---
title: "Ligand Field Theory and Colour"
book: "University Chemistry — Year 2"
subject: chemistry
language: en
chapter: 19
exercises: 12
source: https://one-course.com/books/chemistry/3/en/chapter/19-ligand-field-theory-and-colour
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 19 — Ligand Field Theory and Colour

Ruby is red and emerald is green, yet both owe their colour to the same ion, $\ce{Cr^3+}$, a few of which replace aluminium in corundum, $\ce{Al2O3}$, or in beryl. Copper sulfate crystals are deep blue, but the powder left after heating them is white. In each case the colour comes from electrons jumping between d orbitals whose energies the surrounding atoms have pulled apart — by an amount that depends on which atoms surround the ion and how. This chapter computes that splitting, first with point charges, then with [molecular orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-lcao); it explains colours, magnetism, and the [18-electron rule](https://one-course.com/books/chemistry/3/en/chapter/18-transition-metals-and-coordination-complexes#def-b2-coordination-complexes-electron-count) of the previous chapter.

**You already know.**

[Chapter 18](https://one-course.com/books/chemistry/3/en/chapter/18-transition-metals-and-coordination-complexes#ch-b2-coordination-complexes): $d^n$ configurations, geometries, electron counts. [Chapter 13](https://one-course.com/books/chemistry/3/en/chapter/13-atomic-orbitals-shapes-energies-and-sizes#ch-b2-atomic-orbitals): the shapes of the five d orbitals. [Chapter 15](https://one-course.com/books/chemistry/3/en/chapter/15-fragment-orbitals-and-polyatomic-molecules#ch-b2-fragment-orbitals): [fragment orbitals](https://one-course.com/books/chemistry/3/en/chapter/15-fragment-orbitals-and-polyatomic-molecules#def-b2-fragment-orbitals-fragment). The Year 1 volume: absorbance, complementary colours, unpaired electrons, Hund’s rule.

![A ruby crystal (public domain); an emerald in its rock (Didier Descouens, CC BY-SA 3.0); crystals of copper(II) sulfate pentahydrate (Stephanb, CC BY-SA 3.0). Wikimedia Commons. Chromium(III) colours the first two, the aqua complex of copper(II) the third.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/img-d4f502ee6a50.jpg)

![A ruby crystal (public domain); an emerald in its rock (Didier Descouens, CC BY-SA 3.0); crystals of copper(II) sulfate pentahydrate (Stephanb, CC BY-SA 3.0). Wikimedia Commons. Chromium(III) colours the first two, the aqua complex of copper(II) the third.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/img-08a11465b190.jpg)

![A ruby crystal (public domain); an emerald in its rock (Didier Descouens, CC BY-SA 3.0); crystals of copper(II) sulfate pentahydrate (Stephanb, CC BY-SA 3.0). Wikimedia Commons. Chromium(III) colours the first two, the aqua complex of copper(II) the third.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/img-04f2a8bcbeec.jpg)

*A ruby crystal (public domain); an emerald in its rock (Didier Descouens, CC BY-SA 3.0); crystals of copper(II) sulfate pentahydrate (Stephanb, CC BY-SA 3.0). Wikimedia Commons. Chromium(III) colours the first two, the aqua complex of copper(II) the third.*

## 19.1 The crystal-field model

**Definition 19.1 (Crystal field).**

The *crystal-field model* treats the ligands of a complex as negative point charges (or the negative ends of dipoles) around the metal ion, and computes how they change the energies of its d orbitals. In an octahedral field, the d orbitals split into two sets separated by the *crystal-field splitting* $\Delta_o$.

Place the six ligands on the axes. The $\mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$ orbitals point their lobes straight at ligands: an electron in them is repelled more and they rise. The $\mathrm d_{xy}$, $\mathrm d_{xz}$, $\mathrm d_{yz}$ orbitals point between the ligands and rise less. The first pair is called $e_g$, the second set of three $t_{2g}$ (names from group theory, in the Year 3 volume).

![Two d orbitals among the four ligands of the xy plane of an octahedron (black dots). The lobes of d_x2-y2 point at the ligands, those of d_xy between them.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/fig-8c87ae6da4d8.svg)

*Two d orbitals among the four ligands of the $xy$ plane of an octahedron (black dots). The lobes of $\mathrm d_{x^2-y^2}$ point at the ligands, those of $\mathrm d_{xy}$ between them.*

**Theorem 19.2 (Barycentre).**

In an octahedral field, measured from the average energy of the d orbitals, the $e_g$ pair lies at $+0.6\Delta_o$ and the $t_{2g}$ set at $-0.4\Delta_o$.

