Why are transition metal ions in water coloured, but sodium in water is not?
The colour of light is determined by its wavelength. White light is made up of electromagnetic radiation (photons) that spans the range 400nm - 700nm, the visible part of the E.M. spectrum. When white light changes to, for example, red when passing through a solution of Ni2+(aq), there has been a corresponding change in the distribution of wavelengths of the light. This change has occurred because some wavelengths of light incident on the solution has interacted with the Ni2+(aq) ions (pure water is colourless). Red light can be produced from white light, by filtering out all green light - green is opposite red on the colour wheel. So the solution may have filtered out the green component of white light.
How could the solution filter out the green light? Where did the light go? We know that light comes in discrete quanta of energy called photons, who's energy is related to the wavelength of light by E=hc/l (I will use l instead of lambda). By conservation of energy, if the green photons 'disappeared', there must have been a corresponding energy transfer. Taking a closer look at the divalent nickel cation, we see that the d sub shell is split by the surrounding water molecules into two new sub shells. It is no coincidence that the energy gap between the two new sub shells, formed by the splitting, corresponds to the energy of a green photon. When white light is shone on Ni2+(aq), some of the green light is absorbed by electrons residing in the the lower energy level of the split d sub shell which are excited to the higher energy level. The energy gaps caused by the splitting of the d sub shell of all first row transition metals corresponds to somewhere in the visible spectrum. For Na+(aq) there are no available electronic excitations which correspond to photons in the visible region of light.
