a
since, the coordination number of \(M n^{2+}\) ion in the complex ion is \(4,\) it will have either tetrahedral \(\left(s p^{3}-\right.\) hybridisation) or square planar \(\left(d s p^{2}-\right.\) hybridisation) geometry. But the fact that the magnetic moment of the complex ion is \(5.9 B M\) suggests that it should have tetrahedral geometry rather than square planar because of the presence of five unpaired electrons in the \(d\) -orbitals.

The number of unpaired electrons can be determined as follows:

As we know that, magnetic moment \(=\sqrt{n(n+2)}\)

\(5.9=\sqrt{n(n+2)}\)

On squaring both sides, the above equation becomes

\((5.9)^{2}=n^{2}+2 n\)

\(n^{2}+2 n-35=0\)

On solving the above equation, we get \(n=5\)