Crystal Field Theory: Complete Notes for NEET, JEE, IIT-JAM, BITSAT, GATE, CSIR-NET, TGT and PGT
A concept-first walkthrough of octahedral splitting, CFSE, spectrochemical series, weak/strong fields, tetrahedral and square planar geometry, and the Jahn-Teller effect — with solved examples and 40 practice MCQs.
Table of Contents
- Historical Background and Basic Postulates
- Shapes and Orientation of the Five d Orbitals
- Crystal Field Splitting in an Octahedral Field
- The Barycentre Rule and the t2g–eg Energy Gap
- Crystal Field Stabilization Energy (CFSE): Definition and Calculation
- Absorption Spectrum and the Origin of Colour
- Weak Field vs Strong Field: High-Spin and Low-Spin Complexes
- Pairing Energy and the Complete CFSE Table (d1–d10)
- Factors Governing the Magnitude of Δo
- The Spectrochemical Series
- Colour, Absorption and the Colour Wheel
- Crystal Field Splitting in a Tetrahedral Field
- Crystal Field Splitting in a Square Planar Field
- Comparing Octahedral, Tetrahedral and Square Planar Splitting
- Applications of CFT: Ionic Radii, Lattice Energy and Hydration Enthalpy
- Applications of CFT: Spinel Structures
- The Jahn-Teller Effect
- Magnetic Moment and Its Use in Determining Electronic Configuration
- Solved Examples and In-Text Questions (from the source unit)
- Practice MCQs (40 Questions with Answers)
1. Historical Background and Basic Postulates
Crystal Field Theory did not begin its life as a theory of coordination compounds at all. Hans Bethe proposed it in 1929 while trying to account for the colour and magnetic behaviour of crystalline metal salts, and the model rested entirely on electrostatics — no covalent bonding was assumed at that stage. John Van Vleck revisited the idea in 1935 and allowed some covalent character into the picture, which brought the theory closer to physical reality without discarding its electrostatic core. Even today, chemists apply the same framework to isolated complex ions, not just to solids, because the essential requirement is only that the ligands sit in a regular geometric arrangement around the metal.
The theory rests on two assumptions, and it is worth being precise about both because exam questions often probe exactly this point.
- Ligands are treated as point negative charges. A ligand's actual size, shape, or orbital character is ignored entirely — what matters is only the charge (or, for a neutral ligand like NH₃ or H₂O, the negative end of its dipole, arising from the lone pair on the donor atom).
- Only electrostatic interaction is considered, and this splits into two components: attraction between the metal's positive nucleus and the ligand's negative charge, and repulsion between the metal's valence-shell electrons and the ligand electrons.
Given these assumptions, CFT asks a single question: how do these point charges affect the energy levels of the metal ion's s, p, d and f orbitals? The s orbital is spherically symmetric, so a spherical field of charges raises its energy uniformly in every direction — no splitting occurs. The same logic applies to a completely filled p subshell in a transition metal ion. The d orbitals are a different story entirely, and that difference is the entire subject of this unit.
2. Shapes and Orientation of the Five d Orbitals
Before splitting can make sense, the geometry of the five d orbitals needs to be fixed firmly in mind. They fall naturally into two groups based on where their lobes point relative to the Cartesian axes.
| Orbital | Lobe orientation |
|---|---|
| dxy | Lobes lie between the x and y axes |
| dxz | Lobes lie between the x and z axes |
| dyz | Lobes lie between the y and z axes |
| dx²−y² | Lobes lie directly along the x and y axes |
| dz² | Two lobes along the z axis, plus a donut-shaped ring around the middle in the xy plane |
So the dxy, dxz and dyz orbitals point between the axes, while dx²−y² and dz² point along the axes. This single geometric fact is the reason the entire splitting pattern in an octahedral field looks the way it does. Worth noting: the dz² orbital is mathematically a combination of dz²−x² and dz²−y², sometimes written as d2z²−x²−y² — this shows up occasionally in derivations but rarely needs to be memorised beyond recognising the notation if it appears.
In an isolated, free metal ion, all five d orbitals are degenerate — same energy, no preference. A perfectly spherical external field keeps them degenerate too, since every orbital experiences the same repulsion regardless of direction. It is only when the field becomes directional, as it does the moment six ligands approach along specific axes, that the degeneracy breaks.
3. Crystal Field Splitting in an Octahedral Field
Picture an octahedral complex ML6: six ligands sit at the corners of a regular octahedron, four of them in one plane and the remaining two above and below it, all approaching along the x, y and z axes. Orbitals whose lobes point directly along these axes will feel a stronger repulsive push from the incoming ligand charges than orbitals whose lobes point between the axes.
That single observation splits the five d orbitals into two sets:
- eg set — dx²−y² and dz², lying along the axes, pushed to higher energy.
- t2g set — dxy, dxz and dyz, lying between the axes, pushed to lower energy relative to the eg set.
The subscript g (gerade) reflects the fact that an octahedral field is centrosymmetric — it has a centre of inversion, and both orbital sets retain that symmetry label. This labelling convention comes up constantly in group theory-flavoured exam questions, so it is worth being comfortable stating it rather than just recognising it.