**Proof.** The field raises the average of the five d energies but does not change it by splitting (the barycentre rule, admitted: the sum of the shifts of a full set of orbitals is fixed by the total charge, not by its arrangement). With $e_g$ at $x$ and $t_{2g}$ at $x - \Delta_o$, the condition $2x + 3(x - \Delta_o) = 0$ gives $x = 0.6\Delta_o$. ∎

## 19.2 High spin and low spin

**Definition 19.3 (Spin states).**

The *pairing energy* $P$ is the energy needed to put a second electron into an occupied orbital rather than into an empty one of the same energy. A complex is *high spin* when its electrons occupy the $e_g$ orbitals before pairing in $t_{2g}$, *low spin* when they pair in $t_{2g}$ first.

**Proposition 19.4 (Choice of the spin state).**

For $d^4$ to $d^7$ octahedral ions, the complex is [low spin](#def-b2-ligand-field-spin) if $\Delta_o > P$ and [high spin](#def-b2-ligand-field-spin) if $\Delta_o < P$. For $d^1$–$d^3$ and $d^8$–$d^{10}$ only one configuration exists.

**Proof.** From $d^3$ ($t_{2g}^3$), the fourth electron either enters $e_g$ (cost $\Delta_o$ above $t_{2g}$) or pairs in $t_{2g}$ (cost $P$). The cheaper wins; the same comparison holds for each electron up to $d^7$. For $d^1$–$d^3$ there is room in $t_{2g}$ without pairing; for $d^8$–$d^{10}$, $t_{2g}$ is full and $e_g$ must be used anyway. ∎

**Definition 19.5 (Crystal-field stabilisation energy).**

The *crystal-field stabilisation energy* (CFSE) of a configuration $t_{2g}^ae_g^b$ is $(-0.4a + 0.6b)\Delta_o$, the energy of the d electrons relative to the barycentre (the pairing energies counted separately).

**Proposition 19.6 (CFSE along the series).**

[High spin](#def-b2-ligand-field-spin), the CFSE is $0$, $-0.4$, $-0.8$, $-1.2$, $-0.6$, $0$, $-0.4$, $-0.8$, $-1.2$, $-0.6$, $0$ (in $\Delta_o$) for $d^0$ to $d^{10}$: it vanishes for $d^0$, $d^5$, $d^{10}$. [Low spin](#def-b2-ligand-field-spin) it reaches $-2.4\Delta_o$ for $d^6$.

**Proof.** Fill the configurations and apply the definition; the values are those of the figure, computed. ∎

![Left: crystal-field stabilisation energy along the d series (computed), high spin (solid) and low spin (dashed). Right: the two configurations of a d6 ion: four unpaired electrons when _o < P (as (Fe(H2O)6)2+), none when _o > P (as (Fe(CN)6)4-).](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/fig-ab9e59c36ee1.svg)

![Left: crystal-field stabilisation energy along the d series (computed), high spin (solid) and low spin (dashed). Right: the two configurations of a d6 ion: four unpaired electrons when _o < P (as (Fe(H2O)6)2+), none when _o > P (as (Fe(CN)6)4-).](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/fig-9f4bc9de3d4e.svg)

*Left: [crystal-field stabilisation energy](#def-b2-ligand-field-cfse) along the d series (computed), [high spin](#def-b2-ligand-field-spin) (solid) and [low spin](#def-b2-ligand-field-spin) (dashed). Right: the two configurations of a $d^6$ ion: four unpaired electrons when $\Delta_o < P$ (as $\ce{[Fe(H2O)6]^2+}$), none when $\Delta_o >
P$ (as $\ce{[Fe(CN)6]^4-}$).*

**Method 19.7 (Spin state, CFSE and unpaired electrons).**

1. Find $n$ from the oxidation state ( $n = g - m$ ).
2. For $d^4$ – $d^7$ , compare $\Delta_o$ and $P$ (or place the ligand in the [spectrochemical series](#def-b2-ligand-field-spectrochemical) ): strong field, [low spin](#def-b2-ligand-field-spin) ; weak field, [high spin](#def-b2-ligand-field-spin) .
3. Fill $t_{2g}$ and $e_g$ accordingly; CFSE $= (-0.4a + 0.6b)\Delta_o$ .
4. Count the unpaired electrons; any unpaired electron makes the complex [paramagnetic](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-magnetism) .

