Coordination Compounds: JEE Advanced PYQ, 2007–2026

Answer: (C)

2.665 g is 0.010 mol of the compound, and 2.87 g is 0.020 mol of AgCl. So two chloride ions per formula unit are free, and one is bound to chromium. With a coordination number of 6, five water molecules fill the remaining places: [Cr(H₂O)₅Cl]Cl₂·H₂O.

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)

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: (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: (D)

[Co(en)₂Cl₂]⁺: trans (achiral) and cis (a pair of optical isomers), which makes 3. [Co(NH₃)₄Cl₂]⁺: cis and trans, 2. [Co(NH₃)₃Cl₃]: fac and mer, 2. [Cr(en)₃]³⁺: one pair of optical isomers, 2. [Pt(NH₃)₂Cl₂]: cis and trans, 2. [Pt(NH₃)(py)ClBr]: either py, Cl or Br can lie trans to NH₃, which makes 3. The total is 3 + 2 + 2 + 2 + 2 + 3 = 14.

Answer: (C)

Linkage isomerism arises when an ambidentate ligand can coordinate through different donor atoms. NO₂⁻ gives M–NO₂ (nitro) or M–ONO (nitrito). SCN⁻ gives M–SCN (thiocyanato) or M–NCS (isothiocyanato). Each ligand's mode is chosen independently, so the isomers are (NO₂, SCN), (NO₂, NCS), (ONO, SCN) and (ONO, NCS). That is 2 × 2 = 4 modes of attachment. Cis/trans arrangement is a separate type of isomerism and is excluded here.

Answer: (A)

Low spin, t₂g⁶: −2.4Δₒ with 3 pairs. High spin, t₂g⁴e_g²: −0.4Δₒ with 1 pair. The low-spin arrangement gains 2.0Δₒ = 40 000 cm⁻¹ and pays for two extra pairs, 2P = 30 000 cm⁻¹: it is lower by 10 000 cm⁻¹.

Answer: (A)

Only anions outside the square bracket (primary valence satisfied) are free in solution and precipitate with the appropriate reagent. Anions inside the bracket satisfy the secondary valence and stay bound. For [Co(NH₃)₅SO₄]Br, SO₄²⁻ is bound, so BaCl₂ gives no BaSO₄. Br⁻ is free, so AgNO₃ gives pale yellow AgBr.