In a bromine atom, you have the electron configuration #[Ar] 4s^2 3d^10 4p^5#.
The only half-filled orbitals could be one of the #bb(4p)# orbitals.
You can convince yourself that since each orbital has a maximum of #2# electrons, and there are one #4s# and five #3d# orbitals, both those subshells are doubly-occupied in each orbital, seeing as how there are #2# electrons in the #4s# orbital and #10# in the #3d# orbitals.
(Hence, the #4s# and #3d# are all full.)
Therefore, we examine the #4p# orbitals and fill them one at a time, then double up afterwards, in accordance with Hund's Rule:
#ul(uarr darr) " " ul(uarr darr) " " ul(uarr color(white)(darr))#
#4p_x " "" " 4p_y " "" " 4p_z#
So now, how many #4p# orbitals are half-filled? At this point you should be able to answer your own question. :-)