## 19.3 Other geometries

**Proposition 19.8 (Tetrahedral and square-planar fields).**

In a tetrahedral field the order is inverted: two orbitals ($e$) below three ($t_2$), with a splitting $\Delta_t \approx \frac49\Delta_o$ for the same metal and ligands; tetrahedral complexes are [high spin](#def-b2-ligand-field-spin). In a square-planar field the $\mathrm d_{x^2-y^2}$ orbital lies far above the others: a $d^8$ ion fills the four lower orbitals and leaves it empty.

**Proof.** *Admitted at this level.* ∎

The factor $4/9$ comes from the geometry of the point charges; with only four ligands, none pointing along the axes, the splitting is too small to overcome the [pairing energy](#def-b2-ligand-field-spin). Removing the two ligands on the $z$ axis of an octahedron lowers the orbitals with a $z$ component and raises nothing: this explains the square-planar $d^8$ complexes of the previous chapter, whose eight electrons all find room below the empty $\mathrm d_{x^2-y^2}$.

![Splitting of the d orbitals in three geometries, schematic (not to the same scale): octahedral, tetrahedral (inverted and smaller), square planar (one orbital far above the others).](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/fig-63a8f86e5215.svg)

*Splitting of the d orbitals in three geometries, schematic (not to the same scale): octahedral, tetrahedral (inverted and smaller), square planar (one orbital far above the others).*

## 19.4 Colour and the spectrochemical series

**Definition 19.9 (d–d transition).**

A *d–d transition* is the promotion of an electron from a lower to a higher d level of a complex by absorption of light; for a $d^1$ ion, from $t_{2g}$ to $e_g$, at the energy $\Delta_o$.

**Proposition 19.10 (Splitting from colour).**

If a complex absorbs most strongly at the wavelength $\lambda_{\max}$ by a [d–d transition](#def-b2-ligand-field-d-d) of energy $\Delta_o$, then $\Delta_o = N_Ahc/\lambda_{\max}$ per mole, or $\tilde\nu = 1/\lambda_{\max}$ in wavenumbers; the colour seen is the complement of the colour absorbed.

**Proof.** A photon of wavelength $\lambda$ carries $hc/\lambda$; one mole of photons, $N_Ahc/\lambda$. White light minus the absorbed band appears in the complementary colour. ∎

**Method 19.11 (From an absorption band to a colour).**

1. Convert $\lambda_{\max}$ to $\tilde\nu = 1/\lambda_{\max}$ ( $\mathrm{cm}^{-1}$ ) and to $N_Ahc/\lambda_{\max}$ ( $\mathrm{kJ}/\mathrm{mol}$ ): $500\,\mathrm{nm} \leftrightarrow 20\,000\,\mathrm{cm}^{-1}  \leftrightarrow 239\,\mathrm{kJ}/\mathrm{mol}$ .
2. Find the colour absorbed (violet 400–430, blue 430–490, green 490–560, yellow 560–590, orange 590–620, red 620–700 nm, approximately).
3. The colour seen is opposite on the colour wheel.

![A colour wheel: the colour seen is roughly the complement, diametrically opposite, of the band absorbed.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/fig-8b8105eef9d7.svg)

*A colour wheel: the colour seen is roughly the complement, diametrically opposite, of the band absorbed.*

**Definition 19.12 (Spectrochemical series).**

The *spectrochemical series* orders ligands by the splitting they produce for a given metal ion:

$$
\ce{I-} < \ce{Br-} < \ce{Cl-} < \ce{F-} < \ce{OH-} < \ce{H2O} < \ce{NH3} < \text{en} <
  \ce{NO2-} < \ce{CN-} < \ce{CO} .
$$

For a given ligand, $\Delta_o$ also grows with the charge of the metal ion and down a group (4d and 5d complexes, larger than 3d, are almost always [low spin](#def-b2-ligand-field-spin)).

The halides are at the weak end and $\ce{CN-}$ and $\ce{CO}$ at the strong end, which the [crystal-field model](#def-b2-ligand-field-crystal-field) cannot explain: it would put the charged $\ce{F-}$ above the neutral $\ce{CO}$. The explanation needs orbitals.

