Worked examples — Ligand Field Theory (LFT) and MO description (overview)

This page is the drill hall for Ligand Field Theory. The parent note built the why; here we hammer every kind of question an exam or the real world can throw at you. Before every solution I ask you to Forecast — guess the answer first, then check yourself. That guess is where the learning happens.

Everything rests on one equation the parent earned for us:

Two knobs move ${\Delta }_o$:

1. σ-donation pushes $e_g^{\ast }$ up → ${\Delta }_o$ grows.
2. π-interaction moves $t_{2g}$: a π-donor pushes it up (gap shrinks); a π-acceptor pulls it down (gap grows).

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The scenario matrix

Every LFT problem is one (or a blend) of the case-classes below. The figure is the visual map — each labelled box is one cell of the matrix, colour-coded, with an arrow to the example that hits it. Read the figure first, then the table repeats it in words as a checklist.

How to read Figure s01 (the scenario map): each plum box is one case class (a cell). Boxes on the left cover how ${\Delta }_o$ is built (σ, π-donor, π-acceptor); boxes on the right cover what we do with it (spin, degenerate/zero, limits, colour, real-world, exam twist). The small orange tag on each box names the worked example that covers it — so by the end every box is ticked.

Cell	Case class	The stress test	Example
A	Pure σ-donor (no π)	${\Delta }_o$ set by σ overlap alone	Ex 1
B	π-donor ligand (filled π)	$t_{2g}$ rises, gap shrinks	Ex 2
C	π-acceptor ligand (empty π*)	$t_{2g}$ drops, gap grows	Ex 3
D	Spin decision: ${\Delta }_o$ vs $P$	high-spin ⇄ low-spin boundary	Ex 4
E	Degenerate / zero cases	$d^0,d^{10}$, exactly-full sets	Ex 5
F	Limiting behaviour	${\Delta }_o\to 0$ and ${\Delta }_o\to \infty$	Ex 6
G	Colour / energy conversion	${\Delta }_o=hc/\lambda$, real numbers	Ex 7
H	Real-world word problem	why blood/haemoglobin cares about CO	Ex 8
I	Exam-style twist	rank a whole series, spot the trap	Ex 9

Figure s01 also shows, on its right side, the energy-level engine the first three examples share: burnt-orange $e_g^{\ast }$ on top, teal $t_{2g}$ below, and the plum ${\Delta }_o$ arrow between them, drawn for the three field cases (π-donor / σ-only / π-acceptor). Keep glancing back at it — every example just asks which teal or orange bar moves and by how much.

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Worked examples

Example 1 — Pure σ-donor [Cell A]

Forecast: Guess — with only the σ knob turned, is $t_{2g}$ moved at all? Where do 6 electrons go?

1. Identify the ligand type. ${\text{NH}}_3$ donates one lone pair through σ, has no low-lying filled or empty π orbital. Why this step? The whole answer depends on which of the two knobs is live. Here only σ (middle column of Figure s01).
2. σ-donation raises $e_g^{\ast }$. The lone pairs overlap head-on with $d_{z^2},d_{x^2-y^2}$; the antibonding partner $e_g^{\ast }$ climbs. Why this step? Rising $e_g^{\ast }$ is the only thing setting ${\Delta }_o$ when π is dead — clean and large.
3. $t_{2g}$ stays put. No π means no partner for $d_{xy},d_{yz},d_{xz}$ → they stay non-bonding at the barycentre (our zero line, defined above). Why this step? ${\Delta }_o=E(e_g^{\ast })-E(t_{2g})$; if $t_{2g}$ doesn't move, the gap is purely the $e_g^{\ast }$ rise.
4. Fill $d^6$. Large clean gap $\Rightarrow {\Delta }_o>P\Rightarrow$ low spin: $t_{2g}^6{e}_g^{\ast 0}$. Why this step? When climbing the gap costs more than pairing (${\Delta }_o>P$), every electron would rather pair up low; six electrons exactly fill the three $t_{2g}$ orbitals in pairs, leaving $e_g^{\ast }$ empty.

Verify: Low-spin $d^6$ has $6$ paired electrons → $n=0$ unpaired, so the ion is diamagnetic. Magnetic moment $\mu =\sqrt{n(n+2)}{\mu }_B$ with $n=0$ gives $\mu =0$ — matches the observed diamagnetism of $[{\text{Co(NH}}_3{)}_6{]}^{3+}$. ✔

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Example 2 — π-donor shrinks the gap [Cell B]

Forecast: Which line moves that Example 1's line did not — and does the gap widen or narrow?

