d and f Block Elements: Application Problems (3)

Question

d and f Block Elements: application problems for JEE+NEET

doubts.ai · Accepted Answer

Solution — Step by Step - Fe (Z = 26): [Ar]\,3d^6 4s^2. Fe^{2+}: lose 2 electrons from 4s first → 3d^6. - Mn (Z = 25): [Ar]\,3d^5 4s^2. Mn^{2+}: 3d^5. - Zn (Z = 30): [Ar]\,3d^{10} 4s^2. Zn^{2+}: 3d^{10}. For high-spin (free ion / weak-field environment) configurations, fill the 5 d-orbitals singly first, then pair. - 3d^6: 5 singly + 1 paired = 4 unpaired. - 3d^5: all singly = 5 unpaired. - 3d^{10}: all paired = 0 unpaired. - Fe^{2+} (n = 4): = \sqrt{4 6 = 24 4.90 BM - Mn^{2+} (n = 5): = \sqrt{5 7 = 35 5.92 BM - Zn^{2+} (n = 0): = 0 BM (diamagnetic) Mn^{2+} has the highest magnetic moment because its 3d^5 configuration places one electron in each of the five d-orbitals — maximum unpaired electrons. Final answers: (	ext{Fe}^{2+}) \mathbf{4.90} BM, (	ext{Mn}^{2+}) \mathbf{5.92} BM, (	ext{Zn}^{2+}) = 0 BM. Mn^{2+} is most paramagnetic. Why This Works The spin-only formula assumes that orbital angular momentum doesn't contribute (a good approximation for first-row transition metal ions in octahedral or tetrahedral fields). It's an underestimate for ions where the orbital contribution is significant (e.g., Co^{2+}, Fe^{3+} in some complexes). The trick to remembering electron-loss order: 4s electrons leave before 3d . Even though 3d fills after 4s (Aufbau), the energies cross once the 3d shell is partially populated. Alternative Method Memorise this short table for M^{2+} ions of the first transition row: Ion d^n n unpaired (BM) ----- ------- ------------- ----------- Sc²⁺ d^1 1 1.73 Ti²⁺ d^2 2 2.83 V²⁺ d^3 3 3.87 Cr²⁺ d^4 4 4.90 Mn²⁺ d^5 5 5.92 Fe²⁺ d^6 4 4.90 Co²⁺ d^7 3 3.87 Ni²⁺ d^8 2 2.83 Cu²⁺ d^9 1 1.73 Zn²⁺ d^{10} 0 0 The values are symmetric around Mn²⁺ — a useful self-check. Common Mistake Removing 3d electrons before 4s electrons when forming a cation. The correct rule is 4s electrons leave first for transition metals. So Fe (3d^6 4s^2) becomes Fe²⁺ (3d^6), not (3d^4 4s^2). This single mistake derails the entire question. NEET asks magnetic moment questions almost every year — typically with 4–5 ion choices. Memorising the n(n+2) values for n = 1 to 5 saves time.

Ion	$d^n$	$n$ unpaired	$\mu$ (BM)
Sc²⁺	$d^1$	$1$	$1.73$
Ti²⁺	$d^2$	$2$	$2.83$
V²⁺	$d^3$	$3$	$3.87$
Cr²⁺	$d^4$	$4$	$4.90$
Mn²⁺	$d^5$	$5$	$5.92$
Fe²⁺	$d^6$	$4$	$4.90$
Co²⁺	$d^7$	$3$	$3.87$
Ni²⁺	$d^8$	$2$	$2.83$
Cu²⁺	$d^9$	$1$	$1.73$
Zn²⁺	$d^{10}$	$0$	$0$

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