4. The Barycentre Rule and the t2g–eg Energy Gap
Here's a subtlety that trips up a lot of students: simply redistributing a spherical charge into an octahedral arrangement cannot change the total energy of the system. So if the eg orbitals rise in energy, the t2g orbitals must fall by a compensating amount — the "centre of gravity," or barycentre, of the five orbitals stays fixed.
Because there are three t2g orbitals and only two eg orbitals, conservation of the barycentre forces the energy changes into a 6:4 ratio (referenced to the average energy in a hypothetical spherical field):
Each of the two eg orbitals is raised by 0.6Δo → total rise = 1.2Δo
The two exactly cancel, which is the mathematical statement of barycentre conservation. The separation between the t2g and eg levels is called Δo (delta-octahedral), also historically written as 10Dq — Δo is now the preferred notation since it specifies the geometry unambiguously. Typical Δo values fall in the range of 100–250 kJ mol−1, and they're determined experimentally from the electronic (UV-visible) spectrum of the complex, not calculated from first principles.
5. Crystal Field Stabilization Energy (CFSE): Definition and Calculation
Once the splitting pattern is fixed, CFSE tells you how much energy a complex "gains" by having its electrons occupy the lower t2g level rather than being spread uniformly in a hypothetical spherical field. Take the simplest case: [Ti(H2O)6]3+, a d1 system. In the ground state, that single electron occupies one of the three t2g orbitals, giving a configuration written t2g1eg0. Since each electron in t2g lowers the energy by 0.4Δo, the CFSE here is simply −0.4Δo.
Extend this to two electrons in t2g: CFSE = 2 × (−0.4Δo) = −0.8Δo. The pattern generalises cleanly — for any configuration, multiply the number of t2g electrons by −0.4Δo and the number of eg electrons by +0.6Δo, then add.
Worked example — three d electrons (d3): With three electrons in t2g and none in eg (t2g3eg0), CFSE = 3 × (−0.4Δo) = −1.2Δo. This is exactly the configuration Cr3+ or V2+ complexes adopt, and it is unambiguous because only one arrangement is possible for d1, d2, d3, d8, d9 and d10 systems — pairing considerations only become relevant from d4 onward.
6. Absorption Spectrum and the Origin of Colour
Apart from the d10 configuration, every other d-electron count leaves the eg set incompletely filled, meaning an electron can absorb a photon and jump from t2g to eg provided the photon's energy matches Δo exactly. This condition is written hν = Δo, and because Δo for most transition metal complexes falls in the visible region of the electromagnetic spectrum, this single transition is the root cause of the colours transition metal complexes display.
[Ti(H2O)6]3+ is the textbook case because it has only one d electron, so there's no ambiguity about which transition is occurring: t2g1eg0 → t2g0eg1. Experimentally, this absorption occurs at roughly 20,400 cm−1, giving Δo = 20,400 cm−1 directly, and hence CFSE = −(2/5) × 20,400 ≈ −8,160 cm−1, which works out to about −98 kJ mol−1.
If a complex absorbs the red portion of white light, the remaining blue light passes through (or is reflected), and the compound appears blue to the eye — the colour observed is always complementary to the colour absorbed, never the same colour.
7. Weak Field vs Strong Field: High-Spin and Low-Spin Complexes
The real decision-making begins at d4. The fourth electron has two options: pair up inside a t2g orbital (paying a "pairing energy" cost, P) or move into an empty eg orbital (paying a destabilization cost of Δo). Whichever option costs less energy wins.
If Δo < P → electron enters eg (stays unpaired) → configuration t2g3eg1 → high-spin complex
Low-spin complexes have fewer unpaired electrons and arise from strong-field ligands, which is why "low-spin" and "strong-field complex" are used almost interchangeably. High-spin complexes, with more unpaired electrons, come from weak-field ligands. When P and Δo happen to be equal, the two states exist in equilibrium — a genuinely borderline situation that shows up occasionally in spin-crossover chemistry.
Pairing energy P is an intrinsic property of the metal ion, obtained from its atomic spectrum, while Δo depends on the metal, the ligand, and the geometry, obtained from the complex's molecular spectrum. Once both are known for a given complex, the spin state and CFSE follow immediately.
| dn | Ion | Pairing energy P (cm−1) | Complex | Δo (cm−1) |
|---|---|---|---|---|
| d4 | Cr2+ | 20,425 | [Cr(H2O)6]2+ | 13,900 |
| Mn3+ | 25,215 | [Mn(H2O)6]3+ | 21,000 | |
| d5 | Mn2+ | 23,825 | [Mn(H2O)6]2+ | 7,800 |
| Fe3+ | 29,875 | [Fe(H2O)6]3+ | 13,700 | |
| d6 | Fe2+ | 19,150 | [Fe(H2O)6]2+ | 10,400 |
| [Fe(CN)6]4− | 33,000 | |||
| Co3+ | 23,625 | [Co(H2O)6]3+ | 18,200 | |
| [Co(NH3)6]3+ | 22,900 | |||
| [Co(CN)6]3− | 33,500 | |||
| d7 | Co2+ | 20,800 | [Co(H2O)6]2+ | 9,300 |
Notice [Fe(H2O)6]3+: P = 29,875 cm−1 while Δo is only 13,700 cm−1. Since P >> Δo, this complex is unambiguously high-spin, with all five d electrons unpaired — a result worth internalising because it recurs constantly across NEET and JEE question banks.