1. ${\text{F}}^{-}$ is a π-donor: it carries filled, low-energy 2p lone pairs of $t_{2g}$ symmetry. Why this step? Filled ligand π + $t_{2g}$ symmetry match = a live π interaction (dead in Ex 1). This is the left column of Figure s01.
2. Filled–filled π push raises $t_{2g}$. Two filled sets can't both go down; the metal $t_{2g}$ (antibonding partner) rises. Why this step? Raising $t_{2g}$ eats into the gap: ${\Delta }_o=E(e_g^{\ast })-E(t_{2g}\uparrow )$ shrinks.
3. Small gap $\Rightarrow {\Delta }_o<P\Rightarrow$ high spin $t_{2g}^4{e}_g^{\ast 2}$. Why this step? Now climbing the gap (${\Delta }_o$) is cheaper than paying the pairing cost $P$, so electrons spread out into $e_g^{\ast }$ singly before doubling up — that is the high-spin arrangement.

Verify: High-spin $d^6$: $t_{2g}^4{e}_g^{\ast 2}$ → 4 up, then 1 down pairs, leaving unpaired count $n=4$. $\mu =\sqrt{4\cdot 6}=\sqrt{24}\approx 4.90{\mu }_B$ — matches the measured $\sim 4.9{\mu }_B$ paramagnetism. ✔ Same metal, same $d$-count, opposite spin — that is the π-donor knob at work.

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Example 3 — π-acceptor grows the gap [Cell C]

Forecast: Point-charge intuition says the negative ion wins. Predict which knob flips that.

1. CO offers empty, low-lying π* of $t_{2g}$ symmetry. Why this step? Empty π* is the mirror image of Ex 2's filled π — the interaction now runs the other way. This is the right column of Figure s01.
2. Metal $t_{2g}$ (filled) donates into CO π* — back-bonding. The metal set drops. Why this step? Lowering $t_{2g}$ widens ${\Delta }_o$ from the bottom.
3. σ-donation still raises $e_g^{\ast }$. So the gap opens from both ends: $e_g^{\ast }\uparrow$ and $t_{2g}\downarrow$. Why this step? Two independent knobs pushing the same way makes CO's ${\Delta }_o$ the largest in the series — no ionic charge required.

Verify: Spectrochemical ordering places ${\text{F}}^{-}<{\text{CN}}^{-}<\text{CO}$ — a neutral molecule outranking a negative ion, impossible in pure electrostatics but forced by the back-bonding knob. ✔ (See Back-bonding and π-Acceptor Ligands, Spectrochemical Series.)

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Example 4 — The spin decision boundary [Cell D]

Forecast: Guess which one is high-spin. The rule: compare ${\Delta }_o$ against $P$.

1. State the rule. Promote to $e_g^{\ast }$ if it costs less than pairing: high spin when ${\Delta }_o<P$, low spin when ${\Delta }_o>P$. Why this step? This single inequality — cost of climbing (${\Delta }_o$) versus cost of pairing ($P$) — is the entire spin decision (see High-spin vs Low-spin Complexes).
2. Ligand X: ${\Delta }_o=20,000<P=25,000$ → high spin. $d^5$ high spin = $t_{2g}^3{e}_g^{\ast 2}$ → $n=5$. Why this step? Climbing the small gap is cheaper than paying $P$, so all five electrons go into separate orbitals, all unpaired.
3. Ligand Y: ${\Delta }_o=30,000>P=25,000$ → low spin. $d^5$ low spin = $t_{2g}^5{e}_g^{\ast 0}$ → $n=1$. Why this step? Now the gap is dearer than $P$, so electrons pair up inside $t_{2g}$ rather than climb; five in three orbitals leaves exactly one unpaired.

Verify: ${\mu }_X=\sqrt{5\cdot 7}=\sqrt{35}\approx 5.92{\mu }_B$; ${\mu }_Y=\sqrt{1\cdot 3}=\sqrt{3}\approx 1.73{\mu }_B$. Two very different magnetic moments from the same $d^5$ ion — exactly what the ${\Delta }_o$ vs $P$ contest predicts. ✔

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Example 5 — Degenerate & zero cases [Cell E]

Forecast: Which of these can possibly be "high- vs low-spin"? Which can show a d–d colour?

First fix the capacities so we never miscount: $t_{2g}$ is 3 orbitals → holds 6 electrons; $e_g^{\ast }$ is 2 orbitals → holds 4 electrons; together $6+4=10$, the full $d$ shell.