## 19.5 The ligand field

**Definition 19.13 (Ligand field theory).**

*Ligand field theory* describes a complex with [molecular orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-lcao) built from the valence orbitals of the metal (its nine $n$d, $(n+1)$s and $(n+1)$p orbitals) and the donor orbitals of the ligands.

**Proposition 19.14 (σ\sigmaσ bonding in an octahedral complex).**

With six ligands each giving one $\sigma$ donor orbital, six bonding [molecular orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-lcao) form from the metal’s s, three p and the two $e_g$ d orbitals; the three $t_{2g}$ orbitals, which point between the ligands, stay nonbonding; above them lie the antibonding $e_g^*$ orbitals. The ligand electrons fill the six [bonding orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-bonding); the metal’s d electrons go into $t_{2g}$ and $e_g^*$, separated by $\Delta_o$. Bonding plus [nonbonding orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-bonding) can hold $12 + 6 = 18$ electrons.

**Argument.** By the fragment method of [Chapter 15](https://one-course.com/books/chemistry/3/en/chapter/15-fragment-orbitals-and-polyatomic-molecules#ch-b2-fragment-orbitals), each metal orbital interacts with the combination of ligand orbitals of the same symmetry: s with the totally symmetric one, each p with a pair of opposite ligands, $\mathrm d_{z^2}$ and $\mathrm
d_{x^2-y^2}$ with the combinations pointing along their lobes; no $\sigma$ combination matches the $t_{2g}$ orbitals (their overlap with any $\sigma$ donor pointing along an axis is zero by symmetry). The names of the combinations come from group theory, admitted. Filling all the bonding and [nonbonding orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-bonding), and none of the antibonding ones, takes 18 electrons — the [18-electron rule](https://one-course.com/books/chemistry/3/en/chapter/18-transition-metals-and-coordination-complexes#def-b2-coordination-complexes-electron-count). ∎

![Molecular orbitals of an octahedral complex with six donors, schematic. The ligands’ twelve electrons fill the six bonding orbitals; the metal’s d electrons occupy the nonbonding t_2g and the antibonding e_g* levels, separated by _o — the crystal-field picture recovered. Up to 18 electrons fill the bonding and nonbonding levels.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/fig-4c5943c876e7.svg)

*[Molecular orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-lcao) of an octahedral complex with six $\sigma$ donors, schematic. The ligands’ twelve electrons fill the six [bonding orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-bonding); the metal’s d electrons occupy the nonbonding $t_{2g}$ and the antibonding $e_g^*$ levels, separated by $\Delta_o$ — the crystal-field picture recovered. Up to 18 electrons fill the bonding and nonbonding levels.*

**Definition 19.15 (π\piπ-donor and π\piπ-acceptor ligands).**

A *$\pi$-donor ligand* has filled orbitals of $\pi$ symmetry with respect to the metal–ligand bond (lone pairs of $\ce{Cl-}$, $\ce{OH-}$); a *$\pi$-acceptor ligand* has empty ones of low energy (the $\pi^*$ orbitals of $\ce{CO}$ and $\ce{CN-}$). The transfer of electron density from filled metal d orbitals into the empty $\pi^*$ orbitals of a $\pi$-acceptor ligand is called *back-donation*.

**Proposition 19.16 (π\piπ effects on the splitting).**

$\pi$-donor ligands decrease $\Delta_o$; $\pi$-acceptor ligands increase it.

**Proof.** The $t_{2g}$ orbitals have the right symmetry to overlap with ligand $\pi$ orbitals. A filled ligand $\pi$ orbital lies below $t_{2g}$: by [Theorem 17.1](https://one-course.com/books/chemistry/3/en/chapter/17-frontier-orbitals-and-reactivity#thm-b2-frontier-orbitals-perturbation) the interaction pushes $t_{2g}$ up, towards $e_g^*$, and $\Delta_o$ shrinks. An empty $\pi^*$ orbital lies above $t_{2g}$: the interaction pushes $t_{2g}$ down, and $\Delta_o$ grows. The second case transfers metal d density into the ligand’s $\pi^*$: [back-donation](#def-b2-ligand-field-pi-ligands). ∎