8. Pairing Energy and the Complete CFSE Table (d1–d10)
Putting weak-field and strong-field behaviour together across every possible d-electron count gives the master table that most exam papers eventually draw from in one form or another.
| dn | Weak field (high spin) | Strong field (low spin) | ||||
|---|---|---|---|---|---|---|
| Configuration | Unpaired e− | CFSE | Configuration | Unpaired e− | CFSE | |
| d1 | t2g1, eg0 | 1 | −0.4Δo | t2g1, eg0 | 1 | −0.4Δo |
| d2 | t2g2, eg0 | 2 | −0.8Δo | t2g2, eg0 | 2 | −0.8Δo |
| d3 | t2g3, eg0 | 3 | −1.2Δo | t2g3, eg0 | 3 | −1.2Δo |
| d4 | t2g3, eg1 | 4 | −0.6Δo | t2g4, eg0 | 2 | −1.6Δo + P |
| d5 | t2g3, eg2 | 5 | 0.0Δo | t2g5, eg0 | 1 | −2.0Δo + 2P |
| d6 | t2g4, eg2 | 4 | −0.4Δo | t2g6, eg0 | 0 | −2.4Δo + 2P |
| d7 | t2g5, eg2 | 3 | −0.8Δo | t2g6, eg1 | 1 | −1.8Δo + P |
| d8 | t2g6, eg2 | 2 | −1.2Δo | t2g6, eg2 | 2 | −1.2Δo |
| d9 | t2g6, eg3 | 1 | −0.6Δo | t2g6, eg3 | 1 | −0.6Δo |
| d10 | t2g6, eg4 | 0 | 0.0Δo | t2g6, eg4 | 0 | 0.0Δo |
Two things are worth flagging about this table. First, d1, d2, d3, d8, d9 and d10 have only one possible configuration — weak-field and strong-field columns are identical, so spin ambiguity simply doesn't arise for them. Second, the pairing-energy penalty (+P, +2P) only enters the strong-field CFSE from d4 onward, and it accounts for the extra electron-pairing forced by strong ligands beyond whatever pairing already exists in the free ion.
9. Factors Governing the Magnitude of Δo
Four trends describe how Δo responds to changes in the metal ion, and each one has shown up as a standalone exam question at some point.
- Oxidation state of the metal: for a fixed ligand, Δo rises sharply with increasing metal oxidation state. [Fe(H2O)6]2+ has a smaller Δo than [Fe(H2O)6]3+, and similarly [Co(NH3)6]2+ (Δ ≈ 10,200 cm−1) sits well below [Co(NH3)6]3+ (Δ ≈ 22,870 cm−1). A higher charge pulls ligands in closer and intensifies the electrostatic field they exert.
- Position within a transition series: moving from 3d to 4d to 5d metals (same ligand, same geometry), Δo increases by roughly 30–50% at each step. The hexaammine series illustrates this cleanly: [Co(NH3)6]3+ ≈ 22,900–23,000 cm−1, [Rh(NH3)6]3+ ≈ 34,000–34,100 cm−1, [Ir(NH3)6]3+ ≈ 41,000–41,200 cm−1. Larger d orbitals in heavier metals extend further outward and interact more strongly with ligand charge.
- Geometry of the complex: tetrahedral complexes ML4 show a substantially smaller splitting than octahedral ML6 complexes of the same metal and ligand — this is covered quantitatively in Section 12.
- Comparable Δo along a transition series for a fixed ligand and oxidation state: for a given ligand, Δo doesn't vary hugely across the first-row metal ions in the same oxidation state; hexaaqua Mn2+ and Ni2+ complexes, for instance, have broadly comparable Δo.
Putting oxidation-state and metal-identity effects together, the source unit gives this composite ordering of increasing Δo:
10. The Spectrochemical Series
While the four factors above describe how the metal ion affects Δo, ligand identity is just as decisive — arguably more so for predicting spin state. Arranging common ligands by increasing field strength gives the spectrochemical series, named for the fact that the ordering was determined from spectral (absorption) data rather than derived theoretically:
As field strength increases along this series, the frequency of light absorbed by the complex increases too — which is exactly why complexes of the same metal ion, same oxidation state, and same geometry can display completely different colours depending purely on which ligand is attached. Four octahedral Co3+ complexes make the point vividly:
| Complex | Δo (cm−1) | Colour |
|---|---|---|
| [Co(C2O4)3]3− | 18,000 | Dark green |
| [Co(H2O)6]3+ | 18,200 | Blue |
| [Co(NH3)6]3+ | 22,900 | Golden-brown |
| [Co(CN)6]3− | 33,500 | Yellow |
The series also has predictive value for spin state. If a metal ion is known to give a low-spin complex with NH3, it is guaranteed to give a low-spin complex with CN− too, since CN− sits further along the series and produces a stronger field. Likewise, a low-spin complex with ethylenediamine (en) guarantees low-spin behaviour with bipyridyl or cyanide, which lie further right. These predictions, however, are qualitative — the order can occasionally reverse for two ligands sitting close together in the series, so treat it as a strong guide rather than an absolute law.