1. $d^0$: no $d$ electrons at all. $t_{2g}^0{e}_g^{\ast 0}$. Why this step? With nothing to place, "high vs low spin" is undefined and there is no d–d transition → colourless (e.g. ${\text{TiO}}_2$ white).
2. $d^{10}$: all 10 slots filled → $t_{2g}^6{e}_g^{\ast 4}$ (using the capacities just fixed). Why this step? A completely full set means no vacancy to jump into → no d–d transition → colourless (e.g. ${\text{Zn}}^{2+}$ salts are white).
3. $d^3$: $t_{2g}^3{e}_g^{\ast 0}$ — half-filled lower set, only one arrangement possible regardless of ${\Delta }_o$. Why this step? High-spin and low-spin coincide for $d^3$: the spin question is degenerate/trivial. Still coloured (a vacancy in $e_g^{\ast }$ exists to absorb into).

Verify: Unpaired counts: $d^0\to n=0$ (diamagnetic), $d^{10}\to n=0$ (diamagnetic), $d^3\to n=3$, $\mu =\sqrt{3\cdot 5}=\sqrt{15}\approx 3.87{\mu }_B$ — the textbook value for ${\text{Cr}}^{3+}$. ✔ Colour: only $d^3$ (with an empty $e_g^{\ast }$ to absorb into) is coloured. ✔

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Example 6 — Limiting behaviour [Cell F]

Forecast: Sketch the two extremes before reading. Where do all 6 electrons end up in each?

1. ${\Delta }_o\to 0$ (free-ion limit). The gap closes; $t_{2g}$ and $e_g^{\ast }$ merge into one 5-fold set. Why this step? With no cost to occupy $e_g^{\ast }$ (the gap is smaller than any $P$), electrons spread out by Hund's rule → maximum spin. For $d^6$: 4 unpaired ($n=4$), the high-spin ceiling.
2. ${\Delta }_o\to \infty$ (crushing field). The gap is so large that $e_g^{\ast }$ is unreachable; all electrons cram into $t_{2g}$. Why this step? Pairing (cost $P$) is always cheaper than an infinite promotion → low spin. For $d^6$: $t_{2g}^6$, $n=0$.
3. Between the limits the spin flips exactly at ${\Delta }_o=P$ (Example 4's boundary).

Verify: Limit (a) $n=4\Rightarrow \mu =\sqrt{24}\approx 4.90{\mu }_B$; limit (b) $n=0\Rightarrow \mu =0$. The two endpoints bracket every real $d^6$ complex between $0$ and $4.9{\mu }_B$. ✔

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Example 7 — Colour / energy conversion [Cell G]

Forecast: The single $d$ electron jumps $t_{2g}\to e_g^{\ast }$. Guess: is $500\text{nm}$ absorption a big or small ${\Delta }_o$?

1. The absorbed photon energy equals the gap. ${\Delta }_o=h\nu =hc/\lambda$; in wavenumbers $\tilde{\nu }=1/\lambda$. Why this step? The d–d transition energy literally is the MO gap: the electron cannot climb from $t_{2g}$ to $e_g^{\ast }$ unless the photon delivers exactly ${\Delta }_o$ (see d-d Transitions and Colour of Complexes). Here $h$ and $c$ (defined above) just convert wavelength to energy.
2. Convert. $\lambda =500\text{nm}=500\times 10^{-7}\text{cm}=5\times 10^{-5}\text{cm}$. So $\tilde{\nu }=\frac{1}{5\times 10^{-5}\text{cm}}=20,000{\text{cm}}^{-1}.$ Why this step? ${\text{cm}}^{-1}$ (wavenumber, = 1/wavelength-in-cm) is the standard ${\Delta }_o$ unit; now it's directly comparable to Example 4's numbers.
3. Colour. Green light (~500 nm) is absorbed, so the eye receives its complementary colour — magenta (the accurate complement of green; loosely called purple/violet in older texts). Why this step? We see what is transmitted, not what is absorbed; removing the green band from white light leaves the magenta mix.

Verify: $1/(5\times 10^{-5})=20000$ exactly, so ${\Delta }_o=20,000{\text{cm}}^{-1}$. ✔ This is a physically sane mid-range value (typical ${\Delta }_o$ span 10,000–30,000 ${\text{cm}}^{-1}$). ✔ $[{\text{Ti(H}}_2{\text{O)}}_6{]}^{3+}$ is indeed observed as a red-violet/magenta solution. ✔

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Example 8 — Real-world word problem [Cell H]

Forecast: Both ${\text{O}}_2$ and CO can π-accept. Guess which forms the stronger, harder-to-remove bond.

