The number of unpaired electrons in \(Fe^{+2}, Mn^{2+}, Cr^{3+}, V^{5+}\) and \(Cu^{2+}\) are respectively?
Answer & explanation
Correct answer: option 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 = \(1\)