11. Colour, Absorption and the Colour Wheel
The colour a compound displays is governed by which wavelengths of visible light are absorbed and which pass through or are reflected. If a sample absorbs every visible wavelength, no light reaches the eye from it, and it appears black. If it absorbs no visible light at all, it appears white or colourless. Everything interesting happens in between: a compound that absorbs red and blue light, for instance, transmits mainly green, and so appears green — this is precisely the case for aqueous Ni2+ solutions.
Complementary colour pairs (the colour absorbed vs the colour observed) can be read directly off an artist's colour wheel: the three primary colours are red, blue and yellow; combining any two primaries gives a secondary colour (red + blue = purple, blue + yellow = green), and combining all three gives white light. The transition responsible for absorption in a coordination complex — an electron moving from a lower-energy d orbital to a higher-energy one — is called a d–d transition, and its energy corresponds to Δo (or Δt for tetrahedral complexes, introduced next).
Colour depends on two variables above all: the charge on the metal ion and the identity of the ligand. A striking illustration is the series of cobalt(III) ammine complexes — [CoCl(NH3)5](NO3)2, [CoBr(NH3)5](NO3)2, [CoI(NH3)5](NO3)2, [Co(NO2)(NH3)5](NO3)2, [Co(SO4)(NH3)5]NO3 and [Co(CO3)(NH3)5]NO3 — which range visibly from pink through purple, mustard-yellow, orange, salmon and deep red-pink, purely as a function of which single ligand replaces one NH3 group.
12. Crystal Field Splitting in a Tetrahedral Field
Now consider a tetrahedral complex. A useful way to visualise the geometry is to imagine the metal ion sitting at the centre of a cube, with four ligands occupying alternating corners of that cube (the other four corners stay empty). Because this arrangement has no centre of symmetry, the gerade/ungerade (g/u) labels used for octahedral complexes simply don't apply here.
The splitting pattern flips relative to the octahedral case. The two e orbitals (dx²−y² and dz²) now point toward the centres of the cube's faces, while the three t2 orbitals (dxy, dyz, dzx) point toward the centres of the cube's edges — which places t2 orbitals closer to the direction ligands actually approach from. Since none of the d orbitals point directly at any ligand in a tetrahedral arrangement, the destabilization is generally milder than in the octahedral case, but the t2 set still ends up higher in energy than the e set, which is the reverse ordering compared to octahedral t2g/eg.
| dn | Electronic configuration | CFSE |
|---|---|---|
| d1 | e1, t20 | −0.6Δt |
| d2 | e2, t20 | −1.2Δt |
| d3 | e2, t21 | −0.8Δt |
| d4 | e2, t22 | −0.4Δt |
| d5 | e2, t23 | 0.0Δt |
| d6 | e3, t23 | −0.6Δt |
| d7 | e4, t23 | −1.2Δt |
| d8 | e4, t24 | −0.8Δt |
| d9 | e4, t25 | −0.4Δt |
| d10 | e4, t26 | 0.0Δt |
The magnitude of Δt is much smaller than Δo for two compounding reasons: only four ligands are present instead of six (roughly two-thirds the field strength), and the direction of ligand approach doesn't line up with any d orbital (which reduces the field by a further factor of about two-thirds). Multiplying these two reductions together gives the standard approximation:
Because Δt is almost always smaller than the pairing energy P, tetrahedral complexes are essentially always high-spin. Low-spin tetrahedral complexes are rare enough that examiners sometimes ask students to name an exception — Cr[N(SiMe3)2]3(NO) is a documented rare case, though this level of detail is more relevant to CSIR-NET/GATE than to NEET or JEE.
13. Crystal Field Splitting in a Square Planar Field
A square planar complex can be understood as a distorted octahedral complex in which the two ligands along the z-axis have been pulled away entirely. As those two ligands recede, every orbital with a z-component drops in energy (less repulsion along z), while orbitals in the xy-plane rise (the remaining four ligands compress more tightly into that plane). The result is a four-level splitting pattern, from lowest to highest energy: dyz, dxz (degenerate pair) → dz² → dxy → dx²−y², with dx²−y² sitting highest since it points directly at all four remaining ligands.
Square planar geometry shows up overwhelmingly for d8 metal ions combined with strong-field ligands. Ni2+ forms square planar complexes specifically with strong-field ligands like CN− (its complexes with weaker ligands remain tetrahedral or octahedral). Second- and third-row d8 metals behave differently, though — Pd2+, Pt2+, Ir+ and Au3+ adopt square planar geometry essentially regardless of ligand strength, because these heavier metals already generate an intrinsically strong field. [PdCl4]2− is a textbook example: even Cl−, a comparatively weak-field ligand, still produces square planar Pd2+.
14. Comparing Octahedral, Tetrahedral and Square Planar Splitting
Placing all three geometries side by side clarifies how the same five d orbitals respond so differently depending purely on ligand arrangement.
| Geometry | Higher-energy set | Lower-energy set | Typical spin behaviour |
|---|---|---|---|
| Octahedral | eg (dx²−y², dz²) | t2g (dxy, dxz, dyz) | Weak field Δo < P → high spin; strong field Δo > P → low spin |
| Tetrahedral | t2 (dxy, dxz, dyz) | e (dx²−y², dz²) | Almost always high spin (small Δ < Δ< P) |
| Square planar | dx²−y² (highest), then dxy | dz², then dxz/dyz (lowest) | Mostly d8, mostly low spin, strong-field ligands or heavy d8 metals |
Notice that the octahedral and tetrahedral orderings are essentially mirror images of each other — the pair that's stabilised in one geometry is the pair that's destabilised in the other. That inversion follows directly from which orbitals point toward the ligands in each geometry, and it is a fast way to sanity-check any splitting diagram you draw under exam pressure.
