Question

Which from following cations develops lowest value of spin only magnetic moment?

• A. \(\text{V}^{3+}\)
• B. \(\text{Cr}^{3+}\)
• C. \(\text{Mn}^{2+}\)
• D. \(\text{Fe}^{2+}\)

Hint:

For quick comparison, just count unpaired electrons — no need to calculate \(\mu\).

Answer 1

The Correct Option is A

Concept:
The spin-only magnetic moment is given by: \[ \mu = \sqrt{n(n+2)} \ \text{B.M. \] where \(n\) = number of unpaired electrons. Thus, magnetic moment depends only on number of unpaired electrons. Lower the number of unpaired electrons → lower magnetic moment. Step 1: Find electronic configurations of ions.

• V (Z = 23): \([Ar] 3d^3 4s^2\) \(\text{V}^{3+} \rightarrow [Ar] 3d^2\)
• Cr (Z = 24): \([Ar] 3d^5 4s^1\) \(\text{Cr}^{3+} \rightarrow [Ar] 3d^3\)
• Mn (Z = 25): \([Ar] 3d^5 4s^2\) \(\text{Mn}^{2+} \rightarrow [Ar] 3d^5\)
• Fe (Z = 26): \([Ar] 3d^6 4s^2\) \(\text{Fe}^{2+} \rightarrow [Ar] 3d^6\)

Step 2: Count unpaired electrons.

• \(\text{V}^{3+} (3d^2)\) → 2 unpaired electrons
• \(\text{Cr}^{3+} (3d^3)\) → 3 unpaired electrons
• \(\text{Mn}^{2+} (3d^5)\) → 5 unpaired electrons
• \(\text{Fe}^{2+} (3d^6)\) → 4 unpaired electrons

Step 3: Compare magnetic moments. \[ \mu \propto \sqrt{n(n+2)} \] So smallest \(n\) gives smallest magnetic moment.
Step 4: Conclusion. \[ \text{V}^{3+} \text{ has minimum unpaired electrons} \Rightarrow \text{lowest magnetic moment} \]