CHEM 361 / Test Bank / Crystal & Ligand Field Theory
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Test Bank — Unit 3

Crystal Field & Ligand Field Theory

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

Crystal Field Splitting

Questions 1–4
Housecroft & Sharpe, Fig. 20.2 Housecroft
Crystal field splitting in an octahedral field
Fig. 20.2 — Crystal field splitting of d orbitals in an octahedral field
Q1

In an octahedral crystal field, the five d orbitals split into two sets. Which orbitals are raised in energy (the eg set) and which are lowered (the t2g set)?

  • a eg: dxy, dxz, dyz;   t2g: d, dx²−y²
  • b eg: dxy, dx²−y²;   t2g: dxz, dyz, d
  • c eg: d, dx²−y²;   t2g: dxy, dxz, dyz
  • d eg: d, dxy;   t2g: dxz, dyz, dx²−y²
Answer: (c)

In an octahedral field, the six ligands lie along the x, y, and z axes. The d and dx²−y² orbitals point directly at the ligands, so they experience greater electrostatic repulsion and are raised in energy (the eg set). The dxy, dxz, and dyz orbitals point between the ligand axes, experience less repulsion, and are stabilized (the t2g set). The energy gap between these two sets is Δoct.

Housecroft & Sharpe, Fig. 20.4 Housecroft
Factors affecting the magnitude of crystal field splitting
Fig. 20.4 — Factors influencing Δoct
Q2

The crystal field splitting parameter Δoct measures the energy separation between the t2g and eg sets. Which combination of factors leads to the largest Δoct?

  • a Low metal oxidation state, first-row metal, weak-field ligand
  • b High metal oxidation state, first-row metal, strong-field ligand
  • c Low metal oxidation state, third-row metal, weak-field ligand
  • d High metal oxidation state, third-row metal, strong-field ligand
Answer: (d)

Three factors increase Δoct: (1) higher metal oxidation state increases the effective nuclear charge, pulling ligands closer; (2) moving down a group (first → second → third row) increases the spatial extent of d orbitals, enhancing metal–ligand overlap; (3) stronger-field ligands (e.g. CN, CO) interact more strongly. Combining all three — high charge, third-row metal, strong-field ligand — gives the largest Δoct.

Housecroft & Sharpe, Fig. 20.7 Housecroft
Comparison of octahedral and tetrahedral crystal field splitting
Fig. 20.7 — Crystal field splitting in octahedral versus tetrahedral geometries
Q3

In a tetrahedral field, Δtet49 Δoct. Which statement best explains why Δtet is smaller?

  • a Tetrahedral complexes have six ligands instead of four, diluting the field
  • b Only four ligands contribute to the field, and none point directly at any d orbital
  • c The d orbitals in tetrahedral geometry are all degenerate
  • d Tetrahedral ligands are always weak-field
Answer: (b)

Two factors make Δtet smaller. First, there are only four ligands instead of six, reducing the total electrostatic interaction. Second, in a tetrahedron, no ligand points directly along the d orbital axes — the interaction is indirect. Combined, these effects give Δtet49 Δoct for the same metal–ligand combination. The splitting pattern is also inverted: the t2 set is raised and the e set is lowered.

Housecroft & Sharpe, Fig. 20.3 Housecroft
Crystal field stabilization energy diagram
Fig. 20.3 — Crystal field stabilization energy (CFSE)
Q4

Calculate the crystal field stabilization energy (CFSE) in terms of Δoct for a d6 ion in (i) high-spin and (ii) low-spin octahedral environments. Which answer is correct?

  • a High-spin: −0.4 Δoct;   Low-spin: −2.4 Δoct + 2P
  • b High-spin: −2.4 Δoct;   Low-spin: −0.4 Δoct + 2P
  • c High-spin: −1.2 Δoct;   Low-spin: −2.4 Δoct
  • d High-spin: 0;   Low-spin: −2.4 Δoct + 3P
Answer: (a)

For d6 high-spin: t2g4eg2. CFSE = 4(−0.4Δ) + 2(+0.6Δ) = −0.4Δoct. For d6 low-spin: t2g6eg0. CFSE = 6(−0.4Δ) = −2.4Δoct, but three extra electron pairings cost 2P (there is already one forced pairing in high-spin, giving a net cost of 2P additional). The low-spin configuration is favored when Δoct is large enough to offset the pairing energy penalty.

