Answer
$[Pt (en)_{2} Cl_{2}] Br_{2}$
Work Step by Step
The formula is written according to the rules of writing the formula for complex compounds.
1. dichlorobis(ethylenediamine)platinum(IV) is the cation and bromide is the anion. Cations are written first then anion.
2. dichlorobis(ethylenediamine) is the ligand whereas platinum(IV) is the metal.
3. dichlorobis(ethylenediamine):
Dichloro means two chlorine atoms, bis(ethylenediamine) means 2 ethylenediamine, First bis(ethylenediamine) is written then dichloride is written. Thus, $[Pt (en)_{2} Cl_{2}]$
4. Platinum (IV) means platinum has +4 charge. The total charge on the cation is:
$+4 + 2 \times 0 + 2 \times (-1)$
$+4-2$
$+2$
4. Bromide is the anion. It has a -1 charge. In order to balance, we need two bromides. Thus, $Br_{2}$
Therefore, the formula of the complex is:
$[Pt (en)_{2} Cl_{2}] Br_{2}$