The calculated magnetic moment using the 'spin only' formula $μ = \sqrt{n(n+2)}$ for $Cr^{2+}$ and $Fe^{2+}$ respectively is

Both 4.89

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)