Predict the formal bond angles about the central atom for the following molecules: TeBr4

Express your answers using three significant figures separated by a comma.

What are the bond angles for an octahedral structure and how many?

Here is a great page for you. For TeBr4 look at the SF4 molecule in the table. For Octahedral, look for the SF6 molecule.

http://en.wikipedia.org/wiki/Molecular_geometry

How do you figure out the number of bond angles for an octahedral? It's more than one answer?

It's listed in the site I gave you.

To determine the bond angles for an octahedral structure, we first need to understand what an octahedral structure is. In an octahedral structure, there are six identical bonding pairs surrounding the central atom, creating a three-dimensional shape with bond angles of 90 degrees.

In the molecule TeBr4, the central atom is Te (tellurium) and it is surrounded by four Br (bromine) atoms. Since there are only four bonding pairs in TeBr4 (instead of six required for an octahedral structure), it does not possess an octahedral structure. Instead, TeBr4 has a square planar molecular geometry.

In a square planar structure, the bond angles are 90 degrees between the central atom and the surrounding atoms. Thus, the bond angles in TeBr4 are all 90 degrees.

Therefore, the bond angles for TeBr4 are 90 degrees.