Coordination Compounds — JEE Main Questions

50 JEE Main practice questions on Coordination Compounds, part of Chemistry. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. Which of the following complexes has the largest crystal field splitting Δ?

  1. [Fe(H₂O)₆]²⁺
  2. [Fe(H₂O)₆]³⁺
  3. [Fe(CN)₆]³⁻
  4. [Fe(CN)₆]⁴⁻

Answer: [Fe(CN)₆]⁴⁻

Crystal field splitting Δ increases with higher oxidation state and stronger ligand field. CN⁻ is a strong-field ligand, while H₂O is weak. Fe²⁺ in [Fe(CN)₆]⁴⁻ has d⁶ configuration, and with strong CN⁻, Δ is large. Fe³⁺ in [Fe(CN)₆]³⁻ has d⁵, but the lower negative charge on the complex reduces Δ due to less metal-ligand orbital overlap. Thus [Fe(CN)₆]⁴⁻ has the largest Δ.

2. Which of the following pairs of compounds are ionisation isomers?

  1. [Pt(NH3)4Cl2]Br2 and [Pt(NH3)4Br2]Cl2
  2. [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2
  3. [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]
  4. [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4

Answer: [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4

Ionisation isomerism occurs when the counter ion and a ligand exchange places, giving different ions in solution. In option (d), [Co(NH3)5SO4]Br gives SO4²⁻ as ligand and Br⁻ as counter ion, while [Co(NH3)5Br]SO4 gives Br⁻ as ligand and SO4²⁻ as counter ion. Both have same molecular formula but produce different ions, confirming ionisation isomerism.

3. In metal carbonyls, the C-O stretching frequency decreases compared to free CO. This is due to:

  1. σ-donation from CO to metal only.
  2. π-back-donation from metal to CO only.
  3. synergic σ-donation and π-back-donation.
  4. increase in C-O bond order.

Answer: synergic σ-donation and π-back-donation.

In metal carbonyls, CO donates σ electrons to the metal, and the metal donates π electrons into the empty π* orbital of CO. This π-back-donation populates the antibonding orbital, weakening the C-O bond and lowering its stretching frequency. The effect is synergic: σ-donation strengthens the M-C bond, and π-back-donation weakens the C-O bond.

4. Which of the following is NOT a limitation of Crystal Field Theory?

  1. It ignores covalent character in metal-ligand bonds.
  2. It predicts the correct magnetic moment for high-spin octahedral complexes.
  3. It cannot account for charge-transfer spectra.
  4. It cannot explain why CO is a strong-field ligand.

Answer: It predicts the correct magnetic moment for high-spin octahedral complexes.

CFT can predict magnetic moments for high-spin complexes by counting unpaired electrons based on splitting patterns. This is a success, not a limitation. The other options are well-known limitations: CFT treats bonds as purely electrostatic, fails to explain CO's strong-field nature, and cannot explain charge-transfer spectra.

5. Which of the following ligands produces the smallest crystal field splitting Δ in an octahedral complex?

  1. CN⁻
  2. H₂O
  3. I⁻
  4. NH₃

Answer: I⁻

According to the spectrochemical series, ligands are ordered by the magnitude of Δ they produce. I⁻ is a weak-field ligand at the extreme left of the series, giving the smallest Δ. CN⁻ and CO are the strongest, NH₃ is moderately strong, and H₂O is intermediate. Thus, I⁻ produces the smallest splitting.

6. For the complex [Cr(NH3)6]³⁺, what is the spin-only magnetic moment in Bohr magnetons?

  1. 3.87
  2. 4.90
  3. 2.83
  4. 5.92

Answer: 3.87

Cr in [Cr(NH3)6]³⁺ is in +3 oxidation state with d³ configuration. NH3 is a moderate field ligand, but for d³, there is no high-spin/low-spin choice; all three electrons occupy t₂g orbitals, giving 3 unpaired electrons. Spin-only magnetic moment μ = √(n(n+2)) = √(3×5) = √15 ≈ 3.87 BM.

