CHEM 361 / Test Bank / Coordination Chemistry
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Test Bank — Unit 3

Coordination Chemistry

Practice problems drawn from Housecroft, Atkins, JD Lee, and others. Select an answer or click "Show Answer" to reveal the explanation.

Werner’s Theory & Coordination Number

Questions 1–4
Housecroft & Sharpe, Chapter 7 Housecroft
Q1

Alfred Werner distinguished between primary valence and secondary valence in coordination compounds. In modern terminology, what do these two concepts correspond to?

  • a Primary valence = coordination number; secondary valence = oxidation state
  • b Primary valence = number of ligands; secondary valence = number of ions
  • c Primary valence = oxidation state; secondary valence = coordination number
  • d Primary valence = total charge; secondary valence = number of donor atoms
Answer: (c)

Werner’s primary valence (Hauptvalenz) corresponds to the modern concept of oxidation state — the number of ionic bonds or charges the metal satisfies. His secondary valence (Nebenvalenz) corresponds to the coordination number — the number of ligands directly bonded to the metal center. For example, in [Co(NH3)6]Cl3, the primary valence is 3+ and the secondary valence (coordination number) is 6.

Housecroft & Sharpe, Fig. 7.2 Housecroft
Common coordination geometries
Fig. 7.2 — Common coordination geometries in transition metal complexes
Q2

What are the most common coordination numbers and their associated geometries in transition metal complexes?

  • a 2 = linear; 4 = tetrahedral only; 6 = trigonal prismatic
  • b 4 = tetrahedral or square planar; 6 = octahedral
  • c 4 = square planar only; 6 = octahedral or trigonal prismatic equally
  • d 3 = trigonal planar; 5 = square pyramidal; 8 = cubic
Answer: (b)

The most frequently encountered coordination numbers are 4 and 6. Coordination number 4 gives rise to either tetrahedral or square planar geometry, while coordination number 6 overwhelmingly adopts octahedral geometry. Tetrahedral is favoured by small, highly charged metals with large ligands, while square planar is characteristic of d8 metal ions (e.g. Pt2+, Pd2+, Ni2+ with strong-field ligands).

Housecroft & Sharpe, Chapter 20 Housecroft
Q3

Both [Ni(CN)4]2− and [NiCl4]2− are four-coordinate Ni(II) complexes. Predict the geometry of each.

  • a Both are tetrahedral
  • b Both are square planar
  • c [Ni(CN)4]2− is tetrahedral; [NiCl4]2− is square planar
  • d [Ni(CN)4]2− is square planar; [NiCl4]2− is tetrahedral
Answer: (d)

Ni(II) is a d8 ion. CN is a strong-field ligand that forces electron pairing, producing a diamagnetic, square planar complex with a large crystal field splitting. Cl is a weak-field ligand that cannot force pairing, so [NiCl4]2− adopts the tetrahedral geometry with two unpaired electrons (paramagnetic). The ligand field strength determines which geometry a d8 complex adopts.

Atkins & Shriver, Chapter 7 Atkins
Q4

Square planar geometry is particularly common for d8 metal complexes (e.g. Pt2+, Pd2+, Au3+). What factors favour square planar over tetrahedral geometry in these systems?

  • a Strong-field ligands, large crystal field splitting, and the energy gain from fully emptying the dx²−y² orbital
  • b Weak-field ligands and high-spin configurations
  • c Small metal ions with low oxidation states
  • d Bulky ligands that force the geometry to flatten
Answer: (a)

In a d8 system, square planar geometry creates a large energy gap between the dx2−y2 orbital (strongly antibonding) and the other four d orbitals. With strong-field ligands, this gap is large enough that all eight electrons occupy the four lower-energy orbitals, leaving dx2−y2 empty. This produces a diamagnetic, 16-electron complex with substantial crystal field stabilisation energy (CFSE). Second- and third-row metals (Pd, Pt, Au) especially favour this geometry because their larger 4d/5d orbitals produce inherently larger splittings.

