Question:

The correct formula of a complex having IUPAC name Tetraamminedibromoplatinum (IV) bromide is _______.

  • \( [\text{PtBr}(\text{NH}_3)_4] \text{Br}_2 \)
  • \( [\text{PtBr}_2(\text{NH}_3)_4] \text{Br} \)
  • \( [\text{PtBr}_2(\text{NH}_3)_4] \text{Br}_2 \)
  • \( [\text{PtBr}(\text{NH}_3)_4] \text{Br} \)
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The Correct Option is C

Solution and Explanation

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: \[ \boxed{[\text{PtBr}_2(\text{NH}_3)_4]\text{Br}_2} \]

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