Why do charge transfer insulators exist?

Why is the anion's p-band located below the lower Hubbard band?

It's all about the competing energetics of electron transfer. In the Mott-Hubbard case, it's easier for electrons to transfer between two adjacent metal sites (on-site Coulomb interaction U); in the charge-transfer case, it's easier from the anion to the metal (charge-transfer energy $\Delta$). U is determined by repulsive/exchange effects between the cation valence electrons. $\Delta$ is tuned by the chemistry between the cation and anion. Below is a simple picture for thinking about these interactions.

Determining the size of the on-site Coulomb interactions U is rather complicated. A cheap approximation is that it's the energy for two metals $M^{m+}$ to go to $M^{(m+1)+}$ and $M^{(m-1)+}$. The LHB therefore reflects the metal electron energy in the former configuration and the UHB reflects the metal electron energy in the latter configuration. So a rough intuitive estimate for U is the difference between the metal's ionization potential and electron affinity.

The charge-transfer gap is simply the electronegativity difference between the cation and anion in the crystal lattice -- technically different from atomic electronegativity, although similar in many respects. Anions are naturally unstable in vacuum. In contrast to cations, they require the surrounding crystal lattice to stabilize their electronic energy, which alters their electronegativity. The electrostatic potential of the surrounding ions (the Madelung potential) reduces their electron energy and increases their electronegativity. This is what brings the p states lower in energy than the metal d states to begin with. Conversely, the Madelung potential destabilizes the transition metal d states, raising them in energy: the cations are surrounded by negatively charged anions, making it more difficult for electrons to occupy those states. Below is a schematic of how the bands are formed from atomic orbitals in a transition metal oxide with octahedral oxygen coordination.

Thus, one can tune the size of the charge-transfer gap $\Delta$ by using the Madelung potentials, the electronegativity of the cation, and the electronegativity of the anion.

Based on trends in ionization potentials/electron affinities and the estimates described above, we can therefore see that late metals or metals with high oxidation states will have high U and small $\Delta$, resulting in the LHB lying below the anion p-band (at least for oxides). On the other hand, early metals or metals with low oxidation states will have lowU and large $\Delta$, which is consistent with Mott-Hubbard behavior.

Technically, hybridization between the metal and anion also plays a critical role. However, there are quite a few papers (one from Zaanen, Sawatzky, and Allen themselves) using this simple ionic interpretation to provide intuitively satisfying understanding.

Reference:   https://www.quora.com/Why-do-charge-transfer-insulators-exist