Colour, magnetism and stability of complexes

Why a solution has a colour

Copper sulphate crystals are blue. Add ammonia solution and the blue turns much deeper. A solution of a zinc salt stays colourless. Where does the colour come from?

White light holds all the colours of the rainbow. A coloured solution takes in some of them and lets the rest through. The colour you see is what is left.

Colour from d-d transitions

In a complex, the d electrons sit in two sets of levels with a gap $\Delta$ between them. A photon whose energy equals the gap can push an electron from the lower set up to the upper set. This is a d-d transition. The solution absorbs that colour of light.

Take $[\text{Ti(H}_2\text{O)}_6]^{3+}$. The $\text{Ti}^{3+}$ ion has one d electron, in the $t_{2g}$ set. The gap matches green light of about $500\ \text{nm}$. The ion absorbs green and lets the blue and the red through. The solution looks purple.

The colour you see is the colour opposite the absorbed one on the colour wheel. A complex that absorbs red looks green. A complex that absorbs orange looks blue.

The wavelength and the splitting

A photon of wavelength $\lambda$ has energy $E = hc/\lambda$. For one mole of photons, multiply by the Avogadro constant, $6.02 \times 10^{23}\ \text{mol}^{-1}$. When the wavelength is written in nanometres, this gives:

For green light of $500\ \text{nm}$, $\Delta = 119600/500 = 239.2\ \text{kJ/mol}$. A shorter wavelength means a larger splitting.

For the d-d band of a $d^1$ or a $d^3$ ion, the photon energy is the splitting $\Delta_o$. In other ions the band also depends on the repulsion between electrons, so we read it only as a guide.

A stronger ligand changes the colour

Replace water by ammonia on $\text{Cr}^{3+}$. The splitting grows, so the band moves to a shorter wavelength.

• $[\text{Cr(H}_2\text{O)}_6]^{3+}$ absorbs at about $575\ \text{nm}$ (yellow) and looks violet.
• $[\text{Cr(NH}_3)_6]^{3+}$ absorbs at about $465\ \text{nm}$ (blue) and looks yellow.

The blue of copper sulphate deepens with ammonia for the same reason. Ammonia is higher than water on the spectrochemical series.

No d-d transition, no colour

An electron can jump only if there is a place for it to go. So:

• A $d^0$ ion, such as $\text{Sc}^{3+}$ or $\text{Ti}^{4+}$, has no d electron to move.
• A $d^{10}$ ion, such as $\text{Zn}^{2+}$ or $\text{Cu}^+$, has no empty d orbital to move to.

Both kinds of complex are colourless. That is why zinc salts are white and why $\text{Cu}^+$ compounds are mostly white.

Some complexes are intensely coloured even though the metal has no d electrons. The permanganate ion, $\text{MnO}_4^-$, is deep purple, and manganese is $d^0$ there. A different process causes this colour: an electron moves from an oxygen atom to the metal. We call it charge transfer.

Set the wavelength to $500\ \text{nm}$ and note the splitting. Then move it to $600\ \text{nm}$. The splitting falls and the colour changes. Use the second slider to see when a $d^6$ ion becomes low spin.

Practise with new numbers

Try it in Python

The program turns the wavelength a complex absorbs into the splitting and the colour you see.

Magnetism from the splitting

The splitting also decides the magnetic behaviour. A strong field gives low spin and few unpaired electrons. A weak field gives high spin and many. A measured moment therefore tells you the spin state.

To go from the moment back to the number of unpaired electrons, remember that $\mu^2 = n(n+2)$. A moment of $2.83\ \text{BM}$ gives $\mu^2 = 8$, so $n = 2$.

An iron(II) complex with $\mu = 0$ is low spin: all six 3d electrons are paired. One with $\mu = 4.90\ \text{BM}$ has four unpaired electrons and is high spin. The magnet shows what the colour alone cannot.

How stable is a complex?

Dissolve a metal salt and add a ligand, and the complex forms step by step. For four ligands:

The overall stability constant $\beta_4$ is the product of the four stepwise constants. A large constant means that the complex forms nearly completely, and that it hardly dissociates. At $298\ \text{K}$ the free energy change follows from the constant:

Here $R = 8.314\ \text{J/(mol K)}$. A larger $\log K$ gives a more negative $\Delta G^\circ$.