1. CO is a strong π-acceptor with empty π* orbitals (Example 3's ligand). Why this step? Back-bonding from Fe $t_{2g}$ into CO π* is a second bond on top of the σ-donation — a synergic double grip.
2. Back-bonding lowers $t_{2g}$ and stabilises the whole complex (larger ${\Delta }_o$, deeper bonding well). Why this step? A deeper well = larger dissociation energy = the ligand won't leave. That is the physical meaning of "poison": it doesn't come off so ${\text{O}}_2$ can't rebind.
3. ${\text{O}}_2$ π-accepts more weakly, so its Fe bond is shallower and reversible — exactly what respiration needs.

Verify (consistency, not a number): The model predicts CO ranks above ${\text{O}}_2$ in field strength / bond strength; experimentally CO's affinity for haemoglobin is ~200× that of ${\text{O}}_2$ — same direction the back-bonding knob demands. ✔

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Example 9 — Exam-style twist: rank the series [Cell I]

Forecast: Write your order first. Where would a charge-only thinker put ${\text{I}}^{-}$ and CO?

1. Sort by π character. π-donors (biggest push-up of $t_{2g}$) → smallest gap; pure σ → middle; π-acceptors → biggest gap. Why this step? The two knobs decide the ordering, not ionic charge — that is the whole thesis of LFT.
2. π-donor strength ${\text{I}}^{-}>{\text{Cl}}^{-}$ (bigger, softer, more polarizable π lone pairs shrink ${\Delta }_o$ more). Why this step? Within π-donors, better π-overlap = more $t_{2g}$ rise = smaller gap, so the softest donor sits lowest.
3. Assemble the order: ${\text{I}}^{-}<{\text{Cl}}^{-}<{\text{H}}_2\text{O}<{\text{NH}}_3<{\text{CN}}^{-}<\text{CO}.$ Why this step? Reading left to right we move from strong π-donors, through σ-only ligands, to strong π-acceptors — the gap grows monotonically along that physical axis.
4. Name the trap. A point-charge (CFT) thinker ranks by charge and puts the anions (${\text{I}}^{-},{\text{Cl}}^{-},{\text{CN}}^{-}$) above neutral ${\text{H}}_2\text{O},{\text{NH}}_3,\text{CO}$. That contradicts experiment — neutral CO is strongest. The trap is forgetting π-bonding.

Verify: This matches the standard Spectrochemical Series fragment quoted in the parent note (${\text{I}}^{-}<{\text{Br}}^{-}<{\text{Cl}}^{-}<{\text{F}}^{-}<{\text{H}}_2\text{O}<{\text{NH}}_3<{\text{CN}}^{-}<\text{CO}$). ✔ The neutral-beats-negative inversion is the fingerprint of covalent LFT over electrostatic Crystal Field Theory (CFT). ✔

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Recall check

Recall Which knob and which line?

A ligand's filled π lone pairs (π-donor) move which line, and up or down? ::: They raise $t_{2g}$ up, shrinking ${\Delta }_o$. A π-acceptor (empty π*) moves which line, and up or down? ::: It lowers $t_{2g}$ (back-bonding), growing ${\Delta }_o$. σ-donation always affects which line? ::: It raises $e_g^{\ast }$ (grows ${\Delta }_o$). What does the star in $e_g^{\ast }$ mean? ::: Antibonding — higher energy than the parent orbitals, out-of-phase overlap. What is the barycentre? ::: The average energy of the five $d$ orbitals; the zero line a non-bonding orbital sits on. What is $P$? ::: Pairing energy: the extra cost of putting a second electron into an already-occupied orbital. For $d^3$, why is the spin question trivial? ::: $t_{2g}^3$ is the only arrangement; high- and low-spin coincide. Convert $\lambda =500\text{ nm}$ absorption to ${\Delta }_o$. ::: $\tilde{\nu }=1/(5\times 10^{-5}\text{ cm})=20,000{\text{ cm}}^{-1}$. Spin rule in one inequality? ::: High spin if ${\Delta }_o<P$; low spin if ${\Delta }_o>P$.

Cross-links: Molecular Orbital Theory · Nephelauxetic Effect · High-spin vs Low-spin Complexes · d-d Transitions and Colour of Complexes.