Answer
The highest occupied subshell for bromine is $4p$
Work Step by Step
The element bromine $(Z=35)$ has $35$ electrons. In its part of the periodic table, the subshells of the electronic states are filled in the sequence
$1s\;2s\;2p\;3s\;3p\;3d\;4s\;4p....$
In the given sequence, the electronic configuration of bromine becomes
$1s^2\;2s^2\;2p^6\;3s^2\;3p^6\;3d^{10}\;4s^2\;4p^5$
Therefore, the highest occupied subshell for bromine is $4p$.
