Ligand Field Theory (LFT) and MO description (overview): Exercises

Before we start, one shared picture. Everything below refers to it: the metal $d$ -electrons live in two levels — the == $t_{2 g}$ == (three orbitals pointing between the ligand axes) and the == $e_{g}^{*}$ == (two orbitals pointing straight at the ligands, pushed up by σ-antibonding). The gap between them is $Δ_{o}$ .

Figure — Ligand Field Theory (LFT) and MO description (overview)

Level 1 — Recognition

L1.1

State, in one line each, what $t_{2 g}$ and $e_{g}^{*}$ are in the MO (LFT) picture of an octahedral σ-only complex.

Recall Solution

== $t_{2 g}$ ==: the metal $d_{x y}, d_{y z}, d_{x z}$ orbitals; they point between the six ligand directions, so they find no σ-LGO partner and stay non-bonding (essentially pure metal $d$ ).
== $e_{g}^{*}$ ==: the antibonding combination of metal $d_{z^{2}}, d_{x^{2} - y^{2}}$ with the $e_{g}$ ligand group orbitals. Their lobes point straight at the ligands, so they overlap strongly and are pushed up in energy.

The gap is $Δ_{o} = E (e_{g}^{*}) - E (t_{2 g})$ . See figure s01.

L1.2

Which symmetry labels do the six σ-donor ligand group orbitals (LGOs) span in $O_{h}$ ?

Recall Solution

$a_{1 g} + e_{g} + t_{1 u}$ The $a_{1 g}$ matches metal $s$ , the $t_{1 u}$ matches metal $p_{x}, p_{y}, p_{z}$ , and the $e_{g}$ matches metal $d_{z^{2}}, d_{x^{2} - y^{2}}$ . There is no $t_{2 g}$ among the σ-LGOs — that is exactly why the metal $t_{2 g}$ is left non-bonding in a σ-only complex.

L1.3

Classify each ligand as π-donor, π-acceptor, or σ-only (no π): $F^{-}$ , $CO$ , $NH_{3}$ , $Cl^{-}$ , $CN^{-}$ .

Recall Solution

Ligand	Class	Reason
$F^{-}$	π-donor	filled, low-energy $p$ lone pairs donate into metal $t_{2 g}$
$Cl^{-}$	π-donor	same idea, weaker
$NH_{3}$	σ-only	N lone pair only; no π orbital to share
$CN^{-}$	π-acceptor	empty low-lying π* accepts from metal $t_{2 g}$
$CO$	π-acceptor	empty low-lying π* — the classic back-bonder

Level 2 — Application

L2.1

An octahedral complex has $Δ_{o} = 250 kJ mol^{- 1}$ . Convert this to the wavelength of light whose photon energy equals $Δ_{o}$ . Use $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ , $h = 6.626 \times 1 0^{- 34} J s$ , $c = 3.00 \times 1 0^{8} m s^{- 1}$ .

Recall Solution

WHY this tool? A $d$ – $d$ transition promotes an electron across the gap, so the absorbed photon carries exactly $Δ_{o}$ . The bridge between energy and wavelength is $E = h ν = h c / λ$ .

Step 1 — per-molecule energy. Divide the molar value by $N_{A}$ :

= 4.152\times10^{-19}\ \text{J}.$$ **Step 2 — invert $E = hc/\lambda$:** $$\lambda = \frac{hc}{E} = \frac{(6.626\times10^{-34})(3.00\times10^{8})}{4.152\times10^{-19}} \approx 4.79\times10^{-7}\ \text{m} = 479\ \text{nm}.$$ That is blue-green light — the complex absorbs there and transmits its complement.

L2.2

Complex A absorbs at $λ_{A} = 600 nm$ , complex B at $λ_{B} = 400 nm$ . Which has the larger $Δ_{o}$ , and by what factor?

Recall Solution

Reasoning: $Δ_{o} = h c / λ$ , so $Δ_{o} \propto 1/ λ$ . Shorter $λ$ = larger gap. B (400 nm) is shorter, so B has the larger $Δ_{o}$ . $\frac{Δ _{o} ( B )}{Δ _{o} ( A )} = \frac{λ _{A}}{λ _{B}} = \frac{600}{400} = 1.5.$ B's field is $1.5 \times$ stronger. See d-d Transitions and Colour of Complexes.

L2.3

Using the frontier gap $Δ_{o} = E (e_{g}^{*}) - E (t_{2 g})$ , describe which way each level moves when you swap an ammonia ligand set for a fluoride set on the same metal, and hence what happens to $Δ_{o}$ .

Recall Solution

$NH_{3}$ (σ-only): raises $e_{g}^{*}$ by σ-antibonding; leaves $t_{2 g}$ non-bonding (unmoved).
$F^{-}$ (π-donor): still raises $e_{g}^{*}$ , but also pushes $t_{2 g}$ UP (filled F π lone pairs donate into it).

Because $t_{2 g}$ climbs toward $e_{g}^{*}$ , the gap $Δ_{o} = E (e_{g}^{*}) - E (t_{2 g})$ shrinks for fluoride. So $Δ_{o} (NH_{3}) > Δ_{o} (F^{-})$ — consistent with the Spectrochemical Series.

