Crystal Field Theory (CFT) + Colour + Magnetism
Coordination Compounds
55
JEE Qs
15%
Hard
75
min
Master electron filling in split d-orbitals for various geometries and ligand field strengths to correctly predict CFSE, color, and magnetic properties.
🧮 Key Formulas
✅ Key Points for JEE
- 1CFT treats metal-ligand interaction as purely electrostatic, leading to d-orbital splitting, ignoring covalent character.
- 2The pattern and magnitude of d-orbital splitting (Δo, Δt, Δsp) depend on complex geometry, central metal ion, and ligand field strength (spectrochemical series).
- 3Electron distribution in split d-orbitals for d4-d7 configurations depends on the relative magnitudes of crystal field splitting energy (Δo/Δt) and pairing energy (P), dictating high spin (weak field) or low spin (strong field) complexes.
- 4The color of transition metal complexes arises from d-d electronic transitions where light of a specific wavelength is absorbed, and the observed color is its complementary color.
- 5Magnetic properties (diamagnetic/paramagnetic) and spin-only magnetic moment are directly determined by the number of unpaired electrons in the split d-orbitals, as predicted by CFT.
⚠️ Common Mistakes
- ✕Incorrectly assigning electrons to split d-orbitals for d4-d7 systems, especially when distinguishing between strong field (low spin) and weak field (high spin) ligands.
- ✕Confusing the absorbed color with the observed (complementary) color of the complex.
- ✕Calculating magnetic moment incorrectly by using the total number of d-electrons instead of the actual number of unpaired electrons after orbital splitting.
- ✕Applying octahedral splitting rules (t2g/eg) to tetrahedral (e/t2) or square planar complexes, and vice-versa.
📝 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