![Effect of ligand π orbitals on the t_2g level, schematic. A filled ligand π orbital below pushes t_2g up (smaller splitting); an empty π* orbital above pulls it down (larger splitting).](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-ligand-field/fig-6abddec08290.svg)

*Effect of ligand $\pi$ orbitals on the $t_{2g}$ level, schematic. A filled ligand $\pi$ orbital below pushes $t_{2g}$ up (smaller splitting); an empty $\pi^*$ orbital above pulls it down (larger splitting).*

This places $\ce{CO}$ and $\ce{CN-}$, strong $\sigma$ donors and $\pi$ acceptors, at the top of the [spectrochemical series](#def-b2-ligand-field-spectrochemical), and the halides, $\pi$ donors, at the bottom. Metal carbonyls have their $t_{2g}$ orbitals stabilised and filled: the 18-electron complexes of [Chapter 18](https://one-course.com/books/chemistry/3/en/chapter/18-transition-metals-and-coordination-complexes#ch-b2-coordination-complexes).

## 19.6 Exercises

**Exercise 19.1 ★.**

Give the octahedral configuration and CFSE of a $d^3$ ion and of a $d^8$ ion.

**Solution of Exercise 19.1.**

$d^3$: $t_{2g}^3$, CFSE $-1.2\Delta_o$. $d^8$: $t_{2g}^6e_g^2$, CFSE $-2.4 + 1.2 = -1.2\Delta_o$.

**Exercise 19.2 ★.**

How many unpaired electrons do $\ce{[Fe(H2O)6]^2+}$ ([high spin](#def-b2-ligand-field-spin)) and $\ce{[Fe(CN)6]^4-}$ ([low spin](#def-b2-ligand-field-spin)) have?

**Solution of Exercise 19.2.**

$\ce{[Fe(H2O)6]^2+}$, $d^6$ [high spin](#def-b2-ligand-field-spin) $t_{2g}^4e_g^2$: four. $\ce{[Fe(CN)6]^4-}$, [low spin](#def-b2-ligand-field-spin) $t_{2g}^6$: none ([diamagnetic](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-magnetism)).

**Exercise 19.3 ★.**

A complex absorbs most strongly at $600\,\mathrm{nm}$. What colour does it look?

**Solution of Exercise 19.3.**

600 nm is orange light; the complex looks blue (greenish blue).

**Exercise 19.4 ★.**

Order $\ce{H2O}$, $\ce{CN-}$, $\ce{Cl-}$, $\ce{NH3}$ by the splitting they produce.

**Solution of Exercise 19.4.**

$\ce{Cl-} < \ce{H2O} < \ce{NH3} < \ce{CN-}$.

**Exercise 19.5 ★★.**

For a $d^5$ ion, $\Delta_o = 165\,\mathrm{kJ}/\mathrm{mol}$ with water and $390\,\mathrm{kJ}/\mathrm{mol}$ with cyanide, and $P = 300\,\mathrm{kJ}/\mathrm{mol}$ (exercise data). Predict both spin states and the unpaired electrons.

**Solution of Exercise 19.5.**

Water: $\Delta_o < P$, [high spin](#def-b2-ligand-field-spin), $t_{2g}^3e_g^2$, five unpaired electrons. Cyanide: $\Delta_o > P$, [low spin](#def-b2-ligand-field-spin), $t_{2g}^5$, one unpaired electron.

**Exercise 19.6 ★★.**

$\ce{[Co(H2O)6]^2+}$ is pink, $\ce{[CoCl4]^2-}$ deep blue. Explain the change of colour from the change of geometry and ligand.

**Solution of Exercise 19.6.**

In the tetrahedral $\ce{[CoCl4]^2-}$ the splitting is about $4/9$ of an octahedral one, and chloride is a weaker ligand than water: the d–d band moves to longer wavelengths (orange to red), and the complex looks blue. Tetrahedral complexes also absorb more strongly, hence the deep colour.

**Exercise 19.7 ★★.**

Show that square-planar $\ce{[Ni(CN)4]^2-}$ is [diamagnetic](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-magnetism), while tetrahedral $\ce{[NiCl4]^2-}$ has two unpaired electrons.

**Solution of Exercise 19.7.**

$\ce{Ni^2+}$ is $d^8$. Square planar: four lower orbitals hold the eight electrons in pairs, $\mathrm d_{x^2-y^2}$ stays empty: [diamagnetic](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-magnetism). Tetrahedral: $e^4t_2^4$, two unpaired electrons in the $t_2$ set.