High-Spin vs Low-Spin

Questions 5–8
Housecroft & Sharpe, Fig. 20.8 Housecroft
High-spin vs low-spin electron configurations
Fig. 20.8 — High-spin and low-spin d-electron configurations
Q5

Whether an octahedral complex adopts a high-spin or low-spin configuration depends on the competition between two energy quantities. Which are they?

  • a The ionization energy and the electron affinity of the metal
  • b The crystal field splitting energy Δoct and the electron pairing energy P
  • c The bond energy and the lattice energy of the complex
  • d The electronegativity of the metal and the ligand
Answer: (b)

When Δoct < P (pairing energy), electrons prefer to occupy the higher-energy eg orbitals rather than pair up in t2g, giving a high-spin configuration. When Δoct > P, the energy cost of pairing is less than the cost of promoting an electron to eg, so electrons fill t2g first — a low-spin configuration. This competition applies to d4–d7 octahedral complexes.

Housecroft & Sharpe, Fig. 20.10 Housecroft
Spectrochemical series
Fig. 20.10 — The spectrochemical series of ligands
Q6

[CoF6]3− is high-spin but [Co(CN)6]3− is low-spin. Both contain Co3+ (d6). Using the spectrochemical series, explain this difference.

  • a Co3+ has a different oxidation state in the two complexes
  • b CN is a weaker field ligand than F
  • c F is a weak-field ligand (Δoct < P), while CN is a strong-field ligand (Δoct > P)
  • d The fluoride complex is tetrahedral while the cyanide complex is octahedral
Answer: (c)

The spectrochemical series ranks ligands by their ability to split d orbitals: I < Br < Cl < F < H2O < NH3 < en < NO2 < CN < CO. F is near the weak-field end, producing a small Δoct that cannot overcome the pairing energy P — hence high-spin (4 unpaired electrons). CN is near the strong-field end, generating a large Δoct > P, so all six d electrons pair in t2g — low-spin (0 unpaired electrons).

Atkins & Shriver, Chapter 20 Atkins
Q7

A d5 metal ion is placed in an octahedral field. How many unpaired electrons does it have in the high-spin and low-spin configurations, respectively?

  • a High-spin: 3;   Low-spin: 0
  • b High-spin: 5;   Low-spin: 3
  • c High-spin: 4;   Low-spin: 2
  • d High-spin: 5;   Low-spin: 1
Answer: (d)

In high-spin d5: all five d orbitals are singly occupied (t2g3eg2), giving 5 unpaired electrons and maximum spin multiplicity. In low-spin d5: electrons fill t2g first, giving t2g5eg0. Two of the three t2g orbitals hold paired electrons and one holds a single electron, yielding 1 unpaired electron. This dramatic difference (5 vs. 1) is directly detectable by magnetic susceptibility measurements.

JD Lee, Chapter 7 JD Lee
Q8

Tetrahedral complexes are almost always high-spin. Why is a low-spin tetrahedral configuration essentially never observed?

  • a Δtet49 Δoct, which is almost always smaller than the pairing energy P
  • b Tetrahedral complexes have no d-orbital splitting at all
  • c All tetrahedral ligands are weak-field by definition
  • d Tetrahedral geometry only forms with d0 or d10 metals
Answer: (a)

The tetrahedral crystal field splitting Δtet is inherently only about 4/9 the magnitude of Δoct for the same metal and ligand. This reduced splitting is almost never large enough to exceed the electron pairing energy P. Since Δtet < P in virtually all cases, electrons occupy the higher e set before pairing in the lower t2 set, producing a high-spin configuration. Low-spin tetrahedral complexes would require an unusually large Δtet that is rarely achievable.

