The given name is Tetraamminedibromoplatinum(IV) bromide.
Step 1: Identify the ligands and metal oxidation state.
- “Tetraammine” → 4 \( \text{NH}_3 \) ligands (neutral).
- “Dibromo” → 2 \( \text{Br}^- \) ligands (inside the coordination sphere).
- “Platinum(IV)” → Metal oxidation state = +4.
- “Bromide” (at the end of the name) → counter anion(s) outside the coordination sphere.
Step 2: Write the partial complex formula: \[ [\text{PtBr}_2(\text{NH}_3)_4]^{?} \]
Step 3: Determine the charge on the complex ion. Let the overall charge on the coordination sphere be \( x \). \[ x + (2 \times -1) + (4 \times 0) = +4 \] \[ x - 2 = +4 \implies x = +6 \] This would be incorrect since Pt(IV) rarely forms +6 complexes. Let’s reconsider the ligand count relative to oxidation state.
In the complex \( [\text{PtBr}_2(\text{NH}_3)_4] \), platinum has oxidation number: \[ x + (2 \times -1) = +4 \implies x = +6 \] Wait — correction: for a **Pt(IV)** ion, the total oxidation number is +4. So, \[ +4 = x + (-2) \implies x = +6 \] That means two bromides inside (each -1) give the total +4 charge for platinum. The coordination sphere has a net charge of +2.
Therefore, to balance the charge, there must be **two bromide ions** outside the coordination sphere. \[ [\text{PtBr}_2(\text{NH}_3)_4]\text{Br}_2 \]
Hence, the correct formula is: \[ \boxed{[\text{PtBr}_2(\text{NH}_3)_4]\text{Br}_2} \]