15. Applications of CFT: Ionic Radii, Lattice Energy and Hydration Enthalpy
CFT isn't just a bookkeeping tool for colour and magnetism — it explains real deviations from otherwise-smooth periodic trends across the first-row transition metals.
Ionic radii
For a fixed oxidation state, ionic radius is expected to decrease steadily across a transition series as increasing nuclear charge pulls electrons in more tightly. The observed curve, however, isn't smooth — it dips and rises in a way that tracks CFSE almost exactly. Under weak-field conditions, radius starts increasing again once the configuration reaches t2g³eg¹, because the newly occupied eg electron points directly at the ligands and pushes them outward. Under strong-field conditions, the analogous upturn happens later, at t2g⁶eg¹.
Lattice energy
Lattice energy is expected to rise smoothly across a transition series as ionic radii shrink (lattice energy scales roughly as 1/(r⁺ + r⁻)). Ca2+, Mn2+ and Zn2+ — corresponding to d0, d5 and d10 configurations, all with zero CFSE — sit exactly on the expected straight-line trend. Every other metal ion in between deviates above the line by an amount that tracks its CFSE: the deviation grows from d1 to d3, shrinks back down near d5, then grows again toward d8.
Hydration enthalpy
The same logic applies to hydration enthalpy for the process M2+(g) + 6H2O(l) → [M(H2O)6]2+(aq). Stronger electrostatic attraction between the ion and water dipoles should make ΔH more negative in a smooth trend across the series, but the experimental curve shows a distinctive double-humped shape rather than a straight line — again explained by the extra stabilisation CFSE provides at particular d-electron counts, mirroring the ionic radius trend almost point for point.
16. Applications of CFT: Spinel Structures
A spinel is a mixed metal oxide with general formula (A2+)(B3+)2(O2−)4, where A is a divalent metal (often a first-row transition metal) and B is a trivalent metal. Close-packed oxide ions generate a cubic lattice containing both tetrahedral and octahedral holes, and where the A and B ions end up sitting depends directly on their relative CFSE in each geometry.
- Normal spinel: A2+ ions occupy tetrahedral holes, B3+ ions occupy octahedral holes. MgAl2O4 is the reference example — both Mg2+ and Al3+ are d0, so there's no CFSE-based site preference at all, and the ions simply adopt the arrangement favoured by ionic size and charge alone.
- Inverse spinel: half the B3+ ions move into tetrahedral holes, while A2+ ions and the remaining B3+ ions occupy octahedral holes. Magnetite, Fe3O4, written as (Fe3+)tetrahedral(Fe2+, Fe3+)octahedral(O2−)4, is the classic inverse spinel. Since O2− is a weak-field ligand, both iron ions remain high-spin: Fe3+ is d5 (CFSE = 0 in either geometry, hence no site preference — it's the one that ends up in the tetrahedral hole), while Fe2+ is d6 and gains CFSE = −4 Dq by sitting in an octahedral hole.
The difference in CFSE that a given dn ion experiences between octahedral and tetrahedral coordination is called the Octahedral Site Stabilization Energy (OSSE), and it is exactly this quantity that determines normal vs inverse spinel structure. Co3O4 makes the point sharply: Co3+ is low-spin d6 in an octahedral hole, with CFSE = −24 Dq — a stabilisation so large that Co3+ strongly prefers the octahedral site over the high-spin arrangement O2− would otherwise favour, forcing Co3O4 into a normal spinel structure rather than an inverse one.
17. The Jahn-Teller Effect
The Jahn-Teller theorem states that in a nonlinear molecule, if degenerate orbitals are occupied asymmetrically, the molecule will distort to remove that degeneracy, lowering both its symmetry and its overall energy. In octahedral complexes, this becomes relevant whenever the t2g or eg set is unevenly filled — but the effect is far more pronounced when the degeneracy sits in eg, because eg orbitals point directly at the ligands, so any asymmetry there has a much larger geometric consequence than the same asymmetry in t2g (whose orbitals don't point at the ligands at all).
Configurations with eg1 or eg3 occupancy — high-spin d4, low-spin d7, and d9 — are the ones that show significant, experimentally measurable distortion. Two distortion modes are possible:
- Axial elongation (z-out): orbitals with a z-component are stabilised (lowered), while orbitals without a z-component are destabilised (raised).
- Axial compression (z-in): the reverse — orbitals without a z-component are stabilised, those with a z-component are destabilised.
The theorem predicts that distortion occurs but not which type. In practice, axial elongation is far more common than axial compression, for a straightforward structural reason: elongation only weakens two bonds (the axial ones), whereas compression effectively weakens four (the equatorial ones), making elongation the lower-energy option in most real complexes.
Worked reasoning — high-spin d4: configuration t2g3eg1. If the single eg electron occupies dz², electron density concentrates between the metal and the two ligands along z, so those two bonds experience extra repulsion and elongate. If instead the electron occupies dx²−y², elongation happens along x and y instead. The same reasoning extends to low-spin d7 (t2g6eg1). For d9 (two electrons in one eg orbital, one in the other), whichever orbital holds the pair of electrons determines the axis of elongation.
