Coordination Chemistry Practice Questions (with Worked Answers)

Transition-metal complexes pack a lot of bookkeeping into one formula. Before you can say anything about colour, magnetism or shape, you need the oxidation state of the metal, its d-electron count and its coordination number, and an error in any of those spreads through the rest of the answer. This set of fourteen questions trains that bookkeeping and then uses it: naming, high- and low-spin configurations, spin-only magnetic moments, and the three kinds of stereoisomerism that come up most often. Marks are shown in brackets. The answer key shows every step.

Tools you will need

• Oxidation state of the metal: overall charge of the complex minus the sum of the ligand charges. Neutral ligands: H₂O, NH₃, CO, en (ethane-1,2-diamine). Anionic: Cl⁻, F⁻, OH⁻, CN⁻, C₂O₄²⁻ (oxalate).
• d-electron count: group number of the metal minus its oxidation state (for the first row: Ti 4, V 5, Cr 6, Mn 7, Fe 8, Co 9, Ni 10, Cu 11, Zn 12).
• Coordination number: the number of donor atoms bonded to the metal, not the number of ligands. Each en counts as two.
• Spin-only magnetic moment: μ = √(n(n + 2)) BM, where n is the number of unpaired electrons.
• IUPAC 2005 ligand names: aqua (H₂O), ammine (NH₃), chlorido (Cl⁻), fluorido (F⁻), cyanido (CN⁻), hydroxido (OH⁻), carbonyl (CO). Ligands are listed alphabetically before the metal, ignoring multiplying prefixes; an anionic complex takes an “-ate” ending.

The theory behind all of these is in naming coordination compounds, crystal field theory and isomerism in complexes.

Questions

Question 1 (4 marks) Give the oxidation state of the metal and its coordination number in each of: (a) K₃[Fe(CN)₆] (b) [Co(NH₃)₅Cl]Cl₂ (c) [Pt(NH₃)₂Cl₂] (d) [Cr(en)₂(H₂O)₂]³⁺

Question 2 (4 marks) Give the IUPAC name of each compound: (a) [Co(NH₃)₆]Cl₃ (b) K₄[Fe(CN)₆] (c) [CrCl₂(H₂O)₄]Cl (d) Na₂[CoCl₄]

Question 3 (3 marks) Write the formula of: (a) the diamminesilver(I) ion (b) tetraamminedichloridocobalt(III) chloride (c) potassium hexafluoridocobaltate(III)

Question 4 (4 marks) State the number of d electrons on the metal in: (a) [Ti(H₂O)₆]³⁺ (b) [Cr(H₂O)₆]³⁺ (c) [Ni(NH₃)₆]²⁺ (d) [Cu(H₂O)₆]²⁺

Question 5 (3 marks) Explain why [Ti(H₂O)₆]³⁺ is coloured but [Zn(H₂O)₆]²⁺ is colourless.

Question 6 (5 marks) [Fe(H₂O)₆]²⁺ is high spin and [Fe(CN)₆]⁴⁻ is low spin. (a) Give the d-electron configuration (t₂g and e_g occupancy) for each. (2) (b) Calculate the spin-only magnetic moment of each. (2) (c) Explain, in terms of the ligands, why the two complexes differ. (1)

Question 7 (4 marks) Compare [FeF₆]³⁻ (high spin) with [Fe(CN)₆]³⁻ (low spin): give the number of unpaired electrons and the spin-only magnetic moment of each.

Question 8 (2 marks) A chromium complex has a measured magnetic moment of 3.87 BM. How many unpaired electrons does it have, and what is the most likely oxidation state of chromium in an octahedral complex?

Question 9 (4 marks) [Ni(CN)₄]²⁻ is diamagnetic, while [NiCl₄]²⁻ is paramagnetic with two unpaired electrons. Suggest the shape of each complex and explain the difference in magnetism.

Question 10 (3 marks) Draw (or describe) the geometric isomers of [Co(NH₃)₄Cl₂]⁺. State whether either is chiral.

Question 11 (3 marks) [Co(NH₃)₃Cl₃] exists as two isomers. Name the type of isomerism, identify the two forms and describe how they differ.

