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

Step 1: Understanding the Concept:
The "spin-only" magnetic moment depends on the number of unpaired electrons (\(n\)) in the d-subshell of the transition metal ion. Fewer unpaired electrons result in a lower magnetic moment.
Step 2: Key Formula or Approach:
\[ \mu = \sqrt{n(n+2)}\text{ B.M.} \] Step 3: Detailed Explanation:
Let's find the number of unpaired electrons (\(n\)) for each cation:
- (A) \(\text{V}^{3+}\): Vanadium (\(Z=23\)) is \([Ar] 3d^3 4s^2\). \(\text{V}^{3+}\) is \([Ar] 3d^2\). \(\mathbf{n = 2}\).
- (B) \(\text{Cr}^{3+}\): Chromium (\(Z=24\)) is \([Ar] 3d^5 4s^1\). \(\text{Cr}^{3+}\) is \([Ar] 3d^3\). \(\mathbf{n = 3}\).
- (C) \(\text{Mn}^{2+}\): Manganese (\(Z=25\)) is \([Ar] 3d^5 4s^2\). \(\text{Mn}^{2+}\) is \([Ar] 3d^5\). \(\mathbf{n = 5}\).
- (D) \(\text{Fe}^{2+}\): Iron (\(Z=26\)) is \([Ar] 3d^6 4s^2\). \(\text{Fe}^{2+}\) is \([Ar] 3d^6\). Unpaired electrons: $4$ (since $10-6=4$). \(\mathbf{n = 4}\).
Among the options, \(\text{V}^{3+}\) has the minimum number of unpaired electrons (\(n=2\)), so it will have the lowest magnetic moment.
Step 4: Final Answer:
\(\text{V}^{3+}\) has the lowest magnetic moment.