Level 3 — Analysis

L3.1

$[CoF_{6}]^{3 -}$ is high-spin ( $t_{2 g}^{4} e_{g}^{* 2}$ ) while $[Co(NH_{3})_{6}]^{3 +}$ is low-spin ( $t_{2 g}^{6} e_{g}^{* 0}$ ). Both are $Co^{3 +}$ ( $d^{6}$ ). Explain the difference purely from the MO level shifts — do not just quote "strong/weak field."

Recall Solution

The rule: electrons pair up in $t_{2 g}$ (low spin) only if the gap $Δ_{o}$ exceeds the pairing energy $P$ ; otherwise they spread into $e_{g}^{*}$ (high spin).

$NH_{3}$ : σ-only. $e_{g}^{*}$ is pushed high, $t_{2 g}$ stays put → large $Δ_{o}$ . Here $Δ_{o} > P$ , so all six $d$ -electrons cram into $t_{2 g}$ : $t_{2 g}^{6} e_{g}^{* 0}$ , low spin.
$F^{-}$ : π-donor. Its filled π lone pairs donate into $t_{2 g}$ , shoving it up toward $e_{g}^{*}$ → small $Δ_{o}$ . Now $Δ_{o} < P$ , so the 5th and 6th electrons prefer the higher $e_{g}^{*}$ over pairing: $t_{2 g}^{4} e_{g}^{* 2}$ , high spin.

The single deciding factor is the π-donation raising $t_{2 g}$ for fluoride. See High-spin vs Low-spin Complexes.

L3.2

Compute the crystal/ligand field stabilisation energy (LFSE) for both configurations of L3.1 in units of $Δ_{o}$ (take $t_{2 g}$ at $- 0.4 Δ_{o}$ and $e_{g}^{*}$ at $+ 0.6 Δ_{o}$ relative to the barycentre), ignoring pairing energy corrections. Which is more stabilised per unit $Δ_{o}$ ?

Recall Solution

WHY these numbers? The barycentre (average) must stay fixed: three $t_{2 g}$ each at $- 0.4 Δ_{o}$ and two $e_{g}^{*}$ each at $+ 0.6 Δ_{o}$ give $3 (- 0.4) + 2 (+ 0.6) = 0$ . Good.

Low-spin $t_{2 g}^{6} e_{g}^{* 0}$ : $LFSE = 6 (- 0.4 Δ_{o}) + 0 (+ 0.6 Δ_{o}) = - 2.4 Δ_{o} .$
High-spin $t_{2 g}^{4} e_{g}^{* 2}$ : $LFSE = 4 (- 0.4 Δ_{o}) + 2 (+ 0.6 Δ_{o}) = - 1.6 Δ_{o} + 1.2 Δ_{o} = - 0.4 Δ_{o} .$

Per unit of its own $Δ_{o}$ the low-spin config is far more stabilised ( $- 2.4$ vs $- 0.4$ ). But remember: the fluoride's $Δ_{o}$ is itself small, and low-spin would cost extra pairing energy $P$ — which is why it stays high-spin anyway.

L3.3

A ligand is both a strong σ-donor and a π-acceptor (e.g. $CN^{-}$ ). Trace what each character does to the two frontier levels, and conclude why such ligands sit at the strong-field end.

Recall Solution

σ-donation: raises $e_{g}^{*}$ (pushes the top of the gap up).
π-acceptance (back-bonding): the empty ligand π* accepts from metal $t_{2 g}$ , lowering $t_{2 g}$ (pushes the bottom of the gap down). See Back-bonding and π-Acceptor Ligands.

Both effects widen $Δ_{o} = E (e_{g}^{*}) - E (t_{2 g})$ from both ends simultaneously — top rises, bottom sinks. That double action is why π-acceptors dominate the strong-field end of the Spectrochemical Series, even when neutral.

Level 4 — Synthesis

L4.1

CFT (point-charge model) would rank the most negative ligand as the strongest-field one. Yet neutral $CO$ outsplits $F^{-}$ . Build the full LFT argument, naming which effect CFT is blind to and which experimental series it must reproduce.

Recall Solution

Step 1 — What CFT predicts. Point charges repel $d$ -electrons; a bigger negative charge repels harder, so CFT ranks $F^{-} > CO$ in field strength.

Step 2 — What is observed. The Spectrochemical Series runs $F^{-} < H_{2} O < NH_{3} < CN^{-} < CO,$ i.e. exactly backwards at the ends: neutral CO splits most.

Step 3 — What CFT is blind to. Covalent π-bonding. CFT has no orbitals, only charges, so it cannot see back-bonding at all (via Molecular Orbital Theory).

Step 4 — The LFT fix. $F^{-}$ is a π-donor → raises $t_{2 g}$ → shrinks $Δ_{o}$ . $CO$ is a π-acceptor → lowers $t_{2 g}$ and its strong σ-donation raises $e_{g}^{*}$ → widens $Δ_{o}$ from both sides. So CO wins despite being neutral. Charge is not the driver; orbital overlap and π-character are. This is the headline proof LFT beats Crystal Field Theory (CFT).

