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Question: Answered & Verified by Expert
In $\left[\mathrm{CO}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, there are three unpaired electrons present. The $\mu_{\text {calculated }}$ is 3.87 BM which is quite different from the $\mu_{\text {experimental }}$ of 4.40 BM .
This is because of
ChemistryCoordination CompoundsJIPMERJIPMER 2016
Options:
  • A increase in number of unpaired electrons
  • B some contribution of the orbital motion of the electron to the magnetic moment
  • C change in orbital spin of the electron
  • D d-d transition
Solution:
1021 Upvotes Verified Answer
The correct answer is: some contribution of the orbital motion of the electron to the magnetic moment
In $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, there are three unpaired electrons present as,

$\mu_{\mathrm{cal}}=\sqrt{n(n+1)} \mathrm{BM}=\sqrt{3(3+1)} \mathrm{BM}=3.87 \mathrm{BM}$
As, the $t_{2 g}$ orbital is unsymmetrically filled hence due to orbital motion contribution, magnetic moment goes with the value 4.40 BM

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