Ligands & Chelation

Questions 5–8
Housecroft & Sharpe, Fig. 7.4 Housecroft
Examples of mono-, bi-, and polydentate ligands
Fig. 7.4 — Denticity of selected ligands
Q5

Ligands are classified by how many donor atoms they use to bind a metal centre. Which of the following correctly defines monodentate, bidentate, and polydentate ligands with appropriate examples?

  • a Monodentate = two donors (e.g. en); bidentate = one donor (e.g. NH3); polydentate = three or more (e.g. EDTA)
  • b Monodentate = one donor (e.g. NH3, Cl); bidentate = two donors (e.g. ethylenediamine); polydentate = three or more donors (e.g. EDTA)
  • c Monodentate = one donor (e.g. EDTA); bidentate = two donors (e.g. CO); polydentate = any bridging ligand
  • d Monodentate = binds one metal; bidentate = bridges two metals; polydentate = bridges three metals
Answer: (b)

Denticity refers to the number of donor atoms a single ligand uses to bind one metal centre. Monodentate ligands (NH3, Cl, H2O) donate through one atom. Bidentate ligands like ethylenediamine (en) use two nitrogen donors, forming a chelate ring. Polydentate ligands have three or more donor atoms — EDTA, for instance, is hexadentate with six donors. This classification is distinct from bridging, which describes a ligand connecting two or more metal centres.

Housecroft & Sharpe, Chapter 7 Housecroft
Q6

The chelate effect describes the enhanced stability of complexes containing chelating (multidentate) ligands compared to analogous complexes with monodentate ligands. What is the primary thermodynamic origin of this effect?

  • a Chelating ligands form stronger individual M–L bonds
  • b Chelating ligands have a more favourable enthalpy of binding
  • c Replacement of monodentate ligands by chelates releases more free molecules, increasing entropy (ΔS > 0)
  • d Chelating ligands are always stronger σ-donors than monodentate analogues
Answer: (c)

The chelate effect is primarily entropic in origin. When a bidentate ligand like ethylenediamine (en) replaces two monodentate NH3 molecules, the reaction [M(NH3)2]n+ + en → [M(en)]n+ + 2 NH3 goes from 2 particles on the left to 3 on the right. This net increase in the number of free molecules produces a favourable entropy change (ΔS > 0), making ΔG more negative. The enthalpy change is typically similar since the M–N bond strengths are comparable, so entropy is the dominant driving force.

Housecroft & Sharpe, Fig. 7.5 Housecroft
Structure of EDTA and its metal complex
Fig. 7.5 — EDTA and its coordination to a metal centre
Q7

EDTA (ethylenediaminetetraacetic acid) is described as a hexadentate ligand. How many donor atoms does it possess, and how does it coordinate to a metal centre?

  • a Six donors: two nitrogen atoms and four carboxylate oxygen atoms, wrapping around the metal to occupy all six sites in an octahedral geometry
  • b Four donors: four carboxylate oxygen atoms only, leaving two sites for water
  • c Six donors: six oxygen atoms from the four carboxylate groups
  • d Five donors: two nitrogen atoms and three carboxylate oxygen atoms
Answer: (a)

EDTA4− coordinates through six donor atoms: two amine nitrogen atoms from the ethylenediamine backbone and four oxygen atoms, one from each of the four carboxylate groups. This allows a single EDTA molecule to occupy all six coordination sites around an octahedral metal centre, forming five chelate rings. The resulting 1:1 metal-EDTA complex is extremely stable, which is why EDTA is widely used in analytical chemistry and as a sequestering agent for metal ions.

JD Lee, Chapter 7 JD Lee
Q8

The thiocyanate ion (SCN) is an ambidentate ligand — it can bind through either sulfur (forming M–SCN, thiocyanato) or nitrogen (forming M–NCS, isothiocyanato). How does HSAB theory predict which end binds?

