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: Understand the electronic configuration of Cr(III).
Chromium (Cr) has the electron configuration \( [Ar] 3d^5 4s^1 \). In the +3 oxidation state (Cr(III)), it loses 3 electrons, leaving the configuration \( [Ar] 3d^3 \). This means Cr(III) has 3 unpaired electrons.
Step 2: Use the formula for the spin-only magnetic moment.
The formula for the spin-only magnetic moment is: \[ \mu = \sqrt{n(n + 2)} \] where \( n \) is the number of unpaired electrons. For Cr(III), \( n = 3 \), so: \[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \, \text{BM} \] Step 3: Conclusion.
Therefore, the spin magnetic moment of Cr(III) is approximately 3.87 Bohr magnetons, which corresponds to option (A).