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

• A

  increase in number of unpaired electrons
• Correct answer
  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

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

Hence, the correct answer is option: 'B'.