The number of unpaired electrons in \(Fe^{+2}, Mn^{2+}, Cr^{3+}, V^{5+}\) and \(Cu^{2+}\) are respectively?

\(4, 5, 3, 0, 1\)

The correct answer is option 1. \(4, 5, 3, 0, 1\)

The ground-state electronic configuration of iron\((Fe)\) is

The configuration of \(Fe^{2+}\) is

The number of unpaired electrons = \(4\)

The ground-state electronic configuration of manganese\((Mn)\) is

The configuration of \(Mn^{2+}\) is

The number of unpaired electrons = \(5\)

The ground-state electronic configuration of chromium\((Cr)\) is

The configuration of \(Cr^{3+}\) is

The number of unpaired electrons = \(3\)

The ground-state electronic configuration of vanadium \((V)\) is

The configuration of \(V^{5+}\) is

The number of unpaired electrons = \(0\)

The ground-state electronic configuration of copper \((Cu)\) is

The configuration of \(Cu^{5+}\) is

The number of unpaired electrons = \(1\)