Arrange these transition metal ions, based on decreasing number of unpaired electrons.
A. \(Fe^{2+}\)
B. \(Mn^{2+}\)
C. \(Co^{2+}\)
D. \(Sc^{3+}\)
E. \(Cu^{2+}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. B > A > C > E > D.
The number of unpaired electrons in transition metal ions is related to their electronic configurations. Let us analyze the given ions:
A. \(Fe^{2+}\): In the ground state, the electronic configuration of neutral Fe is \([Ar] 3d^6 4s^2\). When it loses two electrons to form \(Fe^{2+}\), the 4s electrons are lost, resulting in an electronic configuration of \([Ar] 3d^6\). This has four unpaired electrons.
B. \(Mn^{2+}\): The neutral Mn has an electronic configuration of \([Ar] 3d^5 4s^2\). When it loses two electrons to form \(Mn^{2+}\), it becomes \([Ar] 3d^5\), which has five unpaired electrons.
C. \(Co^{2+}\): Neutral Co has an electronic configuration of \([Ar] 3d^7 4s^2\). When it loses two electrons to form \(Co^{2+}\), it becomes \([Ar] 3d^7\), which has three unpaired electrons.
D. \(Sc^{3+}\): Scandium (\(Sc^{3+}\)) loses three electrons from its neutral state of \([Ar] 3d^1 4s^2\), resulting in the electronic configuration of \([Ar] 3d^0\). This has no unpaired electrons.
E. \(Cu^{2+}\): The neutral Cu has an electronic configuration of \([Ar] 3d^9 4s^2\). When it loses two electrons to form \(Cu^{2+}\), it becomes \([Ar] 3d^9\), which has one unpaired electron.
So, the correct answer is 2. B > C > A > D > E.