In this question, we are given four ions, two cations and two anions, and we need to write the empirical formula, which is the smallest possible formula in terms of the quantity of atoms. The ions are:
Ca2+, V5+, Br-, S2-
One way of doing this is by matching their charges, because in a neutral compound, we need to have the total charge of the compound to be equal to 0, therefore we will have:
Ca and Br = CaBr2, two bromides to match the +2 charge of Calcium, this is Calcium bromide
Ca and S = CaS, this is Calcium sulfide
V and Br = VBr5, this is Vanadium (V) bromide
V and S = V2S5, this is Vanadium (V) sulfide