Answer
Tetraaqua(oxalato)platinum(IV) bromide.
Work Step by Step
1. In $[Pt(H_{2}O)_{4}(C_{2}O_{4})]Br_{2}$ , $[Pt(H_{2}O)_{4}(C_{2}O_{4})]$is the cation whereas $Br_{2}$ is the anion. Cations are named first then are anions.
2. In order to name the cation, Ligand is named first then is the metal. Prefixes are added and Ligands are alphabetically named. $(H_{2}O)_{4}$ is tetraaqua $(C_{2}O_{4})$ is oxalato
3. Pt is platinum.
4. In order to find the oxidation number on the metal, we add the charges and equal it to the charge on the complex, in this case zero. Also, Water has no charge, oxalate has $-2$ charge, and bromide has $2\times(-1)$. Thus,
$x-2+2\times(-1)=0$
$x-2-2=0$
$x-4=0$
$x=+4$
The charge is +4 on the metal. Hence, the name of the complex is Tetraaqua(oxalato)platinum(IV) bromide.
