hybridization - Color of chromate and permanganate

I've heard quite a few times that the chromate and permanganate have a $d^3s$ configuration. Also, their colors arise due to a rapid switching of electrons between the oxygen and metal atoms.

I don't really understand the 'rapid switching' part--it's obvious why it can give color, but I fail to see why there is a need for such a switching-what's so special about $\mathrm{Cr}$ and $\mathrm{Mn}$? (I also do not know what the switching exactly is)

An explanation of $d^3s$ would be appreciated, though not necessary.

Answer

By "rapid switching" they technically mean Ligand-to-Metal Charge Transfer (LMCT). A more modern framework is Ligand Field Theory. I would have to teach a class to fully explain it in those terms, but I'll try to explain it in terms of this hybridization you brought up.

A chemical bond implies a higher probability of finding electrons between two bound nuclei. Atomic orbitals describe the electron density for individual nuclei. A bond between two nuclei necessitates some physical overlap of the relevant atomic orbitals.

It is in this way that we say that the linear combination of atomic orbitals, say two consisting of a $2s$ and $2p$ orbital, results in a molecular orbital that describes the molecule. We would write the wave function for the complex as:

$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p)$$

The amount of "mixing" is just a matter of adjusting the coefficients, $C_n$. And so we might say that the bond in this made-up molecule is $sp$ hybridized. $sp^3$ hybridized would mean the following kind of wave function: $$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p) \pm C_3 \psi(2p) \pm C_4 \psi(2p)$$

Similarly, $sd^3$ could mean a wave function of the following nature:

$$\Psi = C_1 \psi( (n+1) s) \pm C_2 \psi(n d) \pm C_3 \psi(nd) \pm C_4 \psi(nd) $$ where $n=3$ in the case of Mn or Cr.

If one performs the proper molecular orbital calculations on the valence $3d$, $4s$, and $4p$ for manganese or chromium and the $2s$ and $2p$ for oxygen in tetrahedral symmetry then you can draw the molecular orbital energy diagram for your complex once you match your eigenvalues to the correct trace of the appropriate transformation matrix... and you can hear a bunch more jargon on it, or I can explain the main point of our models: why we think MnO$_4$ $^-$ is purple.

The bonds in, specifically permanganate (I have never actually done the calculations on dichromate, although they should be similar in principle) are between Mn$^{+7}$ and four O$^{-2}$ in the geometry shown here. The electronegativity of oxygen clearly dictates most of the electron density and, as a result, has them sitting on ligands (oxygen) instead of the metal. An electronically excited state can be reached with light absorption in the range of 500-600 nm light due to the relative weakness of the interactions between the ligand and metal. If they were stronger interactions then it would take more energy to promote electrons into a higher energy state and the color would be shifted into the UV. A colorwheel tells you that light absorption in the 500-600nm range should be approximately violet, which is what we see for the color of this particular complex. Anyway, this excited state means some electrons temporarily move from the ligand to the metal, resulting in a LMCT band.