Magnetic moment of \(Co^{2+}\) will be

3.87 BM

The correct answer is option 1. 3.87 BM.

The atomic number of \(Co\) is 27 and its electronic configuration will be :

\(_{27}Co = [_{18}Ar]4s^23d^7\)

In \(+2\) oxidation state the configuration will be

\(Co^{2+}=[_{18}Ar]4s^03d^7\)

The number of unpaired electrons, \(n = 3\)

Thus, magnetic moment,

\(\mu = \sqrt{n(n + 2)} BM\)

or, \(\mu = \sqrt{3(3 + 2)} BM\)

or, \(\mu = \sqrt{15} BM\)

or, \(\mu = 3.87 BM\)