Jahn-Teller Effect & Distortions

Questions 9–12
Housecroft & Sharpe, Fig. 20.11 Housecroft
Jahn-Teller distortion of an octahedral complex
Fig. 20.11 — Jahn-Teller distortion in octahedral d-metal complexes
Q9

The Jahn-Teller theorem states that any non-linear molecule with an orbitally degenerate ground state will undergo a geometric distortion. Which octahedral d-electron configurations show the strongest Jahn-Teller distortions?

  • a d3 and d8 (half-filled and filled t2g)
  • b d4 (high-spin) and d9 (unequal occupation of eg)
  • c d5 high-spin and d10 (spherically symmetric)
  • d d1 and d6 low-spin only
Answer: (b)

Strong Jahn-Teller distortions occur when the eg orbitals are unevenly occupied, because these orbitals point directly at the ligands and unequal occupation creates a large driving force for distortion. High-spin d4 (t2g3eg1) and d9 (t2g6eg3) both have one eg orbital singly occupied and the other empty or doubly occupied. Low-spin d7 also shows a strong effect. Uneven t2g occupation produces weaker distortions.

Housecroft & Sharpe, Fig. 20.12 Housecroft
Tetragonal elongation in Cu2+ complexes
Fig. 20.12 — Tetragonal distortion in Cu2+ complexes
Q10

Cu2+ complexes (d9) typically display tetragonal elongation, where the two axial bonds are longer than the four equatorial bonds. Use the Jahn-Teller effect to explain this observation.

  • a Cu2+ is too small for regular octahedral coordination
  • b The axial ligands are always different from the equatorial ligands
  • c The d9 configuration has an unevenly filled eg set (d2 dx²−y²1 or vice versa), and distortion lifts the degeneracy to lower the total energy
  • d Crystal packing forces compress the equatorial plane
Answer: (c)

Cu2+ is d9 with configuration t2g6eg3. The three electrons in the doubly degenerate eg set cannot occupy both orbitals equally. In a tetragonal elongation, the axial ligands move away, lowering d relative to dx²−y². The extra electron occupies the now-lower d, removing the degeneracy and lowering the total energy. This is a textbook example of a strong Jahn-Teller distortion.

Atkins & Shriver, Chapter 20 Atkins
From octahedral to square planar via tetragonal distortion
Fig. 20.13 — Progression from octahedral to square planar geometry through tetragonal elongation
Q11

Square planar geometry can be viewed as the limiting case of tetragonal distortion of an octahedral complex. What happens in this extreme limit?

  • a All six ligands move into the equatorial plane
  • b The four equatorial ligands are removed, leaving two axial ligands
  • c The t2g and eg sets merge back into a single degenerate set
  • d The two axial ligands are removed entirely, leaving four ligands in the equatorial plane with a further splitting of d orbitals
Answer: (d)

Starting from a regular octahedron, tetragonal elongation progressively lengthens the two axial bonds. In the extreme limit, the axial ligands are effectively removed, leaving four ligands in the xy plane — a square planar geometry. This causes extensive further splitting of the d orbitals: dx²−y² becomes highest in energy (pointing directly at the four remaining ligands), while d drops significantly. The resulting d orbital order from low to high energy is approximately dxz,dyz < d < dxy < dx²−y².

Housecroft & Sharpe, Fig. 20.14 Housecroft
Square planar complexes of Pd2+ and Pt2+
Fig. 20.14 — Square planar d8 complexes
Q12

Most Pd2+ and Pt2+ complexes adopt square planar geometry rather than tetrahedral. Why is this the case?

