Question

Although \(Cr^{3+}\)and \(Co^{2+}\) ions have the same number of unpaired electrons, the magnetic moment of \(Cr^{2+}\) is 3.87 B.M. and that of \(Co^{2+}\)is 4.87 𝐵.𝑀. Why?

Answer 1

The electronic configuration of Cr in its +3 oxidation state is \(Cr^{3+}=\left [Ar \right ]3d^3\)and the electronic configuration of 𝐶𝑜 in its +2 oxidation state is \(Co^{2+}=\left [ Ar \right ]3d^7\).

The number of unpaired electrons in both \(Cr^{3+}and~~Co^{2+}\) is the same i.e., 3.

The total magnetic moment depends upon the total spin quantum number (S) and the resultant orbital angular momentum (L);

\(\mu=\sqrt{4S(S+1)+L(L+1)}\)B.M

\(Cr^{3+}\) has symmetrical electronic configuration \((t_{2g}^{3}e_{g}^{o})\) due to the presence of three unpaired electrons. Thus, there is no orbital contribution present.

Whereas \(Co^{2+}\) has unsymmetrical electronic configuration \((t_{2g}^{5}e_{g}^{2})\) with both paired and unpaired electrons. Thus, \(Co^{2+}\) shows appreciable orbital contribution.

Thus, even though \(Cr^3+ ~{\rm and}~ Co^{2+}\) ions have the same number of unpaired electrons, the overall magnetic moment \(Cr^{3+}\) is less i.e., 3.87 B.M. and that of \(Co^{2+}\) is more i.e., 4.87 B.M.