Why is chromate coloured but tungstate colourless?
Both complexes (and many more such as permanganate, sulphate, …) share the same general MO scheme which I am going to shamelessly copy from my older answer:


Figure 1: Qualitative MO scheme of a tetrahedric complex with &#x3c3; and &#x3c0; bonding between metal and ligands. Double vertical lines represent electron pairs.

The intense colour of permanganate and chromate is due to LMCT transitions: ligand-to-metal charge transfer. Essentially, an electron from the HOMO (a $\mathrm{t_1}$ orbital that can be thought of a ligand-centred p-type orbital) is excited to the LUMO (an $\mathrm{e}$-type orbital that can be thought of one of the metal’s empty d orbitals) by absorbing a photon from the visible range; we then see the complementary colour, meaning that chromate absorbs blue light and looks yellow to our eyes.

The energies of LUMO and HOMO – or rather, the energy difference between the two – determines the wavelength of the absorbed photon and thus the colour that we see. For both permanganate and chromate, the wavelength is well within the visible spectrum resulting in the strongly coloured complexes. In cases such as sulphate or tungstate, the absorption wavelength is in the ultraviolet range which cannot be perceived by human eyes; therefore, the solution appears uncoloured.

Without computer calculations, it is probably impossible to accurately predict the ultimate wavelength but note that such an absorption is practically always present if the corresponding orbitals are available although usually in the ultraviolet range. An obvious counterexample would be any zinc complex in which the metal has a $\mathrm d^{10}$ configuration and the LUMO is thus the much further removed $\mathrm a_1$ orbital; therefore, zinc complexes are typically colourless and do not display such an LMCT band.