Question

Calculate the spin-only magnetic moment of a \( Mn^{2+} \) ion (Atomic number = 25).

• A. \( 3.87 \, BM \)
• B. \( 4.90 \, BM \)
• C. \( 5.92 \, BM \)
• D. \( 6.93 \, BM \)

Hint:

Always remove electrons from the \(4s\) orbital before the \(3d\) orbital when forming transition metal cations.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Magnetic moment is a measure of the magnetism caused by unpaired electrons in an atom or ion. We need to find this for the Manganese(II) ion.
Step 2: Key Formula or Approach:
Spin-only magnetic moment \( \mu = \sqrt{n(n+2)} \) Bohr Magnetons (BM), where \( n \) is the number of unpaired electrons.
Step 3: Detailed Explanation:
Atomic number of \( Mn = 25 \).
Electronic configuration of \( Mn \): \( [Ar] 3d^5 4s^2 \).
For \( Mn^{2+} \), two electrons are removed (from the \( 4s \) orbital):
\( Mn^{2+} \) configuration: \( [Ar] 3d^5 \).
In the \( 3d \) subshell, there are 5 orbitals. According to Hund's rule, 5 electrons will occupy these orbitals singly.
Number of unpaired electrons (\( n \)) \( = 5 \).
\[ \mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} \]
\[ \mu \approx 5.92 \, BM \]
Step 4: Final Answer:
The magnetic moment is \( 5.92 \, BM \).