Question

The spin only magnetic moment of Fe\(^{3+}\) ion (in BM) is approximately.

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

The spin-only magnetic moment formula applies when there is no contribution from orbital motion of electrons, which is true for transition metal ions like Fe\(^{3+}\).

Answer 1

The Correct Option is C

Step 1: {Electronic Configuration of Fe}
The electronic configuration of Fe is \( {Fe} = [{Ar}] 3d^6 4s^2 \). For Fe\(^{3+}\), it loses three electrons, resulting in the configuration: \[ {Fe}^{3+} = [{Ar}] 3d^5 \] Step 2: {Number of Unpaired Electrons}
For Fe\(^{3+}\), the number of unpaired electrons, \( n \), is 5 (since it has 5 electrons in the 3d orbitals). Step 3: {Spin Only Magnetic Moment Formula}
The formula for calculating the spin-only magnetic moment is: \[ \mu_s = \sqrt{n(n+2)} { BM} \] where \(n = 5\). Therefore: \[ \mu_s = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} { BM} \approx 6 { BM} \] Thus, the correct answer is (C).

Answer 2

Step 1: Determine the electron configuration of Fe³⁺
Atomic number of Fe (iron) = 26
Electronic configuration of Fe: [Ar] 3d⁶ 4s²
Fe³⁺ means loss of 3 electrons: 2 from 4s and 1 from 3d
So, Fe³⁺ configuration = [Ar] 3d⁵

Step 2: Count the number of unpaired electrons
3d⁵ configuration means 5 electrons in the d-orbitals
These electrons occupy all five d-orbitals singly (Hund’s rule), so there are 5 unpaired electrons

Step 3: Use the spin-only formula for magnetic moment
Magnetic moment (μ) = √(n(n+2)) BM
Where n = number of unpaired electrons
n = 5
μ = √(5(5+2)) = √35 ≈ 5.92 BM

Step 4: Final Answer
The spin-only magnetic moment of Fe³⁺ is approximately:
6 BM