7. Which of the following complexes is coloured due to charge transfer and not due to d-d transition?

  1. KMnO4
  2. [Sc(H2O)6]3+
  3. [Cu(NH3)2]+
  4. [Zn(H2O)6]2+

Answer: KMnO4

KMnO4 contains Mn in +7 oxidation state (d0). No d-d transition is possible, yet it is deep purple. The colour arises from ligand-to-metal charge transfer (LMCT) from oxygen to Mn. Options a, b, c are colourless because they are d10 or d0 complexes where d-d transitions are absent.

8. Which of the following best describes an ambidentate ligand?

  1. A ligand that has only one donor atom and forms a single coordinate bond.
  2. A ligand that binds through two donor atoms simultaneously to form a ring.
  3. A ligand that can bind through two different donor atoms, but only one at a time.
  4. A ligand that can bind through the same donor atom in two different ways.

Answer: A ligand that can bind through two different donor atoms, but only one at a time.

An ambidentate ligand has two different donor atoms but can use only one at a time to form a coordinate bond. Examples include NO2- (nitro vs nitrito) and SCN- (thiocyanato vs isothiocyanato). This is distinct from chelating ligands which use both donor atoms simultaneously.

9. Which complex is used as an anticancer drug and why is the trans-isomer inactive?

  1. [Pt(NH3)4]Cl2; it has higher stability
  2. trans-[Pt(NH3)2Cl2]; cis cannot bind to DNA
  3. cis-[Pt(NH3)2Cl2]; trans cannot bind to adjacent DNA bases
  4. K[PtCl3(NH3)]; it is more soluble

Answer: cis-[Pt(NH3)2Cl2]; trans cannot bind to adjacent DNA bases

Cisplatin (cis-[Pt(NH3)2Cl2]) is used as an anticancer drug because its two Cl ligands are on the same side, allowing it to bind to adjacent guanine bases on DNA, preventing replication. The trans-isomer has Cl atoms opposite, so it cannot bind effectively and is inactive.

10. In the complex [Co(NH3)5Cl]Cl2, what is the coordination number of cobalt?

  1. 5
  2. 3
  3. 4
  4. 6

Answer: 6

The coordination number is the number of donor atoms directly bonded to the metal. In [Co(NH3)5Cl]^2+, there are five NH3 ligands (each donating through N) and one Cl- ligand (donating through Cl), giving a total of 6 donor atoms. Hence coordination number is 6.

11. A complex absorbs light at 500 nm and appears purple. Which of the following statements is correct?

  1. The complex absorbs yellow light and has Δ ≈ 210 kJ/mol.
  2. The complex absorbs blue-green light and has Δ ≈ 240 kJ/mol.
  3. The complex absorbs red light and has Δ ≈ 180 kJ/mol.
  4. The complex absorbs blue light and has Δ ≈ 240 kJ/mol.

Answer: The complex absorbs blue-green light and has Δ ≈ 240 kJ/mol.

Purple is complementary to yellow-green, but 500 nm corresponds to blue-green light. Using E = hc/λ, Δ = (6.626×10⁻³⁴ J·s × 3×10⁸ m/s) / (500×10⁻⁹ m) = 3.98×10⁻¹⁹ J = 239 kJ/mol ≈ 240 kJ/mol. Option b correctly states absorbs blue-green light and Δ ≈ 240 kJ/mol.

12. A yellow cobalt complex CoCl3·4NH3 gives 1 mole of AgCl with AgNO3 and has molar conductivity corresponding to 2 ions. What is its structural formula?

  1. [Co(NH3)4Cl]Cl2
  2. [Co(NH3)4Cl2]Cl
  3. [Co(NH3)4]Cl3
  4. [Co(NH3)4Cl3]

Answer: [Co(NH3)4Cl2]Cl

Only Cl⁻ outside the coordination sphere are precipitated by AgNO3. Since 1 Cl⁻ is precipitated, 2 Cl⁻ are inside the sphere. Conductivity shows 2 ions total, so the complex ion is [Co(NH3)4Cl2]⁺ and the counter ion is Cl⁻. Thus the formula is [Co(NH3)4Cl2]Cl.

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