L4.2

$[Cr(H_{2} O)_{6}]^{3 +}$ absorbs near $575 nm$ ; $[Cr(CN)_{6}]^{3 -}$ absorbs near $375 nm$ . (a) Which has the larger $Δ_{o}$ ? (b) Convert both to $Δ_{o}$ in $kJ mol^{- 1}$ . (c) Explain the shift with the π-picture.

Recall Solution

(a) Shorter $λ$ → larger gap. $375 < 575 nm$ , so the CN⁻ complex has the larger $Δ_{o}$ .

(b) Use $Δ_{o} = N_{A} h c / λ$ .

{575\times10^{-9}} \approx 2.08\times10^{5}\ \text{J mol}^{-1} = 208\ \text{kJ mol}^{-1}.$$ $$\Delta_o(\text{CN}^-) = \frac{(6.022\times10^{23})(6.626\times10^{-34})(3.00\times10^{8})} {375\times10^{-9}} \approx 3.19\times10^{5}\ \text{J mol}^{-1} = 319\ \text{kJ mol}^{-1}.$$ **(c)** $\text{CN}^-$ is a π-acceptor: back-bonding **lowers** $t_{2g}$ while its σ-donation raises $e_g^*$, giving a bigger $\Delta_o$. A bigger gap → higher-energy (shorter-$\lambda$) absorption, hence the blue-shift from 575 nm to 375 nm.

Level 5 — Mastery

L5.1

The Nephelauxetic Effect shows electron–electron repulsion parameters (Racah $B$ ) decrease when a free ion is put into a complex. Argue, from the LFT/MO viewpoint, why this decrease is direct evidence of covalency, and why pure CFT can never predict it.

Recall Solution

Step 1 — What $B$ measures. The Racah parameter $B$ scales the repulsion between two $d$ -electrons. Smaller $B$ = electrons feel each other less = the $d$ -cloud is more spread out.

Step 2 — What LFT says spreads it. In LFT the "metal $d$ " MOs ( $t_{2 g}$ , $e_{g}^{*}$ ) are actually mixtures of metal $d$ and ligand orbitals. An electron in such an MO spends part of its time out on the ligands — the cloud has expanded ("nephelauxetic" = cloud-expanding). More expansion → smaller $B$ . This delocalisation is covalency.

Step 3 — Why CFT fails. In Crystal Field Theory (CFT) the $d$ -orbitals are pure metal, untouched point-charge orbitals — they cannot delocalise onto ligands by construction. So CFT predicts $B$ unchanged. The observed drop can only come from orbital mixing → LFT.

L5.2

A student claims: "Since $t_{2 g}$ is non-bonding, its energy is fixed and only $e_{g}^{*}$ ever moves, so every change in $Δ_{o}$ across the spectrochemical series is due to $e_{g}^{*}$ ." Identify the precise flaw and give a counterexample.

Recall Solution

The flaw: $t_{2 g}$ is non-bonding only in the σ-only special case. The moment a ligand has π-orbitals of $t_{2 g}$ symmetry, $t_{2 g}$ becomes bonding/antibonding and moves:

π-donor ( $F^{-}, Cl^{-}$ ): $t_{2 g}$ shifts up → $Δ_{o}$ shrinks.
π-acceptor ( $CO, CN^{-}$ ): $t_{2 g}$ shifts down → $Δ_{o}$ grows.

Counterexample: the entire strong end of the series (CN⁻, CO) works by lowering $t_{2 g}$ , not by raising $e_{g}^{*}$ further. So the change in $Δ_{o}$ there is dominated by $t_{2 g}$ motion — the exact opposite of the claim.

L5.3

Rank $Δ_{o}$ for the three complexes and justify each move by naming the level(s) that shift: $[FeF_{6}]^{3 -}$ , $[Fe(H_{2} O)_{6}]^{3 +}$ , $[Fe(CN)_{6}]^{3 -}$ (all $Fe^{3 +}$ ).

Recall Solution

Ranking (smallest → largest): $[FeF_{6}]^{3 -} < [Fe(H_{2} O)_{6}]^{3 +} < [Fe(CN)_{6}]^{3 -} .$

$F^{-}$ (π-donor): pushes $t_{2 g}$ up → smallest gap → high-spin.
$H_{2} O$ (weak/σ-dominant): $t_{2 g}$ roughly non-bonding, moderate $e_{g}^{*}$ → middle.
$CN^{-}$ (π-acceptor): back-bonding pulls $t_{2 g}$ down while σ raises $e_{g}^{*}$ → largest gap → low-spin.

Matches the Spectrochemical Series order for these three.

Recall One-line self-check before you leave

$Δ_{o} = E (e_{g}^{*}) - E (t_{2 g})$ ; σ-donation raises $e_{g}^{*}$ ; π-donors raise $t_{2 g}$ (small gap); π-acceptors lower $t_{2 g}$ (big gap); $Δ_{o} = h c / λ$ links the gap to colour.