Question

$Pt{Cl}_4.6H_{2}O$ can exist as a hydrated complex; $1m$ aqueous solution has the depression in freezing point of ${3.72}^{o}C$. Assume $100\%$ ionization and $K_f\left(H_{2}O\right)={1.86}^o{mol}^{-1}kg$, then the complex is:

• A. $\left[Pt{\left(H_{2}O\right)}_6\right]{Cl}_4$
• B. $\left[Pt{\left(H_{2}O\right)}_4{Cl}_2\right]{Cl}_2\cdot 2H_{2}O$
• C. $\left[Pt{\left(H_{2}O\right)}_3{Cl}_3\right]Cl\cdot 3H_{2}O$
• D. $\left[Pt{\left(H_{2}O\right)}_2{Cl}_4\right]\cdot 4H_{2}O$

Answer 1

The correct option is C $\left[Pt{\left(H_{2}O\right)}_3{Cl}_3\right]Cl\cdot 3H_{2}O$

$\Delta T_f=Molality\times K_f\times i$

$⟹3.72=1\times 1.86\times i$
$⟹i=2$ , hence complex is binary.
Also, $CN$ of $Pt$ is 6 and complex is hydrated.
so complex is $\left[Pt{\left(H_{2}O\right)}_3{Cl}_3\right]Cl\cdot 3H_{2}O$