Question 12 (4 marks) (a) Explain why [Co(en)₃]³⁺ shows optical isomerism. (2) (b) Which of cis-[Co(en)₂Cl₂]⁺ and trans-[Co(en)₂Cl₂]⁺ is chiral? Explain. (2)

Question 13 (3 marks) Square-planar [Pt(NH₃)₂Cl₂] has two isomers, but a tetrahedral complex of formula [MA₂B₂] has only one. Explain.

Question 14 (5 marks) (a) Write an equation for the reaction of [Cu(H₂O)₆]²⁺ with excess concentrated hydrochloric acid, and state the change in coordination number and shape. (3) (b) The equilibrium [Ni(NH₃)₆]²⁺ + 3en ⇌ [Ni(en)₃]²⁺ + 6NH₃ lies far to the right. Explain why, using entropy. (2)

Total: 50 marks

Answer key

Question 1 (a) CN⁻ × 6 = −6; complex charge is −3 (balanced by 3K⁺). Fe = −3 − (−6) = +3. CN 6. (1) (b) The complex ion is [Co(NH₃)₅Cl]²⁺ (two Cl⁻ outside the bracket). Co + 0 + (−1) = +2, so Co = +3. CN 6. (1) (c) Neutral: Pt + 0 + 2(−1) = 0, so Pt = +2. CN 4. (1) (d) en and H₂O are neutral, so Cr = +3. Two bidentate en give 4 donor atoms plus 2 waters: CN 6. (1)

Question 2 (a) Hexaamminecobalt(III) chloride. (1) (b) Potassium hexacyanidoferrate(II). Four K⁺ balance a 4− complex; 6 CN⁻ give −6, so Fe is +2. The anion takes “ferrate”. (1) (c) Tetraaquadichloridochromium(III) chloride. Alphabetical order: aqua before chlorido. Charge: Cr + 4(0) + 2(−1) = +1, so Cr = +3. (1) (d) Sodium tetrachloridocobaltate(II). Complex charge −2, 4 Cl⁻ give −4, so Co = +2. (1)

Older textbooks write “chloro” and “cyano”. Many exam boards still accept them, but the current IUPAC names are chlorido and cyanido.

Question 3 (a) [Ag(NH₃)₂]⁺ (1) (b) [Co(NH₃)₄Cl₂]Cl (IUPAC ordering gives [CoCl₂(NH₃)₄]Cl; both are accepted.) Co(III) + 2Cl⁻ gives a 1+ ion, so one Cl⁻ counter-ion. (1) (c) K₃[CoF₆]. Co³⁺ + 6F⁻ = 3−, balanced by 3K⁺. (1)

Question 4 (a) Ti group 4, +3 → d¹ (1) (b) Cr group 6, +3 → d³ (1) (c) Ni group 10, +2 → d⁸ (1) (d) Cu group 11, +2 → d⁹ (1)

Remember that transition-metal ions lose their 4s electrons first, so Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s². See transition metal electron configurations.

Question 5 (3 marks)

• In an octahedral complex the ligands split the d orbitals into a lower t₂g set and a higher e_g set (1).
• Ti³⁺ is d¹: the electron can absorb a photon of visible light and be promoted from t₂g to e_g. The complementary colour is transmitted, so the solution looks violet (1).
• Zn²⁺ is d¹⁰: both sets are full, so no d–d transition is possible and no visible light is absorbed (1).

Question 6 (a) Both are Fe²⁺, d⁶. High spin [Fe(H₂O)₆]²⁺: t₂g⁴ e_g² (1). Low spin [Fe(CN)₆]⁴⁻: t₂g⁶ e_g⁰ (1). (b) High spin: 4 unpaired, μ = √(4 × 6) = √24 = 4.90 BM (1). Low spin: 0 unpaired, μ = 0, so the complex is diamagnetic (1). (c) CN⁻ is a strong-field ligand that produces a large splitting Δₒ, bigger than the pairing energy, so electrons pair in t₂g. H₂O is weaker, Δₒ is smaller than the pairing energy, and electrons spread out into e_g (1). See high spin vs low spin.

Question 7 (4 marks) Both are Fe³⁺, d⁵.

• [FeF₆]³⁻, high spin, t₂g³ e_g²: 5 unpaired (1); μ = √(5 × 7) = √35 = 5.92 BM (1).
• [Fe(CN)₆]³⁻, low spin, t₂g⁵: 1 unpaired (1); μ = √(1 × 3) = √3 = 1.73 BM (1).

