Question

$1 \times 10^{-3} m$ solution of $Pt \left( NH _{3}\right)_{4} Cl _{4}$ in $H _{2} O$ shows depression in freezing point by $0.0054^{\circ} C$. The structure of the compound will be [Given, $K_{f}\left(H_{2} O\right)=1.860 \,km ^{-1}$ ]

• A. $ [Pt(NH_{3})_{4}]Cl_{4} $
• B. $ [Pt(NH_{3})_{3}]Cl_{3} $
• C. $ [Pt(NH_{3})_{2}Cl_{2}]Cl_{2} $
• D. $ [Pt(NH_{3})Cl_{3}]Cl $

Answer 1

Answer: (C)

$\Delta T_{f}=i K_{f} m$
or $i=\frac{\Delta T_{f}}{K_{f} \times m}=\frac{0.0054}{1.86 \times 1 \times 10^{-3}}=2.9$
$=i \approx 3$ Thus, each molecule of the complex dissociates to form three ions.
This can be so only if the formula of the complex is $\left[ Pt \left( NH _{3}\right)_{2} Cl _{2}\right] Cl _{2}$.