electron = 1<br/><br/>$\mu = \sqrt {n(n + 2)} = \sqrt {1(1 + 2)} = \sqrt 3 = 1.73$<br/><br/>where n = no. of unpaired e<sup>$-$</sup><br/><br/>$\therefore$ $\mu$ = 1.73 BM<br/><br/>(b) Cul $\Rightarrow$ Cu<sup>+</sup> $-$[Ar]3d<sup>10</sup>, Unpaired electron = 0<br/><br/>I<sup>$-$</sup> $\to$ [Xe], unpaired electron = 0<br/><br/>Therefore, $\mu$ = 0<br/><br/>(c) [Cu(NH<sub>3</sub>)<sub>4</sub>]Cl<sub>2</sub><br/><br/>Cu<sup>2+</sup> $\to$ [Ar]3d<sup>9</sup><br/><br/>Unpaired electron = 1,<br/><br/>Therefore, $\mu$ = 1.73 BM<br/><br/>(d) Molecular orbital configuration of $O_2^ -$ (17 electrons) 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/>Unpaired electron = 1<br/><br/>Therefore, $\mu$ = 1.73 BM<br/><br/>Therefore, Cul does not have magnetic moment of 1.73 BM.
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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