Three things make a complex more stable:

• A highly charged, small metal ion. It attracts the ligands more strongly.
• A strongly basic ligand. It is a better electron pair donor.
• The chelate effect. A ligand that binds through two or more atoms and forms rings gives a more stable complex than the same number of single-atom ligands.

Here is an example of the chelate effect. The complex $[\text{Ni(NH}_3)_6]^{2+}$ has $\log K$ of about $8.7$. The complex $[\text{Ni(en)}_3]^{2+}$ has six nitrogen atoms around nickel too, but $\log K$ is about $18.3$. Its constant is more than a billion times larger. The reason is disorder. Each en molecule replaces two water molecules, so the number of free particles rises. More particles means more disorder, and the free energy falls.

Stability has two faces. A complex can be stable in energy but quick to swap its ligands, which we call labile. Another can be unstable in energy but slow to change, which we call inert. The ion $[\text{Co(NH}_3)_6]^{3+}$ is of the second kind. In acid it ought to fall apart, yet it lasts for days.

Practise with new numbers

A look at metal carbonyls

Carbon monoxide bonds to metals in two ways at once. The carbon atom gives its lone pair to an empty orbital of the metal, which is the usual coordinate bond. The metal then gives some of its own d electrons back, into an empty antibonding orbital of CO. This is back bonding.

Each bond helps the other. The metal-carbon bond becomes stronger, and the carbon-oxygen bond a little weaker. We call this a synergic bond. It explains why CO is such a strong field ligand.

Three examples: $[\text{Ni(CO)}_4]$ is tetrahedral, $[\text{Fe(CO)}_5]$ is trigonal bipyramidal and $[\text{Cr(CO)}_6]$ is octahedral. In every one the metal is in the zero oxidation state.

Where complexes are used

• In living things. Haemoglobin holds iron, chlorophyll holds magnesium and vitamin $\text{B}_{12}$ holds cobalt.
• In medicine. The complex cis-$[\text{Pt(NH}_3)_2\text{Cl}_2]$, called cisplatin, treats cancer. The ligand EDTA binds lead ions in lead poisoning.
• In analysis. Nickel(II) ions give a scarlet red solid with dimethylglyoxime. Iron(III) ions and thiocyanate ions give a blood red colour. EDTA is used to titrate the hardness of water.
• In extracting metals. Silver and gold dissolve in cyanide solution as $[\text{Ag(CN)}_2]^-$ and $[\text{Au(CN)}_2]^-$. Nickel is purified through the volatile $[\text{Ni(CO)}_4]$.
• In photography. Sodium thiosulphate dissolves unused silver bromide from the film as $[\text{Ag(S}_2\text{O}_3)_2]^{3-}$.
• In electroplating. The $[\text{Ag(CN)}_2]^-$ ion releases silver slowly, which gives a smooth coat.
• In industry. Complexes of rhodium and titanium act as catalysts, for example in the making of polythene.

Common mistakes

• Thinking the colour you see is the colour absorbed. It is the opposite colour on the wheel.
• Saying a longer wavelength means a larger splitting. The splitting grows as the wavelength gets shorter.
• Expecting colour from $d^0$ and $d^{10}$ ions. They have no d-d transition.
• Taking a small $K$ for a stable complex. A large stability constant means more stable.
• Forgetting the minus sign in $\Delta G^\circ = -5.706 \times \log K$.

Remember these

• Colour comes from a d-d transition. The colour seen is opposite the colour absorbed.
• $\Delta = 119600/\lambda$ in kJ/mol, with $\lambda$ in nm.
• $d^0$ and $d^{10}$ complexes are colourless.
• A larger $K$ means a more stable complex. Chelates are the most stable.
• CO bonds by donation and by back bonding together.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A complex absorbs red light. What colour does its solution show?

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

3. Ligand X gives a complex that absorbs at $450\ \text{nm}$. Ligand Y gives one that absorbs at $600\ \text{nm}$. Which is higher on the spectrochemical series?

4. Why is $[\text{Ni(en)}_3]^{2+}$ much more stable than $[\text{Ni(NH}_3)_6]^{2+}$?