Question

What is the value of spin only magnetic moment for \( Mn^{2+} \) in BM?

• A. 3.87
• B. 4.9
• C. 1.73
• D. 5.92

Hint:

If $n=5$, the magnetic moment is always "5 point something." Specifically, $\sqrt{35} = 5.92$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The spin-only magnetic moment depends on the number of unpaired electrons in the metal ion.
Step 2: Key Formula or Approach:
1. Write electronic configuration of \( Mn \) and \( Mn^{2+} \).
2. Find number of unpaired electrons (\( n \)).
3. Use formula: \( \mu = \sqrt{n(n+2)} \) Bohr Magnetons (BM).
Step 3: Detailed Explanation:
Atomic number of \( Mn = 25 \).
Ground state electronic configuration: \( [Ar] 3d^{5} 4s^{2} \).
Configuration of \( Mn^{2+} \): \( [Ar] 3d^{5} \).
In the \( 3d \) subshell, there are 5 orbitals. According to Hund's rule, 5 electrons will occupy these singly.
So, number of unpaired electrons \( n = 5 \).
Calculating magnetic moment:
\[ \mu = \sqrt{5(5+2)} \]
\[ \mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.916 \text{ BM} \]
Rounding to two decimal places, we get 5.92 BM.
Step 4: Final Answer:
The spin-only magnetic moment for \( Mn^{2+} \) is 5.92 BM.