X-ray crystallography confirms these distortions directly. CuF2 (Cu2+ is d9) is the textbook case: it shows two long axial Cu–F bonds (227 pm) and four shorter, mutually equal equatorial bonds (193 pm) — exactly the asymmetric bond-length pattern the Jahn-Teller theorem predicts.
| Configuration | High-spin J.T. distortion | Low-spin J.T. distortion |
|---|---|---|
| d1, d2 | Weak | Weak |
| d3 | — | — |
| d4 | Strong | Weak |
| d5 | — | Weak |
| d6 | Weak | — |
| d7 | Weak | Strong |
| d8 | — | — |
| d9 | Strong | Strong |
| d10 | — | — |
18. Magnetic Moment and Its Use in Determining Electronic Configuration
The spin-only magnetic moment formula connects a measurable magnetic property directly back to the number of unpaired electrons, which makes it one of the most practically useful tools CFT hands you for confirming high-spin vs low-spin assignments experimentally.
Compounds are diamagnetic if repelled by a magnetic field (no unpaired electrons) and paramagnetic if attracted into one (unpaired electrons present). Comparing calculated spin-only values against experiment for several first-row ions shows the formula's reliability:
| Ion | Unpaired e− (n) | S | μ calc. (μB) | μ expt. (μB) |
|---|---|---|---|---|
| Ti3+ | 1 | ½ | 1.73 | 1.7–1.8 |
| V3+ | 2 | 1 | 2.83 | 2.7–2.9 |
| Cr3+ | 3 | 1½ | 3.87 | 3.8 |
| Mn3+ | 4 | 2 | 4.90 | 4.8–4.9 |
| Fe3+ | 5 | 2½ | 5.92 | 5.9 |
Worked example: A Co(II) complex shows a magnetic moment of 4.0 μB. Co2+ is d7, which allows two possible octahedral configurations: high-spin t2g5eg2 (S = 1½, 3 unpaired electrons) or low-spin t2g6eg1 (S = ½, 1 unpaired electron). The spin-only moments for these come out to 3.87 μB and 1.73 μB respectively. Since the observed value (4.0 μB) sits close to 3.87 and nowhere near 1.73, the complex must be high-spin, t2g5eg2.
19. Solved Examples and In-Text Questions (from the source unit)
These are the self-assessment and terminal questions carried directly from the unit, worked through fully rather than left as bare answers.
SAQ 1. Draw the dx²−y² orbital with proper axes. What can you say about the direction of its lobes?
Answer: The dx²−y² orbital has four lobes lying directly along the positive and negative x and y axes (no lobes along z). This is precisely why it belongs to the eg set in an octahedral field — its lobes point straight at incoming ligands along those axes.
SAQ 2. What is meant by the barycentre in crystal field splitting?
Answer: The barycentre is the "centre of gravity" of the d orbital energies — the average energy level that is conserved when the five degenerate orbitals split into t2g and eg sets. The concept is general and applies to any coordination geometry, not just octahedral.
SAQ 3. Calculate the CFSE of an octahedral complex with three electrons in the d orbitals.
Answer: With three electrons and only one possible configuration (t2g3eg0), CFSE = 3 × (−0.4Δo) = −1.2Δo.
SAQ 4. What would be the CFSE for an octahedral complex of a d7 ion in weak field and strong field?
Answer: Weak field (high spin): t2g5eg2, giving CFSE = −5(0.4Δo) + 2(0.6Δo) = −0.8Δo. Strong field (low spin): t2g6eg1, giving CFSE = [−6(0.4Δo) + 1(0.6Δo)] + P = −1.8Δo + P.
SAQ 5. If pairing energy P for Fe3+ is 29,875 cm−1 and Δo for [Fe(H2O)6]3+ is 13,700 cm−1, find: (i) whether the complex is high spin or low spin, (ii) the number of unpaired electrons, (iii) whether the complex is coloured.
Answer: (i) Since P >> Δo, the complex is high spin. (ii) Fe3+ (d5) in the high-spin state adopts t2g3eg2, giving 5 unpaired electrons. (iii) Yes — the eg orbitals aren't completely filled, so electrons can be excited from t2g to eg by absorbing visible light, and the complex will therefore appear coloured.
Terminal Question 1. [CoA6]3+ is red; [CoB6]3+ is green. Which ligand produces the larger Δo?
Answer: Ligand A (giving the red complex) produces the larger Δo. A red complex absorbs higher-frequency (blue/green) light, which corresponds to a larger energy gap between t2g and eg.
Terminal Question 2. Is [CoF6]4− more likely to be low-spin or high-spin?
Answer: High-spin, since F− sits low on the spectrochemical series and is not a strong-field ligand.
Terminal Question 3. Sketch d-orbital energy levels for [Fe(H2O)6]2+ and [Fe(CN)6]3− and give the number of unpaired electrons in each.
Answer: [Fe(H2O)6]2+: Fe2+ is d6, weak-field H2O gives high-spin t2g4eg2, with 4 unpaired electrons. [Fe(CN)6]3−: Fe3+ is d5, strong-field CN− gives low-spin t2g5eg0, with just 1 unpaired electron.