  • a Hard metals always bind through S; soft metals through N
  • b The binding mode depends only on the oxidation state, not the metal identity
  • c Both ends bind equally to all metals since SCN is symmetric
  • d Soft metals (e.g. Pt2+, Pd2+) bind through the soft S atom; hard metals (e.g. Cr3+, Co3+) bind through the hard N atom
Answer: (d)

HSAB (Hard-Soft Acid-Base) theory predicts that hard acids prefer hard bases and soft acids prefer soft bases. In SCN, the nitrogen end is a harder donor (small, electronegative, low polarisability), while the sulfur end is a softer donor (large, polarisable, lower electronegativity). Therefore, hard metal centres like Cr3+ and Co3+ preferentially bind through nitrogen (isothiocyanato, M–NCS), while soft metals like Pt2+ and Pd2+ bind through sulfur (thiocyanato, M–SCN).

Stability Constants

Questions 9–12
Housecroft & Sharpe, Chapter 7 Housecroft
Q9

What is the difference between a stepwise stability constant (Kn) and an overall stability constant (βn) for a metal complex?

  • a Kn is for the full reaction; βn is for a single ligand addition step
  • b Kn is the equilibrium constant for adding the nth ligand; βn = K1 × K2 × … × Kn
  • c Kn and βn are the same quantity with different symbols
  • d βn = K1 + K2 + … + Kn
Answer: (b)

The stepwise stability constant Kn describes the equilibrium for adding a single ligand to a complex that already has (n−1) ligands: [MLn−1] + L → [MLn]. The overall stability constant βn describes the complete formation from the aqua ion: M + nL → [MLn]. These are related by βn = K1 × K2 × … × Kn, which is the product (not sum) of all stepwise constants. Larger values of βn indicate greater thermodynamic stability.

Atkins & Shriver, Chapter 7 Atkins
Q10

Predict which complex is more stable: [Cu(NH3)4]2+ or [Cu(en)2]2+ (where en = ethylenediamine). Explain your reasoning.

  • a [Cu(NH3)4]2+ is more stable because NH3 is a stronger σ-donor
  • b They are equally stable since both have four Cu–N bonds
  • c [Cu(en)2]2+ is more stable due to the chelate effect — the favourable entropy of displacing monodentate ligands
  • d [Cu(NH3)4]2+ is more stable because it has a higher coordination number
Answer: (c)

Both complexes contain four Cu–N bonds with similar individual bond strengths, so the enthalpy contributions are comparable. However, the reaction replacing four NH3 with two en molecules releases two extra free particles into solution (4 NH3 released, 2 en consumed). This net increase in disorder produces a favourable entropy change. Experimentally, log β for [Cu(en)2]2+ is significantly larger than for [Cu(NH3)4]2+, directly demonstrating the chelate effect.

Housecroft & Sharpe, Fig. 7.7 Housecroft
Irving-Williams series plot
Fig. 7.7 — The Irving–Williams series of stability constants
Q11

The Irving–Williams series shows the trend in stability constants for first-row divalent metal complexes: Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+ > Zn2+. Why does Cu2+ sit at the maximum rather than Zn2+?

  • a Cu2+ (d9) benefits from Jahn–Teller distortion and large CFSE, while Zn2+ (d10) has zero CFSE
  • b Cu2+ has a higher oxidation state than Zn2+
  • c Zn2+ is too large to form strong bonds
  • d Cu2+ forms more covalent bonds due to its higher electronegativity
Answer: (a)

The Irving–Williams series reflects the combined effects of decreasing ionic radius across the period (increasing electrostatic attraction) and crystal field stabilisation energy (CFSE). From Mn2+ to Cu2+, CFSE generally increases. Cu2+ (d9) additionally benefits from a Jahn–Teller distortion that further stabilises the complex. Zn2+ (d10) has a completely filled d shell with zero CFSE, causing a sharp drop in stability despite its small ionic radius. This explains the characteristic “tent-shaped” plot.

Atkins & Shriver, Chapter 7 Atkins
Q12

What is the macrocyclic effect, and how does it compare to the chelate effect?

