Question:

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

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If $n=5$, the magnetic moment is always "5 point something." Specifically, $\sqrt{35} = 5.92$.
Updated On: Apr 30, 2026
  • 3.87
  • 4.9
  • 1.73
  • 5.92
Show Solution
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The Correct Option is D

Solution and Explanation

Step 1: Configuration
Mn ($Z=25$) is $[Ar] 3d^5 4s^2$. $Mn^{2+}$ is $[Ar] 3d^5$.
Step 2: Unpaired Electrons
In $3d^5$, there are 5 unpaired electrons ($n = 5$).
Step 3: Magnetic Moment Formula
$\mu = \sqrt{n(n+2)}$ BM. $\mu = \sqrt{5(5+2)} = \sqrt{35}$.
Step 4: Calculation
$\sqrt{35} \approx 5.92$ BM.
Final Answer:(D)
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