Question

What is the spin magnetic moment of Cr(III) in Bohr Magnetons?

• A. 3.87 BM
• B. 2.83 BM
• C. 5.92 BM
• D. 4.90 BM

Hint:

To calculate the spin-only magnetic moment, use the formula \( \mu = \sqrt{n(n + 2)} \), where \( n \) is the number of unpaired electrons.

Answer 1

The Correct Option is A

Step 1: Neutral chromium is [Ar]3d\(^5\)4s\(^1\); transition metals always lose their 4s electrons before any 3d electrons when forming a cation.
Step 2: Removing the single 4s electron and two 3d electrons to reach the +3 state leaves 3d\(^3\), three unpaired electrons sitting in separate d-orbitals by Hund's rule.
Step 3: Spin-only moment: \[ \mu = \sqrt{n(n+2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \text{ BM} \], matching option (A).