The metal ion (in gaseous state) with lowest spin-only magnetic moment value is :
Solution
<table style="width:100%;text-align:left;min-width:460px">
<thead>
<tr>
<th></th>
<th>Valence shell configuration</th>
<th>Unpaired electrons</th>
</tr>
</thead>
<tbody>
<tr>
<td>V<sup>2+</sup></td>
<td>3d<sup>3</sup>4s<sup>0</sup></td>
<td>n = 3</td>
</tr>
<tr>
<td>Ni<sup>2+</sup></td>
<td>3d<sup>8</sup>4s<sup>0</sup></td>
<td>n = 2</td>
</tr>
<tr>
<td>Cr<sup>2+</sup></td>
<td>3d<sup>4</sup>4s<sup>0</sup></td>
<td>n = 4</td>
</tr>
<tr>
<td>Fe<sup>2+</sup></td>
<td>3d<sup>6</sup>4s<sup>0</sup></td>
<td>n = 4</td>
</tr>
</tbody>
</table><br/><br/>
Since Ni<sup>2+</sup> has least number of unpaired electrons.
Hence Ni<sup>2+</sup> will have lowest spin only magnetic
moment Value.
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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