Coordination Compounds: JEE Main PYQ, 2020–2025

Answer: (A)

Crystal field theory treats the metal ion as a positive centre and the ligands as negative point charges or dipoles. The d orbitals lying closer to the approaching ligands are destabilised by electrostatic repulsion, and this produces the splitting. There is no covalent bonding or orbital overlap in the model. Valence bond theory and molecular orbital theory use the opposite assumption, so they are not the correct description here.

Answer: (A)

Describe each isomer by which ligand pairs are trans to each other. (1) All three pairs trans. (2), (3), (4) One pair trans and the other two pairs cis, with the trans pair being NH₃, Cl or Br in turn. (5) All three pairs cis. That gives 5 geometrical isomers. Only the all-cis isomer lacks a mirror plane, so it exists as two enantiomers and the total stereoisomer count is 6. The question asks for geometrical isomers, so the answer is 5.

Answer: (A)

Rule: for an anionic complex, the cation is named first, with no multiplicative prefix, and the metal name ends in -ate. Some metals use Latin stems, so copper becomes cuprate. The oxidation state is found from charge balance: three K⁺ give a complex ion of 3−, so x + 4(−1) = −3, giving x = +1. The name is potassium tetracyanidocuprate(I).

Answer: (A)

Crystal field theory says the five d orbitals split into t2g (3 orbitals) and eg (2 orbitals) with the average energy unchanged. If t2g drops by x, then 3x + 2y = 0 for the total shift to cancel, and the gap is Δo = y - x. This gives x = -0.4Δo and y = +0.6Δo. Remember that t2g is lower, since its orbitals point between the ligands, and eg is higher, since its orbitals point at them.

Answer: (A)

Ligands inside the coordination sphere (square brackets) do not dissociate in solution. Counter-ions outside the brackets do. In [Cr(H₂O)₅Cl]Cl₂·H₂O one Cl is coordinated and two Cl⁻ are free. Each free Cl⁻ gives AgCl with Ag⁺, so 1 mol of the compound gives 2 mol AgCl. The lattice water has no effect on this.

Answer: (A)

Step 1: en spans two cis positions. Of the four remaining positions, two lie trans to the en nitrogens and two are mutually trans to each other (perpendicular to the plane of en). Step 2: Place the two Cl⁻ ligands. (i) Both Cl⁻ in the mutually trans pair gives trans-Cl₂ and is achiral. (ii) Both Cl⁻ in the positions trans to the en nitrogens leaves the NH₃ ligands mutually trans and is achiral. (iii) One Cl⁻ trans to an en nitrogen and the other in the mutually trans pair has no mirror plane, so it is chiral and exists as d and l forms. Step 3: This gives 3 geometrical isomers, one of them chiral, so the total is 1 + 1 + 2 = 4 stereoisomers.

Answer: (B)

With x for nickel and 0 for each CO: x + 4 × 0 = 0, so x = 0.

Answer: (A)

The ion outside the bracket is Cl⁻, so the complex is the cation [Pt(NH₃)₂Cl(NH₂CH₃)]⁺, and the metal keeps the plain name platinum. Alphabetical order ignores multiplying prefixes: ammine, chlorido, methylamine. The organic ligand name is placed in parentheses. Oxidation state: x + 0 + (−1) + 0 = +1, so x = +2. The name is diamminechlorido(methylamine)platinum(II) chloride.

Answer: (A)

Werner's principle is that ions inside the coordination sphere do not give their ordinary tests, and the number of ions in solution is found from colligative properties. The absence of tests for Co³⁺ and NO₂⁻ shows that all six nitrite ions (3 from Co(NO₂)₃ and 3 from 3KNO₂) are bound to Co, giving a coordination number of 6. The complex ion is [Co(NO₂)₆]³⁻, which is balanced by 3 K⁺, so the compound is K₃[Co(NO₂)₆] and gives 1 + 3 = 4 ions (i = 4). ΔTf = i·Kf·m = 4 × 1.86 × 0.020 ≈ 0.149 K.

Answer: (A)

Principle: oxidation state equals the metal charge that balances the overall charge given neutral and charged ligands. Carbonyl (CO) is a neutral ligand, so it contributes zero charge. For [Fe(CO)₅] with overall charge 0: x + 5(0) = 0, so Fe is 0. Naming: five CO ligands give 'penta' with the ligand name 'carbonyl', then the metal 'iron'. Since the complex is neutral, use the plain metal name, giving pentacarbonyliron(0).