This is in a chart and I don't know what I'm supposed to do
H-O-Br with two lone pairs on the central O. There are thus 4 electron rich areas with two bonds and 2 lone pairs.
Electron pair geometry = tetrahedral
Molecular geometry = bent (109.5 degree angle)
Here's an image of the shape of HOBr:
You get to that diagram by writing the Lewis structure (satisfying the bonding requirements for H, O, and Br as well as determining the central atom) of HOBr and then reasoning out the distribution of the bonding and non-bonding electron pairs using the ideas in VSEPR.
The electron pair geometry is tetrahedral and the molecular shape is angular (or bent). The VSEPR shorthand would be AX2E2.
Edit: the electron pair geometry takes all electrons around the central atom into consideration while the molecular shape looks only at the distribution of the atoms in space (ignoring the non-bonding electrons).