The calculated magnetic moment using the 'spin only' formula $μ = \sqrt{n(n+2)}$ for $Cr^{2+}$ and $Fe^{2+}$ respectively is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Both 4.89
We can calculate magnetic moment using the spin-only formula:
$\mu = \sqrt{n(n+2)} \, \text{BM}$
where n = number of unpaired electrons.
Step 1: Cr²⁺
- Cr atomic number = 24 → electron configuration: [Ar] 3d⁵ 4s¹
- Cr²⁺ → lose 2 electrons → [Ar] 3d⁴
- Number of unpaired electrons, n = 4
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM}$
Step 2: Fe²⁺
- Fe atomic number = 26 → [Ar] 3d⁶ 4s²
- Fe²⁺ → lose 2 electrons → [Ar] 3d⁶
- Number of unpaired electrons in high-spin octahedral → n = 4
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM}$
- If low-spin (strong field ligand), n = 0 → μ = 0 BM
Assuming high-spin (common for free ions in aqueous solution)