Isomerism โ Geometric, optical, ionization, linkage
Coordination Compounds
15
JEE Qs
8%
Hard
75
min
Practice drawing structures for each type of isomerism for various common complex formulas to develop a systematic approach to identify and count all possible isomers.
๐งฎ Key Formulas
โ Key Points for JEE
- 1Systematically draw all possible structural arrangements for a given complex before determining isomer types. Rotation of the entire molecule or specific ligands should not be counted as new isomers.
- 2For geometric isomerism, focus on the relative positions of identical ligands. Square planar (e.g., MA2B2) and octahedral (e.g., MA2B4, MA3B3) complexes are key; tetrahedral complexes generally do not show geometric isomerism.
- 3For optical isomerism, determine if a complex is chiral by checking for the presence of a plane of symmetry or a center of symmetry. If either is present, the complex is generally achiral. Cis isomers of bidentate chelates (e.g., cis-[M(AA)2B2]) are often optically active.
- 4Identify ambidentate ligands (e.g., NO2-, SCN-, CN-) as a prerequisite for linkage isomerism. Different bonding modes lead to distinct isomers with different properties.
- 5Ionization isomerism requires interchangeable ligands and counter-ions. Check if exchanging an inner-sphere ligand with an outer-sphere counter-ion results in a different complex formula and different ions in solution.
โ ๏ธ Common Mistakes
- โFailing to identify a plane of symmetry or a center of symmetry, leading to incorrect assessment of optical activity.
- โConfusing geometric isomers with optical isomers, or double-counting structures that are identical through rotation.
- โMissing potential isomers by not considering all possible ligand arrangements or not recognizing ambidentate ligands.
- โIncorrectly applying the rules for different coordination numbers and geometries (e.g., assuming tetrahedral complexes show geometric isomerism).
๐ 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]2โ (B) [Fe(CN)6]3โ (C) [Fe(CN)5NO]2โ (D) [CoCl4]2โ (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, ฮo 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]2โ (paramagnetic) and [Ni(CO)4] (diamagnetic), choose the correct geometry and oxidation state. (1) [NiCl4]2โ: NiII , tetrahedral [Ni(CO)4] : NiII , (2) [NiCl4]2โ : NiII , square planar [Ni(CO)4] : Ni(0) square planar , square planar (3) [NiCl4]2โ: NiII , tetrahedral [Ni(CO)4] : Ni(0), (4) [NiCl4]2โ: 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