  • a The macrocyclic effect is weaker than the chelate effect because cyclic ligands are more rigid
  • b The macrocyclic effect is simply another name for the chelate effect
  • c The macrocyclic effect refers to the instability of macrocyclic complexes due to strain
  • d Macrocyclic ligands form even more stable complexes than open-chain chelates; the effect has both enthalpic (preorganisation) and entropic contributions
Answer: (d)

The macrocyclic effect describes the observation that macrocyclic ligands (e.g. porphyrins, crown ethers, cyclam) form significantly more stable complexes than analogous open-chain polydentate ligands. This extra stabilisation has both enthalpic and entropic origins. Enthalpically, the macrocycle is pre-organised into a shape complementary to the metal ion, reducing the reorganisation energy upon binding. Entropically, the cyclic ligand is more rigid and loses less conformational freedom upon coordination. The macrocyclic effect thus goes beyond the chelate effect, which itself already exceeds monodentate ligand binding.

Metal–Ligand Bonding Overview

Questions 13–15
Housecroft & Sharpe, Fig. 20.1 Housecroft
Orientation of d orbitals in an octahedral field
Fig. 20.1 — The five d orbitals in an octahedral ligand field
Q13

In a σ-only octahedral complex, the five d orbitals split into two sets. Which d orbitals point directly at the ligands and are therefore raised in energy?

  • a dxy, dxz, dyz (the t2g set)
  • b dz2 and dx2−y2 (the eg set)
  • c Only dz2, because it has electron density along the z axis
  • d All five d orbitals are raised equally by the ligand field
Answer: (b)

In an octahedral field, the six ligands lie along the x, y, and z axes. The dz2 and dx2−y2 orbitals (collectively called the eg set) have their lobes pointing directly at the ligands, resulting in strong σ-antibonding interactions that raise their energy. The dxy, dxz, and dyz orbitals (the t2g set) point between the ligands and experience less repulsion, remaining at lower energy. The energy gap between these two sets is Δoct.

Housecroft & Sharpe, Fig. 20.7 Housecroft
Spectrochemical series
Fig. 20.7 — The spectrochemical series of ligands
Q14

The spectrochemical series ranks ligands by the magnitude of crystal field splitting (Δ) they produce. Rank the following ligands from weakest to strongest field: CO, CN, H2O, NH3, Cl, I.

  • a CO < CN < NH3 < H2O < Cl < I
  • b I < Cl < NH3 < H2O < CN < CO
  • c I < Cl < H2O < NH3 < CN < CO
  • d Cl < I < H2O < NH3 < CO < CN
Answer: (c)

The standard spectrochemical series from weak field to strong field includes: I < Br < Cl < F < H2O < NH3 < en < NO2 < CN < CO. Halides are weak-field π-donors, water and ammonia are intermediate σ-donors, and CN and CO are strong-field ligands that act as π-acceptors. The ability to accept electron density into empty π* orbitals is what makes CO and CN produce the largest Δoct values.

Housecroft & Sharpe, Fig. 20.6 Housecroft
Pi-back-bonding in metal carbonyls
Fig. 20.6 — σ-donation and π-back-bonding in metal–CO bonding
Q15

CO is one of the strongest field ligands in the spectrochemical series. What bonding features make CO such a strong-field ligand?

  • a CO is both a σ-donor (through a lone pair on carbon) and a π-acceptor (accepting metal d-electron density into its empty π* orbitals), which synergistically increases Δoct
  • b CO has a very high electronegativity, making it a strong σ-donor only
  • c CO is a pure π-donor ligand that destabilises the eg orbitals
  • d CO binds through its oxygen atom, which has strong σ-donor ability
Answer: (a)

CO binds to metals through carbon using a lone pair in a σ-donation step (C → M). Simultaneously, filled metal d orbitals (t2g in an octahedral complex) donate electron density back into the empty π* antibonding orbitals of CO — this is π-back-bonding (M → CO). These two processes are synergistic: σ-donation increases electron density on the metal, which enhances π-back-donation, and vice versa. The π-back-bonding stabilises the t2g set while the eg set remains destabilised by σ-antibonding, resulting in a very large Δoct.

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Coordination Chemistry