Question

The calculated magnetic moment of two dipositive ions of 3d series element is 4.9 BM. The ions are

• A. \(Ti^{2+}\) and \(Sc^{2+}\)
• B. \(Mn^{2+}\) and \(Cr^{2+}\)
• C. \(V^{2+}\) and \(Ti^{2+}\)
• D. \(Cr^{2+}\) and \(Fe^{2+}\)
• E. \(Fe^{2+}\) and \(Ni^{2+}\)

Hint:

A quick shortcut: The magnetic moment value X.Y usually implies there are X unpaired electrons.
So, \(4.9 \text{ BM} \implies 4\) unpaired electrons.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The magnetic moment (\(\mu\)) is calculated using the spin-only formula based on the number of unpaired electrons (\(n\)).

Step 2: Key Formula or Approach:
\[ \mu = \sqrt{n(n+2)} \text{ Bohr Magnetons (BM)} \]

Step 3: Detailed Explanation:
For a value of \(\mu = 4.9 \text{ BM}\), the number of unpaired electrons must be \(n = 4\) (since \(\sqrt{4(6)} = \sqrt{24} \approx 4.89\)).
Check the electronic configuration of dipositive ions:
- \(Cr^{2+}\): Atomic No. 24 (\(3d^5 4s^1\)). Ion is \(3d^4\). Unpaired electrons = 4.
- \(Fe^{2+}\): Atomic No. 26 (\(3d^6 4s^2\)). Ion is \(3d^6\). Unpaired electrons = 4 (one pair, four single).
Both ions have 4 unpaired electrons.

Step 4: Final Answer:
The ions are \(Cr^{2+}\) and \(Fe^{2+}\).