Terminal Question 4. Draw crystal field energy diagrams (with high-spin and low-spin configurations where applicable) for octahedral complexes of (a) Mn2+, (b) Cr2+, (c) Co2+.
Answer: (a) Mn2+ is d5: high spin t2g3eg2 (5 unpaired), low spin t2g5eg0 (1 unpaired). (b) Cr2+ is d4: high spin t2g3eg1 (4 unpaired), low spin t2g4eg0 (2 unpaired). (c) Co2+ is d7: high spin t2g5eg2 (3 unpaired), low spin t2g6eg1 (1 unpaired).
20. Practice MCQs (40 Questions with Answers)
These extend beyond the source unit and are written in the style commonly seen across NEET, JEE, BITSAT and GATE-level coordination chemistry sections. Attempt each before checking the answer.
Q1. Crystal Field Theory was first proposed by:
(a) Werner (b) Van Vleck (c) Hans Bethe (d) Pauling
Answer: (c) Hans Bethe (1929)
Q2. In CFT, ligands are treated as:
(a) Covalently bonded neutral atoms (b) Point negative charges (c) Point positive charges (d) Uncharged spheres
Answer: (b) Point negative charges
Q3. In an octahedral field, the orbitals that experience greater repulsion are:
(a) t2g (b) eg (c) Both equally (d) Neither
Answer: (b) eg
Q4. The ratio in which t2g is lowered and eg is raised (conserving the barycentre) is:
(a) 4:6 (b) 6:4 (c) 3:2 (d) 1:1
Answer: (b) 6:4 (t2g lowered by 0.4Δo each, eg raised by 0.6Δo each)
Q5. CFSE of a d3 octahedral complex is:
(a) −0.6Δo (b) −0.8Δo (c) −1.2Δo (d) −1.6Δo
Answer: (c) −1.2Δo
Q6. Which configuration corresponds to a low-spin d6 octahedral complex?
(a) t2g4eg2 (b) t2g6eg0 (c) t2g3eg2 (d) t2g5eg1
Answer: (b) t2g6eg0
Q7. [Fe(CN)6]4− is diamagnetic. This is because:
(a) Fe2+ is high spin d6 (b) Fe2+ is low spin d6 with all electrons paired in t2g (c) Fe is in +3 state (d) CN− is a weak-field ligand
Answer: (b) Fe2+ is low spin d6 with all electrons paired in t2g
Q8. Which ligand lies at the extreme weak-field end of the spectrochemical series?
(a) CN− (b) I− (c) NH3 (d) H2O
Answer: (b) I−
Q9. Which ligand lies at the extreme strong-field end of the spectrochemical series?
(a) F− (b) H2O (c) CN− (d) Cl−
Answer: (c) CN−
Q10. Δt for a tetrahedral field is approximately:
(a) (9/4)Δo (b) (4/9)Δo (c) equal to Δo (d) (1/2)Δo
Answer: (b) (4/9)Δo
Q11. Tetrahedral complexes are almost always:
(a) Low spin (b) High spin (c) Diamagnetic (d) Colourless
Answer: (b) High spin, since Δt is usually smaller than P
Q12. Square planar geometry is most commonly associated with which d-electron count?
(a) d4 (b) d5 (c) d8 (d) d10
Answer: (c) d8
Q13. [PdCl4]2− is square planar even though Cl− is a weak-field ligand because:
(a) Pd is a first-row metal (b) Second/third-row d8 metals adopt square planar geometry regardless of ligand strength (c) Cl− is actually a strong-field ligand (d) Pd2+ is d10
Answer: (b) Second/third-row d8 metals adopt square planar geometry regardless of ligand strength
Q14. The subscript "g" in t2g and eg indicates:
(a) Ground state (b) Gerade (symmetric under inversion) (c) General configuration (d) Geometry type
Answer: (b) Gerade (symmetric under inversion)
Q15. Jahn-Teller distortion is most pronounced for degeneracy in:
(a) t2g orbitals (b) eg orbitals (c) Both equally (d) Neither, since JT applies only to tetrahedral complexes
Answer: (b) eg orbitals, since they point directly at the ligands
Q16. Which configuration shows strong Jahn-Teller distortion in both high-spin and low-spin cases?
(a) d5 (b) d8 (c) d9 (d) d3
Answer: (c) d9 (eg3 configuration in both cases)
Q17. Axial elongation is more common than axial compression because:
(a) It lowers symmetry more (b) It weakens only two bonds rather than four (c) It increases Δo (d) It is thermodynamically forbidden otherwise
Answer: (b) It weakens only two bonds rather than four
Q18. CuF2 shows Jahn-Teller distortion because Cu2+ has the configuration:
(a) d7 (b) d8 (c) d9 (d) d10
Answer: (c) d9
Q19. The spin-only magnetic moment formula is:
(a) μ = n(n+2) (b) μ = √[S(S+1)] μB (c) μ = 2S (d) μ = n/2
Answer: (b) μ = √[S(S+1)] μB
Q20. A complex with μ = 5.92 μB most likely has how many unpaired electrons?