  • a They are d8 ions with large Δ (second/third-row, strong field), making it energetically favorable to leave the high-energy dx²−y² orbital empty
  • b Pd and Pt are too large for tetrahedral coordination
  • c Square planar geometry is always preferred for d8 regardless of the metal
  • d The ligands around Pd2+ and Pt2+ are always bidentate chelates
Answer: (a)

Pd2+ and Pt2+ are d8 ions from the second and third transition rows, which means their d orbitals are spatially extended and produce very large crystal field splittings. In square planar geometry, the dx²−y² orbital is pushed so high in energy that all eight d electrons fill the four lower orbitals, leaving dx²−y² empty. This is strongly stabilized when Δ is large. First-row d8 ions like Ni2+ can be tetrahedral or square planar depending on the ligand field strength.

Ligand Field Theory

Questions 13–15
Housecroft & Sharpe, Fig. 20.15 Housecroft
MO diagram for octahedral complex in ligand field theory
Fig. 20.15 — MO diagram of an octahedral complex (σ-bonding only)
Q13

Ligand field theory (LFT) improves on crystal field theory (CFT) in several ways. What is the key conceptual advance of LFT over CFT?

  • a LFT treats the metal as a point charge in a field of electron-donor ligands
  • b LFT uses a molecular orbital approach, treating metal–ligand bonds as covalent interactions with both σ and π components
  • c LFT only applies to tetrahedral complexes, while CFT covers octahedral
  • d LFT ignores the d orbitals and focuses on ligand-centered orbitals
Answer: (b)

Crystal field theory treats ligands as point charges or dipoles — a purely electrostatic model that ignores covalency entirely. Ligand field theory (LFT) instead constructs molecular orbitals from metal d orbitals and ligand σ/π orbitals, capturing covalent bonding, π back-donation, and the true electronic structure of the complex. LFT explains the spectrochemical series, magnetic properties, and spectroscopic data that CFT cannot fully account for.

Housecroft & Sharpe, Fig. 20.16 Housecroft
Effect of pi-bonding on crystal field splitting
Fig. 20.16 — Effect of π-donor and π-acceptor ligands on Δoct
Q14

In ligand field theory, σ-donor ligands alone produce a baseline Δoct. How do π-acceptor ligands (such as CO and CN) increase Δoct beyond this baseline?

  • a They donate additional electron density into the eg orbitals, pushing them higher
  • b They increase the effective nuclear charge of the metal
  • c Their empty π* orbitals accept electron density from the metal t2g orbitals via back-bonding, lowering the t2g energy and increasing Δoct
  • d They physically compress the octahedron, shortening all bond lengths
Answer: (c)

π-Acceptor ligands like CO and CN possess low-energy empty π* orbitals that interact with the filled metal t2g orbitals. This π back-bonding delocalizes electron density from metal to ligand, stabilizing (lowering) the t2g set. Since the eg level is unaffected by this interaction, the gap Δoct increases. This is why π-acceptor ligands appear at the strong-field end of the spectrochemical series, and conversely, π-donor ligands (like halides) raise t2g and reduce Δoct.

Atkins & Shriver, Chapter 20 Atkins
MO diagram showing pi-backbonding in CO complex
Fig. 20.17 — π-Back-bonding interaction in a metal–carbonyl complex
Q15

Using ligand field theory, explain why CO produces a much larger Δoct than H2O when coordinated to the same metal ion.

  • a CO has a larger dipole moment than H2O
  • b H2O is a π-acceptor, while CO is only a σ-donor
  • c CO is smaller and can approach the metal more closely
  • d CO is a strong σ-donor and a strong π-acceptor (back-bonding into π*), which both raises eg and lowers t2g; H2O is a weak σ-donor and a weak π-donor, which raises t2g
Answer: (d)

CO acts as both a strong σ-donor through its lone pair on carbon and a strong π-acceptor through its low-lying π* orbitals. The σ-donation pushes the eg set up, while the π-acceptance stabilizes the t2g set through back-bonding — both effects increase Δoct. In contrast, H2O is a modest σ-donor with no significant π-acceptor capability; its lone pairs act as weak π-donors that slightly raise t2g, reducing Δoct. This dual σ/π character is why CO sits at the very top of the spectrochemical series.

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Crystal Field & Ligand Field Theory