**Exercise 19.8 ★★.**

A $d^1$ complex absorbs at $500\,\mathrm{nm}$ (exercise data). Compute $\Delta_o$ in $\mathrm{cm}^{-1}$ and $\mathrm{kJ}/\mathrm{mol}$.

**Solution of Exercise 19.8.**

$\tilde\nu = 1/(500 \times 10^{-7}\,\text{cm}) = 20\,000\,\mathrm{cm}^{-1}$; $\Delta_o =
0.1196/(500 \times 10^{-9}) = 2.39 \times 10^{5}\,\mathrm{J}/\mathrm{mol}$, $239\,\mathrm{kJ}/\mathrm{mol}$.

**Exercise 19.9 ★★.**

Why are compounds of $\ce{Zn^2+}$ and $\ce{Sc^3+}$ colourless?

**Solution of Exercise 19.9.**

$\ce{Zn^2+}$ is $d^{10}$ and $\ce{Sc^3+}$ $d^0$: no [d–d transition](#def-b2-ligand-field-d-d) is possible (no empty d level for $\ce{Zn^2+}$, no d electron for $\ce{Sc^3+}$), so no absorption in the visible.

**Exercise 19.10 ★★★.**

Count the electrons in the [molecular orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-lcao) of $\ce{[Co(NH3)6]^3+}$: which levels are filled, and is the complex [diamagnetic](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-magnetism)?

**Solution of Exercise 19.10.**

Co(III) $d^6$ + six ammonia lone pairs: 18 electrons. Twelve fill the six [bonding orbitals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-bonding), six fill the nonbonding $t_{2g}$ (ammonia gives a field strong enough for [low spin](#def-b2-ligand-field-spin) with cobalt(III)); $e_g^*$ is empty. All electrons are paired: [diamagnetic](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-magnetism).

**Exercise 19.11 ★★★.**

Explain why $\ce{CO}$, a neutral and weakly basic molecule, produces a larger splitting than the fluoride ion.

**Solution of Exercise 19.11.**

$\ce{CO}$ is a good $\sigma$ donor and, above all, a $\pi$ acceptor: its empty $\pi^*$ orbitals lower the $t_{2g}$ level by [back-donation](#def-b2-ligand-field-pi-ligands), widening $\Delta_o$. $\ce{F-}$ is a $\pi$ donor, which raises $t_{2g}$ and narrows $\Delta_o$. The point-charge model sees only the charge and gets the order wrong.

**Exercise 19.12 ★★★.**

The hydration enthalpies of the $\ce{M^2+}$ ions from $\ce{Ca^2+}$ to $\ce{Zn^2+}$, plotted against $n$, form two humps above a straight line through $d^0$, $d^5$ and $d^{10}$. Explain the shape with the high-spin CFSE.

**Solution of Exercise 19.12.**

Without a field, the hydration enthalpies would follow a smooth line (ions shrinking along the series). Water gives high-spin aqua complexes whose CFSE is zero for $d^0$, $d^5$, $d^{10}$ and largest for $d^3$ and $d^8$: the extra stabilisation adds two humps, peaking near $\ce{V^2+}$ and $\ce{Ni^2+}$, above the line through $\ce{Ca^2+}$, $\ce{Mn^2+}$, $\ce{Zn^2+}$.

## 19.7 Problem: Red Ruby, Green Emerald

**Problem 19.1.**

Weekend problem — chromium(III) in an octahedral field, the absorption bands of ruby and the colour they leave, the weaker field of emerald, and hydrated and anhydrous copper sulfate

Data stated for the problem (rounded, illustrative): ruby ($\ce{Cr^3+}$ in $\ce{Al2O3}$) absorbs in two bands, near $556\,\mathrm{nm}$ and $405\,\mathrm{nm}$; in emerald ($\ce{Cr^3+}$ in beryl) the first band lies near $620\,\mathrm{nm}$. For a $d^3$ ion the first band corresponds to $\Delta_o$. $N_Ahc = 0.1196\,\mathrm{J}\,\mathrm{m}/\mathrm{mol}$.

