Chemistry · Ch 5 — Coordination Compounds

Colour in Coordination Compounds

5.5.5

Colour in Coordination Compounds

One of the most striking features of transition metal complexes is the sheer range of colours they display. A coloured substance is, physically, one that removes part of the visible spectrum from white light as the light passes through it — what reaches the eye is no longer white, but whatever wavelengths were not absorbed. The colour actually seen is therefore the complementary colour to the one absorbed: the colour generated by the leftover, unabsorbed wavelengths. For example, if a complex absorbs green light, what emerges — and what we see — is red.

(The observed relationship between the specific wavelength absorbed and the resulting colour, for a set of representative coordination entities, is tabulated separately — see the accompanying table card; not reproduced here.)

Table 5.3Relationship between the Wavelength of Light absorbed and the Colour observed in some Coordination Entities

Coordination entity	Wavelength of light absorbed (nm)	Colour of light absorbed	Colour of coordination entity
$[CoCl(NH_3)_5]^{2+}$	535	Yellow	Violet
$[Co(NH_3)_5(H_2O)]^{3+}$	500	Blue Green	Red
$[Co(NH_3)_6]^{3+}$	475	Blue	Yellow Orange
$[Co(CN)_6]^{3-}$	310	Ultraviolet (not in visible region)	Pale Yellow

The CFT explanation: d–d transitions

Crystal Field Theory explains this absorption directly in terms of the $d$ -orbital splitting developed in the previous section. Because the $d$ orbitals of the metal are split into sets of different energy (e.g. $t_{2g}$ and $e_g$ in an octahedral field), an electron can absorb a photon of just the right energy to jump from the lower-energy set to the higher-energy set — a $d$ – $d$ transition.

Consider $[\text{Ti(H}_2\text{O})_6]^{3+}$ , an octahedral complex that appears violet. $\text{Ti}^{3+}$ is a $3d^1$ ion, so in the ground state its single electron sits in the lower $t_{2g}$ level; the next available state is the empty, higher-energy $e_g$ level. When the complex absorbs light in the blue-green region of the visible spectrum, that photon supplies exactly the energy needed to promote the electron:

t_{2g}^{1}e_g^{0} \;\longrightarrow\; t_{2g}^{0}e_g^{1}

Having absorbed the blue-green component of white light, the complex transmits/reflects the remaining wavelengths, which combine to give the violet colour actually observed.

Figure 5.10Transition of an electron in an octahedral field

Fig. 5.10 — Transition of an electron in an octahedral field

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure is an energy-level diagram for a $d^1$ octahedral complex (like $\ce{[Ti(H2O)6]^{3+}}$ ). It shows two horizontal lines representing the two sets of $d$ orbitals split by the crystal field:

The lower line is labelled $t_{2g}$ (the three $d_{xy}, d_{xz}, d_{yz}$ orbitals).
The higher line is labelled $e_g$ (the two $d_{z^2}, d_{x^2-y^2}$ orbitals).

The vertical separation between these two lines is the crystal field splitting energy, denoted $\Delta_o$ (or $\Delta_0$ for octahedral).

On the lower $t_{2g}$ line, a single arrow (representing the one $d$ electron of $\ce{Ti^{3+}}$ ) points upward toward the $e_g$ line. A wavy arrow (representing a photon of light) is drawn from the $t_{2g}$ electron to the $e_g$ level, with the label $h\nu = \Delta_o$ beside it.

What the diagram teaches:

In the ground state, the single $d$ electron occupies the lower-energy $t_{2g}$ set. When the complex absorbs a photon whose energy exactly equals $\Delta_o$ , the electron is promoted to the empty $e_g$ level — this is a $d$ – $d$ transition ( $t_{2g}^1 e_g^0 \rightarrow t_{2g}^0 e_g^1$ ). The colour we see is the complement of the colour absorbed.

Key formula developed from this figure:

h\nu = \Delta_o

where:

$h$ = Planck’s constant ( $6.626 \times 10^{-34}\ \text{J s}$ )
$\nu$ = frequency of the absorbed light (in $\text{s}^{-1}$ or Hz)
$\Delta_o$ = crystal field splitting energy for an octahedral complex (in joules or $\text{cm}^{-1}$ ) …

In short, CFT attributes the colour of coordination compounds to $d$ – $d$ electronic transitions driven by the crystal-field splitting of the $d$ orbitals — something Valence Bond Theory, which treats the $d$ -derived hybrid orbitals as equivalent, has no way of explaining.

No ligand field, no splitting, no colour

If there is no ligand field around the metal, there is no crystal field splitting — and therefore no $d$ – $d$ transition is possible, so the substance is colourless. This is confirmed experimentally: removing the water ligands from $[\text{Ti(H}_2\text{O})_6]\text{Cl}_3$ by heating leaves it colourless, and anhydrous $\text{CuSO}_4$ (with no ligands bound to the copper) is white, whereas hydrated $\text{CuSO}_4\!\cdot\!5\text{H}_2\text{O}$ (with water ligands present) is blue.

