Question

Out of Fe\(^{2+}\), Co\(^{2+}\), Cr\(^{3+}\), Ni\(^{2+}\), the one which shows highest magnetic moment is:

• A. Fe\(^{2+}\)
• B. Co\(^{2+}\)
• C. Cr\(^{3+}\)
• D. Ni\(^{2+}\)

Hint:

When comparing magnetic moments, consider both number of unpaired electrons and pairing energy. Cr\(^{3+}\) with 3 unpaired electrons in stable half-filled \( t_{2g} \) orbitals gives maximum spin-only moment.

Answer 1

The Correct Option is C

Magnetic moment depends on number of unpaired electrons, calculated by: \[ \mu = \sqrt{n(n+2)}\,\text{BM} \] Where \( n \) = number of unpaired electrons. Let's calculate for each: - Fe\(^{2+}\): Atomic number 26 → Fe\(^{2+}\) = 3d\(^6\) → 4 unpaired electrons - Co\(^{2+}\): Atomic number 27 → Co\(^{2+}\) = 3d\(^7\) → 3 unpaired electrons - Cr\(^{3+}\): Atomic number 24 → Cr\(^{3+}\) = 3d\(^3\) → 3 unpaired electrons - Ni\(^{2+}\): Atomic number 28 → Ni\(^{2+}\) = 3d\(^8\) → 2 unpaired electrons But Cr\(^{3+}\) has only 3 electrons all unpaired in \( t_{2g} \) orbitals with no pairing energy losses, thus greater stability and maximum spin-only moment (low pairing). Hence: \[ \mu_{\text{Cr}^{3+}} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\,\text{BM} \] While Fe\(^{2+}\) has 4 unpaired, it experiences some pairing loss in high-spin cases. So Cr\(^{3+}\) shows more stable unpaired electron configuration. \[ \boxed{\text{Cr}^{3+}\ \text{has highest magnetic moment}} \]