Question

Identify the precipitate and its quantity obtained when 1 mole of aqueous solution of Tetraammine dichloroplatinum(IV) bromide is treated with aqueous silver nitrate in excess.

• A. Two moles of silver chloride
• B. Two moles of silver chloride and one mole of silver bromide
• C. One mole of silver bromide
• D. Two moles of silver bromide

Hint:

In reactions where silver nitrate is added to a platinum complex containing bromide ions, silver bromide precipitates out. The number of moles of silver bromide depends on the number of bromide ions in the complex.

Answer 1

The Correct Option is D

Step 1: Understanding the reaction.
When Tetraammine dichloroplatinum(IV) bromide (\([Pt(NH_3)_4Cl_2]Br_2\)) reacts with excess silver nitrate (AgNO₃), it forms a precipitate. The bromide ions (Br⁻) from the platinum complex react with silver ions (Ag⁺) to form silver bromide (AgBr) as a precipitate. Since two bromide ions are present in the complex, two moles of silver bromide are formed.
Step 2: Analyzing the options.
(A) Two moles of silver chloride: Incorrect. This is not the product because chloride ions are not in excess here; bromide ions react.
(B) Two moles of silver chloride and one mole of silver bromide: Incorrect. Only silver bromide forms as the precipitate, not silver chloride.
(C) One mole of silver bromide: Incorrect. Two moles of silver bromide will be formed due to two bromide ions in the platinum complex.
(D) Two moles of silver bromide: Correct — Two moles of silver bromide are formed, one from each bromide ion in the complex.
Step 3: Conclusion.
The correct answer is (D) Two moles of silver bromide, as two bromide ions react with silver ions to form two moles of silver bromide.