$\mathrm{Nd^{2+}}$ = __________
Solution
Nd(60) = [Xe] 4f<sup>4</sup> 5d<sup>0</sup> 6s<sup>2</sup>
<br/><br/>Nd<sup>2+</sup>
= [Xe] 4f<sup>4</sup> 5d<sup>0</sup> 5s<sup>0</sup>
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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