Question

Calculate magnetic moment for \(\text{Fe}^{3+}\)?

• A. \(3.87\) BM
• B. \(2.83\) BM
• C. \(5.92\) BM
• D. \(1.73\) BM

Hint:

Count the unpaired d electrons in the ferric ion and apply the spin-only formula.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The magnetic moment is given by \(\mu=\sqrt{n(n+2)}\) BM, where \(n\) is the number of unpaired electrons.

Step 2: Electronic configuration
\(\text{Fe}\) (Z = 26) is \([\text{Ar}]3d^{6}4s^{2}\). \(\text{Fe}^{3+}\) loses the two 4s electrons and one 3d electron, so it is \([\text{Ar}]3d^{5}\).

Step 3: Unpaired electrons
The five d electrons are all unpaired, so \(n=5\).

Step 4: Calculation
\(\mu=\sqrt{5\times 7}=\sqrt{35}\approx 5.92\ \text{BM}\)

Final Answer:
The magnetic moment of Fe3+ is 5.92 BM. \[ \boxed{\text{(C) }5.92\ \text{BM}} \]