# Predicting color of precipitate

Is there any way of predicting whether a precipitate will be colored? I know that for solutions, transition metal ions with unfilled d-orbitals will have color because of d-orbital splitting in the complex ions. However, this isn't applicable to precipitates because those are just ionic solids, not complex ions.

For example, $\ce{PbI2}$ is a yellow solid, and $\ce{Al(OH)3}$ is white. Is there a way to predict that $\ce{PbI2}$ will be colored while $\ce{Al(OH)3}$ will not be, or is it something we have to memorize to know?

• For the most part you just have to know. You also have to realize that trace impurities can change the color. It also isn't simple since the perceived color also depends on particle size. There is the streak test for minerals which typically produces a different color than the bulk mineral specimen. Wikipedia link to "Streak test" -> en.wikipedia.org/wiki/Streak_%28mineralogy%29 – MaxW Dec 27 '15 at 1:12
• Many examples need to be memorized. However many substances have similar colours to their solutions because they are hydrated crystals, e.g. copper(II) sulphate is a blue precipitate but anhydrous copper(II) sulphate is colourless. Since Al(3+) ions have no partially filled d-orbitals you'd reasonably expect it to have no colour, as most (all?) inorganic colour still involves d-orbital transitions. – Spontification Dec 27 '15 at 4:35

1. Yes, d-orbital splitting is part of the reason for colour. However, $\ce{[FeF6]^3-}$ displays a beautiful d-orbital splitting which could display colour from the energy difference — but doesn’t. There is more to colour than d-orbital splitting; the most important terms are spin-forbidden and Laporte rules.
2. You somehow got the impression that ionic solids have nothing to do with coordination complexes. Please remove this idea from your brain; it is wrong. Ionic solids are equally coordination complexes as present in solution only that the environment and sometimes the composition of the complex is different. E.g, $\ce{CuSO4(aq)}$ is essentially a solvated sulphate ion plus $\ce{[Cu(H2O)6]^2+}$. When precipitated or crystallised as ‘$\ce{CuSO4 . 5H2O}$’, what you actually got was $\ce{[Cu(H2O)4(\mu{-}SO4)2].H2O}$ — clearly a coordination compound.
To answer your question, it depends on the nature of the compound you are precipitating — the colour can derive from cation, anion or their combination. For example, aluminium salts will be colourless unless the anion dictates otherwise. Likewise for zinc salts. For a very large number of salts where the cation is responsible for the colour, these will be the same as in solution. But sometimes ($\ce{PbI2, Fe4[Fe(CN)6]3}$ and others) you just have to know because only this specific combination will give this colour.