**Part I — Chromium(III).**

1. Give the $d^n$ configuration of $\ce{Cr^3+}$ .
2. How is it distributed in an octahedral field?
3. Is there a choice between high and [low spin](#def-b2-ligand-field-spin) ?
4. Compute its CFSE.
5. How many unpaired electrons does it have?
6. Why are chromium(III) complexes kinetically inert (slow to exchange ligands)?

**Part II — Ruby.**

7. What surrounds a $\ce{Cr^3+}$ ion substituted for $\ce{Al^3+}$ in corundum?
8. Convert the first band to wavenumber.
9. Deduce $\Delta_o$ in $\mathrm{kJ}/\mathrm{mol}$ .
10. Which colours are absorbed by the two bands?
11. What remains to be seen?
12. Why are there two bands for a single splitting? (A qualitative answer.)

**Part III — Emerald.**

13. Convert the emerald band to $\Delta_o$ .
14. Compare with ruby: which crystal gives the stronger field?
15. Where does the absorbed colour move, and what colour is transmitted?
16. Suggest a structural reason for the weaker field in beryl.
17. Would the number of unpaired electrons change?

**Part IV — Copper sulfate.**

18. What is the configuration of $\ce{Cu^2+}$ ?
19. In $\ce{CuSO4.5H2O}$ , copper is surrounded by water molecules. Why is the solid blue?
20. What happens to the colour when the water is driven off, and why?
21. Why does a drop of water turn the white powder blue again?
22. Would $\ce{[Cu(NH3)4]^2+}$ absorb at shorter or longer wavelength than the aqua ion?
23. Can a $d^9$ ion be high or [low spin](#def-b2-ligand-field-spin) ?
24. Is $\ce{CuSO4.5H2O}$ [paramagnetic](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-magnetism) ?
25. State $\Delta_o$ of $\ce{Cr^3+}$ in ruby from its first absorption band.

**Solution of Problem 19.1.**

**1.** $d^3$. **2.** $t_{2g}^3$, one electron in each of the three orbitals, parallel spins. **3.** No: three electrons fit in $t_{2g}$ without pairing. **4.** $-1.2\Delta_o$. **5.** Three. **6.** The CFSE is large and the $e_g^*$ orbitals, which point at the ligands, are empty: removing or adding a ligand costs much of that stabilisation. **7.** Six oxide ions at the corners of a slightly distorted octahedron. **8.** $1/(556 \times 10^{-7}) = 17\,990\,\mathrm{cm}^{-1}$, about $1.80 \times 10^{4}\,\mathrm{cm}^{-1}$. **9.** $0.1196/(556 \times 10^{-9}) = 2.15 \times 10^{5}\,\mathrm{J}/\mathrm{mol}$: $215\,\mathrm{kJ}/\mathrm{mol}$. **10.** Yellow-green (556 nm) and violet (405 nm). **11.** Red, with a little blue: the deep red of ruby. **12.** A $d^3$ ion has more than one excited configuration with an electron in $e_g$; electron repulsion gives them different energies, hence several bands (treated in the Year 3 volume). **13.** $0.1196/(620 \times 10^{-9}) = 1.93 \times 10^{5}\,\mathrm{J}/\mathrm{mol}$, $193\,\mathrm{kJ}/\mathrm{mol}$ ($16\,130\,\mathrm{cm}^{-1}$). **14.** Ruby: its splitting is larger. **15.** To orange-red: red is now absorbed, and green (with some blue) is transmitted. **16.** In beryl the chromium sites are slightly larger and the oxide ions farther, and the surroundings differ: a weaker field. **17.** No: a $d^3$ ion keeps three unpaired electrons in any octahedral field. **18.** $d^9$. **19.** The aqua complex absorbs in the red and orange by a [d–d transition](#def-b2-ligand-field-d-d): the transmitted light is blue. **20.** It turns white: without water ligands around copper (sulfate takes their place), the field is weaker and the band moves into the infrared. **21.** Water binds to copper again and the blue aqua complex forms. **22.** At shorter wavelength: ammonia, above water in the series, gives a larger splitting (the deep blue-violet of the ammine complex). **23.** No: $t_{2g}^6e_g^3$ is the only configuration. **24.** Yes: one unpaired electron. **25.** $\boldsymbol{\Delta_o \approx 1.8 \times 10^{4}\,\mathrm{cm}^{-1}}$, about $\boldsymbol{215\,\mathrm{kJ}/\mathrm{mol}}$.