What is the hybridization of chromium in chromate and dichromate ions?
The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:


$\mathrm{t_{2(g)}}: \mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$
$\mathrm{e_{(g)}}: \mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$


In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).

In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for &#x3c0;-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:


Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand &#x3c0; contributions.

The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).

Don’t worry, these schemes will come to haunt you at one point when studying chemistry.