Question

In $\left[CO\right(H_{2}O{)}_6{]}^{2+}$, 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
• 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

Answer 1

Answer: (B)

some contribution of the orbital motion of the electron to the magnetic moment

In $\left[Co\right(H_{2}O{)}_6{]}^{2+}$, there are three unpaired electrons present as,

${\mu }_{cal}=\sqrt{n(n+1)}BM=\sqrt{3(3+1)}BM=3.87BM$

As, the $t_{2g}$ orbital is unsymmetrically filled hence due to orbital motion contribution, magnetic moment goes with the value 4.40 BM

Option B is correct.