Colour in Complexes & Limitations of CFT

Why Coordination Compounds Are Coloured

One of the most striking features of coordination compounds is their colour — the deep blue of $[\text{Cu(H}_2\text{O)}_6]^{2+}$, the violet of $[\text{Ti(H}_2\text{O)}_6]^{3+}$, the orange of dichromate. Crystal Field Theory explains it beautifully.

Recall that the d orbitals are split by $\Delta_o$. A d electron can absorb a photon of visible light whose energy exactly matches $\Delta_o$ and jump from the lower $t_{2g}$ set to the higher $e_g$ set — a d-d transition. The wavelength absorbed is removed from white light, and the colour we see is the complement of the absorbed colour.

Example: $[\text{Ti(H}_2\text{O)}_6]^{3+}$ ($d^1$) absorbs in the green-yellow region (~500 nm) and so appears purple/violet (the complement).

For colour to appear, the ion must have a partially filled d subshell:

• $d^0$ ions (Sc$^{3+}$, Ti$^{4+}$) and $d^{10}$ ions (Zn$^{2+}$, Cu$^+$) are colourless — no d-d transition is possible.

Colour Depends on the Ligand and Metal

Because the colour comes from $\Delta_o$, and $\Delta_o$ depends on the ligand (spectrochemical series), changing the ligand changes the colour:

• $[\text{Cu(H}_2\text{O)}_4]^{2+}$ is pale blue; adding ammonia gives $[\text{Cu(NH}_3)_4]^{2+}$, a much deeper blue — because NH$_3$ is a stronger-field ligand (larger $\Delta_o$, absorbs different wavelength).
• $[\text{Ni(H}_2\text{O)}_6]^{2+}$ (green) → $[\text{Ni(NH}_3)_6]^{2+}$ (blue) → $[\text{Ni(en)}_3]^{2+}$ (violet), as the ligand field strengthens.

The colour also depends on the metal and its oxidation state. Removing all the ligands (or going to $d^0$/$d^{10}$) removes the colour.

[NEET Important] The colour of a complex arises from a d-d transition; its energy ($\Delta_o$) and hence the colour are set by the ligand (spectrochemical series), the metal, and the oxidation state. A stronger-field ligand → larger $\Delta_o$ → absorbs higher-energy (shorter-wavelength) light.

Limitations of Crystal Field Theory

CFT is a big improvement on VBT, but it is still a simplified model and has limitations:

1. It treats the metal-ligand bond as purely electrostatic/ionic and ignores the covalent (orbital-overlap) character of the bond, which is significant in many complexes.
2. It cannot satisfactorily explain why some ligands (like CN$^-$ and CO) are strong field while others (like H$_2$O) are weak — the spectrochemical series is experimental, not predicted by CFT.
3. It does not account for the relative strengths of metal-ligand bonds in a fundamental way.

These shortcomings are addressed by the more advanced Ligand Field / Molecular Orbital Theory (beyond this syllabus).

Key Point: CFT successfully explains colour, magnetism, and the high-spin/low-spin distinction using $\Delta_o$ and the spectrochemical series — but it ignores covalency, which is its main limitation.

Solved Examples

Example 1: Origin of colour

Explain the origin of colour in $[\text{Ti(H}_2\text{O)}_6]^{3+}$.

Solution: Ti$^{3+}$ is $d^1$. The single d electron absorbs visible light (~500 nm) and undergoes a d-d transition from $t_{2g}$ to $e_g$. The complement of the absorbed light makes the complex appear purple/violet.

Example 2: Why colourless

Why is $[\text{Zn(H}_2\text{O)}_6]^{2+}$ colourless?

Solution: Zn$^{2+}$ is $d^{10}$ — the d subshell is completely filled, so no d-d transition is possible. With no absorption of visible light, the complex is colourless.

Example 3: Effect of ligand on colour

Why is $[\text{Cu(NH}_3)_4]^{2+}$ a deeper blue than $[\text{Cu(H}_2\text{O})_4]^{2+}$?

Solution: NH$_3$ is a stronger-field ligand than H$_2$O, so it produces a larger $\Delta_o$. The complex absorbs a different (higher-energy) wavelength, and its complementary colour is a deeper, more intense blue.

Example 4: Predict relative absorption

Two complexes of the same metal differ only in ligand: one has H$_2$O, the other CN$^-$. Which absorbs shorter-wavelength light?

Solution: CN$^-$ is a stronger-field ligand → larger $\Delta_o$ → absorbs higher-energy, shorter-wavelength light than the H$_2$O complex.

Example 5: d0 and d10 colourlessness

Why are both $d^0$ and $d^{10}$ ions colourless?

Solution: A $d^0$ ion has no d electron to excite, and a $d^{10}$ ion has a completely filled d set with no vacancy to receive an electron. In both cases no d-d transition is possible, so they are colourless.

Example 6: Colour change on ligand substitution

Anhydrous CuSO$_4$ is white but hydrated CuSO$_4\cdot5\text{H}_2\text{O}$ is blue. Explain.

Solution: In anhydrous CuSO$_4$ there are no water ligands to set up a crystal field. On hydration, water ligands surround Cu$^{2+}$ ($d^9$), splitting the d orbitals and allowing a d-d transition that gives the blue colour.

Example 7: Limitation of CFT

State one limitation of crystal field theory.

Solution: CFT treats the metal-ligand bond as purely electrostatic and ignores the covalent character (orbital overlap) of the bond, which is significant in many real complexes.

Example 8: Spectrochemical prediction

Does CFT predict the spectrochemical series from first principles?

Solution: No. The order of the spectrochemical series is determined experimentally; CFT cannot derive it (e.g. it cannot explain from electrostatics alone why neutral CO is a stronger-field ligand than the anion F$^-$). This is a limitation of CFT.

Example 9: Colour and oxidation state

How can changing the oxidation state of the metal change the colour?

Solution: A different oxidation state means a different d-electron count and a different effective nuclear charge, which changes $\Delta_o$ and the energy of the d-d transition — and therefore the absorbed wavelength and the observed colour.

Example 10: Identify the coloured ion

Among $[\text{Sc(H}_2\text{O)}_6]^{3+}$, $[\text{V(H}_2\text{O)}_6]^{3+}$ and $[\text{Zn(H}_2\text{O)}_6]^{2+}$, which is coloured?

Solution: $[\text{V(H}_2\text{O)}_6]^{3+}$ — V$^{3+}$ is $d^2$ (partially filled), so it undergoes a d-d transition and is coloured. Sc$^{3+}$ ($d^0$) and Zn$^{2+}$ ($d^{10}$) are colourless.