The ligand itself changes the colour

Because the size of the crystal field splitting depends on the ligand (via the spectrochemical series), changing the ligand on the same metal changes the colour — even though the metal ion and its oxidation state stay the same. This is illustrated by progressively adding the didentate ligand ethane-1,2-diamine (en) to aqueous $[\text{Ni(H}_2\text{O})_6]^{2+}$ , in molar ratios of en : Ni of 1:1, 2:1, and 3:1. The successive substitution steps, with the colour of each species written beneath it, are:

\underset{\text{green}}{[\text{Ni(H}_2\text{O})_6]^{2+}(aq)} + \text{en}\,(aq) = \underset{\text{pale blue}}{[\text{Ni(H}_2\text{O})_4(\text{en})]^{2+}(aq)} + 2\text{H}_2\text{O}

[\text{Ni(H}_2\text{O})_4(\text{en})]^{2+}(aq) + \text{en}\,(aq) = \underset{\text{blue/purple}}{[\text{Ni(H}_2\text{O})_2(\text{en})_2]^{2+}(aq)} + 2\text{H}_2\text{O}

$[\text{Ni(H}_2\text{O})_2(\text{en})_2]^{2+}(aq) + \text{en}\,(aq) = \underset{\text{violet}}{[\text{Ni(en)}_3]^{2+}(aq)} + 2\text{H}_2\text{O}$ As each water pair is successively replaced by en, the colour of the solution shifts visibly through a sequence — from the green of the starting hexaaqua ion, through pale blue and then blue/purple as one and then two en ligands are incorporated, to the violet of the fully substituted $[\text{Ni(en)}_3]^{2+}$ . Each step in this substitution sequence changes the crystal field experienced by the nickel ion, which shifts the energy of the $d$ – $d$ transition and therefore the wavelength absorbed — and so the observed colour changes at every stage.

Figure 5.11Aqueous solutions of complexes of nickel(II) with an increasing number of ethane-1,2-diamine ligands.

Fig. 5.11 — Aqueous solutions of complexes of nickel(II) with an increasing number of ethane-1,2-diamine ligands.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure shows four test tubes lined up side by side, each containing an aqueous solution of a nickel(II) complex. The tubes are labelled with the formula of the complex and show a distinct colour:

First tube: $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ — green
Second tube: $[\text{Ni}(\text{H}_2\text{O})_4(\text{en})]^{2+}$ — pale blue
Third tube: $[\text{Ni}(\text{H}_2\text{O})_2(\text{en})_2]^{2+}$ — blue/purple
Fourth tube: $[\text{Ni}(\text{en})_3]^{2+}$ — violet

The horizontal axis (implicitly) runs from left to right, representing increasing number of ethane-1,2-diamine (en) ligands replacing water molecules. There is no vertical axis — the figure is a photograph of solutions, not a plot. The key physical idea is that as the stronger-field ligand en replaces the weaker-field ligand $\text{H}_2\text{O}$ , the crystal field splitting parameter $\Delta_o$ increases. A larger $\Delta_o$ means the $d$ – $d$ transition absorbs light of shorter wavelength (higher energy), shifting the observed colour from green (absorbs red) through blue to violet (absorbs yellow-green).

The textbook uses this sequence to illustrate the spectrochemical series: ligands are ranked by their ability to split $d$ orbitals. Here, en lies above $\text{H}_2\text{O}$ in the series, so each substitution raises $\Delta_o$ . The colour change is a direct visual demonstration of the relation:

\Delta_o = \frac{hc}{\lambda_{\text{absorbed}}}

where:

$\Delta_o$ = crystal field splitting energy for an octahedral complex (J or eV)
$h$ = Planck’s constant ( $6.626 \times 10^{-34}\ \text{J·s}$ )
$c$ = speed of light ( $3.00 \times 10^8\ \text{m/s}$ )
$\lambda_{\text{absorbed}}$ = wavelength of light absorbed (m or nm) …

Colour in gemstones

The same principle — $d$ – $d$ transitions at transition-metal centres embedded in an otherwise colourless host lattice — accounts for the colour of certain gemstones:

Figure 5.12(a) Ruby found in marble from Mogok, Myanmar; (b) Emerald found in Muzo, Colombia.

Fig. 5.12 — (a) Ruby found in marble from Mogok, Myanmar; (b) Emerald found in Muzo, Colombia.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

Figure 5.12 shows two photographs of gemstones: (a) a red ruby and (b) a green emerald. Both derive their colour from the same transition metal ion — $\text{Cr}^{3+}$ (a $d^3$ system) — but the surrounding crystal lattice differs. In ruby, $\text{Cr}^{3+}$ substitutes for $\text{Al}^{3+}$ in corundum ( $\text{Al}_2\text{O}_3$ ), where it experiences an octahedral ligand field from oxide ions. In emerald, $\text{Cr}^{3+}$ is embedded in beryl (a beryllium aluminium silicate), where the ligand field is also octahedral but the ligands (oxygen atoms from silicate groups) produce a different crystal field splitting energy.

The physical idea is that the colour of a coordination compound arises from $d$ – $d$ transitions of electrons between split $d$ orbitals. The energy difference $\Delta$ (crystal field splitting energy) determines which wavelength of light is absorbed. The complementary colour is observed. For $\text{Cr}^{3+}$ in ruby, $\Delta$ is larger, so it absorbs blue-green light and appears red. In emerald, $\Delta$ is smaller, so it absorbs red light and appears green.

The key formula linking the absorbed wavelength $\lambda$ to the splitting energy is:

\Delta = \frac{hc}{\lambda}

where:

$\Delta$ = crystal field splitting energy (J or eV)
$h$ = Planck’s constant ( $6.626 \times 10^{-34} \ \text{J·s}$ )
$c$ = speed of light ( $3.0 \times 10^8 \ \text{m/s}$ )
$\lambda$ = wavelength of absorbed light (m) …

…

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