The magnetic moment of a transition metal ion is found to be $3.87$ Bohr Magneton (BM). The number of unpaired electrons present in it is:

Open in App

Solution

The magnetic moment of a transition metal ion is found to be $3.87$ Bohr Magneton (BM). The number of unpaired electrons present in it is 3.
$μ = \sqrt{n (n + 2)} = 3.87 B M$
Hence, $n = 3$

Suggest Corrections

Similar questions

\sqrt{15} B M

For first-row transition metal ions the magnetic moment in Bohr magnetons is calculated by the formula.

M^{3 +}

\sqrt{35} B M

M^{3 +}

The magnetic moment of an ion in its +3 oxidation state is 3.85 BM. The number of unpaired electrons present in the ion is _____________.

Join BYJU'S Learning Program