(a) 1 (b) 3 (c) 5 (d) 4
Answer: (c) 5
Q21. For a fixed ligand, Δo generally increases with:
(a) Decreasing oxidation state of the metal (b) Increasing oxidation state of the metal (c) Decreasing atomic number (d) It is independent of oxidation state
Answer: (b) Increasing oxidation state of the metal
Q22. Moving from a 3d to a 4d to a 5d metal (same ligand, same geometry), Δo typically:
(a) Decreases by 30–50% at each step (b) Increases by 30–50% at each step (c) Remains constant (d) Becomes zero
Answer: (b) Increases by 30–50% at each step
Q23. Which of the following has the largest Δo among the hexaammine M(III) series?
(a) [Co(NH3)6]3+ (b) [Rh(NH3)6]3+ (c) [Ir(NH3)6]3+ (d) All are equal
Answer: (c) [Ir(NH3)6]3+ (≈41,000 cm−1)
Q24. A d0 or d10 metal ion complex is expected to be:
(a) Strongly coloured (b) Colourless (c) Paramagnetic (d) Always low-spin
Answer: (b) Colourless, since no d–d transition is possible (no vacant/partially filled d orbital pairing needed for excitation)
Q25. The CFSE typically contributes what fraction of a complex's total binding energy?
(a) 50–60% (b) 2–10% (c) 90–100% (d) It contributes nothing
Answer: (b) 2–10%
Q26. Which ion, with zero CFSE, is expected to fall exactly on the "expected line" for lattice energy across the first transition series?
(a) Fe2+ (b) Mn2+ (c) Ni2+ (d) Cr2+
Answer: (b) Mn2+ (high-spin d5, CFSE = 0)
Q27. In a normal spinel structure, the B3+ ions occupy:
(a) Tetrahedral holes (b) Octahedral holes (c) Both equally (d) Neither
Answer: (b) Octahedral holes
Q28. Fe3O4 (magnetite) is an example of:
(a) Normal spinel (b) Inverse spinel (c) Perovskite (d) Rutile structure
Answer: (b) Inverse spinel
Q29. The difference in CFSE for a dn ion between octahedral and tetrahedral geometry is called:
(a) Ligand Field Stabilization Energy (b) Octahedral Site Stabilization Energy (OSSE) (c) Pairing energy (d) Barycentre energy
Answer: (b) Octahedral Site Stabilization Energy (OSSE)
Q30. Which of the following correctly states the condition for a low-spin octahedral complex?
(a) Δo < P (b) Δo > P (c) Δo = 0 (d) P = 0
Answer: (b) Δo > P
Q31. [Ti(H2O)6]3+ shows a single broad absorption band because:
(a) Ti3+ has 9 d electrons (b) Ti3+ (d1) has only one possible t2g→eg transition (c) Water is colourless (d) Ti3+ is diamagnetic
Answer: (b) Ti3+ (d1) has only one possible t2g→eg transition
Q32. The colour observed in a coordination compound is:
(a) The same as the colour absorbed (b) Complementary to the colour absorbed (c) Unrelated to absorption (d) Always white
Answer: (b) Complementary to the colour absorbed
Q33. Which of these is NOT an assumption of classical Crystal Field Theory?
(a) Ligands are point charges (b) Only electrostatic interactions are considered (c) Metal-ligand bonds have significant covalent character (d) Ligand size and shape are ignored
Answer: (c) Metal-ligand bonds have significant covalent character (this was Van Vleck's later modification, not part of the original electrostatic model)
Q34. [Co(H2O)6]3+ appears blue while [Co(NH3)6]3+ appears golden-brown. This difference arises from:
(a) Different oxidation states of Co (b) Different Δo values due to different ligand field strengths (c) Different geometries (d) Different central metal ions
Answer: (b) Different Δo values due to different ligand field strengths
Q35. For a d4 ion in a strong octahedral field, CFSE equals:
(a) −0.6Δo (b) −1.6Δo + P (c) −1.6Δo (d) −0.6Δo + P
Answer: (b) −1.6Δo + P
Q36. Which orbital set points directly toward the ligands in a tetrahedral complex?
(a) e set (b) t2 set (c) Neither set points directly at ligands (d) Both point directly at ligands
Answer: (c) Neither set points directly at ligands (this is a key structural difference from the octahedral case)
Q37. Which of the following ions would you expect to adopt tetrahedral rather than octahedral geometry?
(a) [Co(NH3)6]3+ (b) [Ni(CN)4]2− (c) [CoCl4]2− (d) [Cr(H2O)6]3+
Answer: (c) [CoCl4]2− (small metal ion, large weak-field ligand, high-spin d7 — classic conditions favouring tetrahedral geometry)
Q38. In square planar splitting derived from an octahedral field, which orbital is raised the most in energy?
(a) dz² (b) dxy (c) dx²−y² (d) dxz/dyz
Answer: (c) dx²−y²
Q39. Which statement about the spectrochemical series is correct?
(a) It can be fully explained by ligand charge alone (b) H2O is a stronger-field ligand than OH− despite OH− carrying a negative charge (c) It varies unpredictably between different metals (d) CN− is a weak-field ligand
Answer: (b) H2O is a stronger-field ligand than OH− despite OH− carrying a negative charge
Q40. The energy required for a d–d transition in most first-row transition metal complexes falls in which region of the electromagnetic spectrum?
(a) Ultraviolet (b) Visible (c) Infrared (d) Radio
Answer: (b) Visible
