We usually count those element that don't have d orbitals as not being able to expand the octet.
So then it would be all of them except Br right?
Don't P and S have expanded octets?
For example SF6 and PF5.
Here is something I found that may prove useful.
In my first response, I believe I was thinking of hybridization. Atoms with no d orbitals can form s and p hybrids but not s, p, and d hybrids (since no d orbitals are there.)
I know I was and that is correct. An expanded octet cannot occur with atoms less than atomic number 10 (neon) because there are no d orbitals. So you are correct; I believe Br is the only one with an atomic number larger than 10. I brought up P and S but those elements ARE greater than 10 AND since they are in the third period, they contain 3d orbitals and extra electrons can fit around them. Therefore, my first response was ok; d orbitals are necessary for hybridization to take place (involving d, s, and p orbitals) AND d orbitals are necessary to accommodate and expanded octet. I hope this helps.