The correct order of the calculated spin-only
magnetic moments of complexs (A) to (D) is:
(A) Ni(CO)4
(B) [Ni(H2O)6]Cl2
(C) Na2[Ni(CN)4]
(D) PdCl2(PPh3)2
Solution
(A) Ni(CO)<sub>4</sub>
<br><br>Ni = 3d<sup>8</sup>4s<sup>2</sup>
<br><br>CO is strong field ligand. So pairing of elections happens.
<br><br>$\therefore$ Number of unpaired electrons = 0
<br><br>$\therefore$ $\mu$<sub>spin</sub> = 0
<br><br>(B) [Ni(H<sub>2</sub>O)<sub>6</sub>]Cl<sub>2</sub>
<br><br>Ni<sup>+2</sup> = 3d<sup>8</sup>4s<sup>0</sup>
<br><br>H<sub>2</sub>O is weak field ligand. So no pairing of electrons happens.
<br><br>Number of unpaired electron = 2
<br><br>$\therefore$ $\mu$<sub>spin</sub> = $\sqrt {n\left( {n + 2} \right)}$ = $\sqrt {2\left( {2 + 2} \right)}$ = $\sqrt 8$ B.M
<br><br>(C) Na<sub>2</sub>[Ni(CN)<sub>4</sub>]
<br><br>Ni<sup>+2</sup> = 3d<sup>8</sup>4s<sup>0</sup>
<br><br>CN<sup>-</sup> is strong field ligand. So pairing of electrons happens.
<br><br>Number of unpaired electron = 0
<br><br>$\therefore$ $\mu$<sub>spin</sub> = 0
<br><br>(D) PdCl<sub>2</sub>(PPh<sub>3</sub>)<sub>2</sub>
<br><br>Pd<sup>2+</sup> = 4d<sup>8</sup>
<br><br>This is dsp<sup>2</sup> complex. And shape is square planar.
<br><br>$\therefore$ $\mu$<sub>spin</sub> = 0
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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