Question

Which of the following ions has the highest magnetic moment?

• A. Zn\(^{2+}\)
• B. Ti\(^{3+}\)
• C. Sc\(^{3+}\)
• D. Mn\(^{2+}\)

Hint:

Magnetic moment order: Mn\(^{2+}\) (5 unpaired) > Ti\(^{3+}\) (1 unpaired) > Zn\(^{2+}\)/Sc\(^{3+}\) (0 unpaired).

Answer 1

The Correct Option is D

Concept: Magnetic moment (\(\mu\)) depends on the number of unpaired electrons (\(n\)): \[ \mu = \sqrt{n(n+2)} \text{ Bohr magnetons} \] Higher the number of unpaired electrons, higher the magnetic moment.

Step 1: Determine electronic configurations and unpaired electrons.
• Zn\(^{2+}\): Atomic number 30. Zn: \([Ar] 3d^{10}4s^2\). Zn\(^{2+}\) loses two 4s electrons → \([Ar] 3d^{10}\). Unpaired electrons = 0.
• Ti\(^{3+}\): Atomic number 22. Ti: \([Ar] 3d^2 4s^2\). Ti\(^{3+}\) loses two 4s electrons and one 3d electron → \([Ar] 3d^1\). Unpaired electrons = 1.
• Sc\(^{3+}\): Atomic number 21. Sc: \([Ar] 3d^1 4s^2\). Sc\(^{3+}\) loses two 4s electrons and one 3d electron → \([Ar] 3d^0\). Unpaired electrons = 0.
• Mn\(^{2+}\): Atomic number 25. Mn: \([Ar] 3d^5 4s^2\). Mn\(^{2+}\) loses two 4s electrons → \([Ar] 3d^5\). Unpaired electrons = 5 (half-filled d-subshell, highly stable).

Step 2: Compare magnetic moments. \[ \mu \propto \sqrt{n(n+2)} \] Mn\(^{2+}\) has 5 unpaired electrons → highest magnetic moment.