Magnetic Properties — Spin-only formula
Coordination Compounds
15
JEE Qs
8%
Hard
60
min
Accurately determine the oxidation state of the central metal ion and then apply Crystal Field Theory to deduce the number of unpaired electrons for correct magnetic moment calculation.
🧮 Key Formulas
✅ Key Points for JEE
- 1The spin-only magnetic moment formula directly correlates the observed magnetic properties (paramagnetism) of a coordination complex with the number of unpaired electrons (n) present in its central metal ion.
- 2A non-zero magnetic moment (µ > 0) indicates paramagnetism due to the presence of unpaired electrons, while a zero magnetic moment (µ = 0) indicates diamagnetism (all electrons are paired).
- 3To use the formula, accurately determine the oxidation state of the central metal ion, its d-electron count, and then apply Crystal Field Theory (CFT) principles (considering strong vs. weak field ligands) to deduce 'n'.
- 4The calculated spin-only magnetic moment helps distinguish between high-spin (more unpaired electrons) and low-spin (fewer or no unpaired electrons) complexes, which is crucial for characterizing their electronic structure.
- 5While the formula is 'spin-only', orbital contribution to magnetic moment is generally ignored in JEE unless specifically mentioned, as it is often small or quenched.
⚠️ Common Mistakes
- ✕Incorrectly determining the oxidation state of the central metal ion, leading to an incorrect d-electron count.
- ✕Misapplying Crystal Field Theory (CFT) principles, especially regarding strong-field (pairing) vs. weak-field (no pairing) ligands, resulting in an wrong 'n'.
- ✕Simple calculation errors when substituting 'n' into the formula or during square root calculations.
- ✕Forgetting to include the unit 'Bohr Magneton (BM)' in the final answer.
📝 Practice Questions
See allQ74.In the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was ______ %.
Q61.The calculated spin-only magnetic moments of K3 [Fe(OH)6] and K4 [Fe(OH)6] respectively are : (1) 3.87 and 4.90 B.M. (2) 4.90 and 5.92 B.M. (3) 4.90 and 4.90 B.M. (4) 5.92 and 4.90 B.M.
Q54.Identify the homoleptic complexes with odd number of d electrons in the central metal : (A) [FeO4]2− (B) [Fe(CN)6]3− (C) [Fe(CN)5NO]2− (D) [CoCl4]2− (E) [Co(H2O)3 F3] Choose the correct answer from the options given below : (1) (A), (B) and (D) only (2) (C) and (E) only (3) (B) and (D) only (4) (A), (C) and (E) only
Q68.In which of the following complexes the CFSE, Δo will be equal to zero? (1) [Fe(en)3]Cl3 (2) K4 [Fe(CN)6] (3) [Fe(NH3)6]Br2 (4) K3 [Fe(SCN)6]
Q70.From the magnetic behaviour of [NiCl4]2− (paramagnetic) and [Ni(CO)4] (diamagnetic), choose the correct geometry and oxidation state. (1) [NiCl4]2−: NiII , tetrahedral [Ni(CO)4] : NiII , (2) [NiCl4]2− : NiII , square planar [Ni(CO)4] : Ni(0) square planar , square planar (3) [NiCl4]2−: NiII , tetrahedral [Ni(CO)4] : Ni(0), (4) [NiCl4]2−: Ni(0), tetrahedral [Ni(CO)4] : Ni(0), tetrahedral square planar
Q53.The correct order of the following complexes in terms of their crystal field stabilization energies is : (1) [Co(NH3)4]2+ < [Co(NH3)6]2+ < [Co(en)3]3+ < [Co(NH3)6]3+ (2) [Co(NH3)6]2+ < [Co(NH3)6]3+ < [Co(NH3)4]2+ < [Co(en)3]3+ (3) [Co(en)3]3+ < [Co(NH3)6]3+ < [Co(NH3)6]2+ < [Co(NH3)4]2+ (4) [Co(NH3)4]2+ < [Co(NH3)6]2+ < [Co(NH3)6]3+ < [Co(en)3]3+
NCERT Chapters
- Class 12 Chemistry Ch 9: Coordination Compounds