Question 8 (2 marks) Solve n(n + 2) = 3.87² ≈ 15.0. n = 3 fits (3 × 5 = 15). Three unpaired electrons (1). Chromium with three d electrons is Cr(III), d³, which gives t₂g³ in any octahedral field (1).

Question 9 (4 marks)

• Ni²⁺ is d⁸ in both complexes (1).
• [Ni(CN)₄]²⁻ is square planar. The strong-field CN⁻ ligands create a large splitting, and the eight electrons pair up in the four lower orbitals, leaving the highest (d_x²−y²) empty: no unpaired electrons, so diamagnetic (1).
• [NiCl₄]²⁻ is tetrahedral (1). Chloride is a weak-field ligand and the tetrahedral splitting is small, so d⁸ is e⁴ t₂⁴ with two unpaired electrons in the upper set (1).

Question 10 (3 marks)

• cis isomer: the two Cl ligands are adjacent, at 90° to each other (1).
• trans isomer: the two Cl ligands are opposite each other, at 180° (1).
• Neither is chiral. Both have mirror planes, because NH₃ and Cl are monodentate and the complex has no “twist” (1).

Question 11 (3 marks) Geometric (cis–trans type) isomerism in an octahedral [MA₃B₃] complex (1).

• fac (facial): the three Cl ligands occupy one triangular face of the octahedron, all mutually at 90° (1).
• mer (meridional): the three Cl ligands lie in a plane that contains the metal, like a line of longitude around the octahedron; two of them are trans (180°) to each other (1).

Question 12 (a) The three bidentate en ligands wrap around the metal like the blades of a propeller, twisting either clockwise or anticlockwise (1). The two forms (labelled Δ and Λ) are non-superimposable mirror images, and the ion has no plane of symmetry, so it is chiral and shows optical isomerism (1). (b) cis is chiral (1). In the trans isomer the two en ligands and the metal lie in one plane, which is a mirror plane, so its mirror image is identical. In the cis isomer the en ligands are twisted relative to each other and there is no mirror plane (1).

Question 13 (3 marks)

• In a square plane, two positions can be adjacent (90°) or opposite (180°), so the two A ligands can be cis or trans: two isomers (1). The cis isomer of [Pt(NH₃)₂Cl₂] is the anticancer drug cisplatin (1).
• In a tetrahedron every position is adjacent to every other (all angles 109.5°), so any arrangement of two A and two B is the same compound: only one form (1).

Question 14 (a) [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O (1) Coordination number falls from 6 to 4 (1). The shape changes from octahedral to tetrahedral (in practice a slightly flattened tetrahedron), because Cl⁻ ions are larger than water molecules and fewer fit round the metal. The colour changes from blue to yellow-green (1). (b) The number of free particles increases from 4 on the left (1 complex + 3 en) to 7 on the right (1 complex + 6 NH₃), so ΔS is positive (1). The Ni–N bonds broken and made are similar, so ΔH is close to zero and ΔG = ΔH − TΔS is negative. This is the chelate effect (1). See chelation explained.

Where students usually drop marks

• Counting ligands instead of donor atoms. [Co(en)₃]³⁺ has three ligands but a coordination number of 6.
• Forgetting the counter-ions. In [Co(NH₃)₅Cl]Cl₂, only one chloride is a ligand, and the charge of the complex ion is 2+, not 0.
• Using the 4s electrons. Transition-metal ions have no 4s electrons; count d electrons from the group number.
• Stating a moment without the working. Always write n first, then μ = √(n(n + 2)), then the value with BM.
• Calling fac/mer “optical”. fac and mer are geometric isomers. Optical isomers need a non-superimposable mirror image.
• Getting the alphabetical order wrong. Ignore “di”, “tri”, “tetra” when sorting: tetraaqua comes before dichlorido.

Key takeaways

• Oxidation state = complex charge minus ligand charges; d count = group number minus oxidation state.
• Coordination number counts donor atoms, so a bidentate ligand counts twice.
• Strong-field ligands (CN⁻, CO) favour low spin; weak-field ligands (F⁻, Cl⁻, H₂O) usually give high spin for first-row ions.
• μ(spin-only) = √(n(n + 2)) BM: 1.73, 2.83, 3.87, 4.90 and 5.92 BM for 1 to 5 unpaired electrons.
• Octahedral MA₄B₂ gives cis/trans, MA₃B₃ gives fac/mer, and [M(en)₃] gives a pair of optical isomers.