Question

The number of chloride ions which would be precipitated when the complex PtCl4.4NH3 is treated with silver nitrate is:(here coordination number of platinum is 6).

• A. Four
• B. One
• C. Three
• D. Two

Answer 1

Answer: (D)

Two

Answer 2

PtCl4.4NH3 º [Pt(NH3)4Cl2]Cl2 $\stackrel { \text { aq. } } { \rightleftharpoons }$ [Pt(NH3)4Cl2]2+ (aq) + 2Cl– (aq).

2Ag+ + 2Cl– $\longrightarrow$ 2AgCl ¯ (white).