Considering that $\Delta$0
> P, the magnetic moment
(in BM) of [Ru(H2O)6]2+ would be _________.
Answer (integer)
0
Solution
Ru(44) : [Kr] 4d<sup>7</sup>5s<sup>1</sup>
<br><br>Ru<sup>+2</sup> = [Kr]4d<sup>6</sup>
<br><br>As $\Delta$<sub>0</sub>
> P,
<br><br>$\therefore$ Pairing of e<sup>–</sup>s will take place.
<br><br>No. of unpaired e<sup>–</sup>s = 0
<br><br>$\therefore$ Magnetic moment = 0 B.M
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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