Crystal Field Theory — Strong vs weak field, CFSE
Coordination Compounds
15
JEE Qs
8%
Hard
75
min
Systematically determine the metal ion's oxidation state, d-electron count, complex geometry, ligand field strength (from spectrochemical series), and then apply the Δ vs P rule to correctly calculate CFSE and determine magnetic properties.
🧮 Key Formulas
✅ Key Points for JEE
- 1Crystal Field Theory (CFT) treats ligands as point charges, causing electrostatic repulsion with metal d-orbitals, leading to their splitting into different energy levels.
- 2In octahedral complexes, d-orbitals split into lower energy t2g (dxy, dyz, dzx) and higher energy eg (dx^2-y^2, dz^2) sets, with the energy difference denoted as Δo (or 10 Dq).
- 3In tetrahedral complexes, d-orbitals split into lower energy e (dx^2-y^2, dz^2) and higher energy t2 (dxy, dyz, dzx) sets, with the energy difference Δt, where Δt ≈ (4/9)Δo.
- 4The spectrochemical series (e.g., I- < Cl- < H2O < NH3 < CN- < CO) ranks ligands based on their ability to cause d-orbital splitting. Strong field ligands cause large Δ, weak field ligands cause small Δ.
- 5For d^4 to d^7 configurations, the spin state (high spin vs. low spin) is determined by comparing Δ (Δo or Δt) with the pairing energy (P). If Δ > P, electrons pair up (low spin); if Δ < P, electrons occupy higher energy orbitals first (high spin).
- 6Crystal Field Stabilization Energy (CFSE) is the net stabilization energy resulting from d-electron distribution in the split orbitals, often expressed in terms of Δ and accounts for pairing energy (P) when applicable.
⚠️ Common Mistakes
- ✕Confusing the d-orbital splitting patterns between octahedral (t2g lower, eg higher) and tetrahedral (e lower, t2 higher) geometries.
- ✕Incorrectly identifying strong vs. weak field ligands or misapplying the spectrochemical series, leading to wrong spin state (high/low spin) determination.
- ✕Errors in calculating the d-electron count for the metal ion, and subsequent incorrect filling of electrons into split orbitals (especially for d^4-d^7 configurations, overlooking the Δ vs P comparison).
- ✕Forgetting to include the additional pairing energy term (N_p * P) in the CFSE calculation for low-spin complexes, or miscounting N_p.
📝 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