1 × 10-3 m solution of Pt ( NH 3)4 Cl 4 in H 2 O shows depression in freezing point by 0.0054° C. The structure of the compound will be [Given, Kf(H2 O)=1.860 km -1 ]

Question

1 × 10-3 m solution of Pt ( NH 3)4 Cl 4 in H 2 O shows depression in freezing point by 0.0054° C. The structure of the compound will be [Given, Kf(H2 O)=1.860 km -1 ] (A)  [Pt(NH3)4]Cl4  (B)  [Pt(NH3)3]Cl3  (C)  [Pt(NH3)2Cl2]Cl2  (D)  [Pt(NH3)Cl3]Cl

Tardigrade Admin · Accepted Answer

Δ Tf=i Kf m or i=(Δ Tf/Kf × m)=(0.0054/1.86 × 1 × 10-3)=2.9 =i ≈ 3 Thus, each molecule of the complex dissociates to form three ions. This can be so only if the formula of the complex is [ Pt ( NH 3)2 Cl 2] Cl 2.

Q. $1 \times 1 0^{- 3} m$ solution of $Pt (N H_{3})_{4} C l_{4}$ in $H_{2} O$ shows depression in freezing point by $0.005 4^{\circ} C$ . The structure of the compound will be [Given, $K_{f} (H_{2} O) = 1.860 k m^{- 1}$ ]

$[Pt (N H_{3})_{4}] C l_{4}$

$[Pt (N H_{3})_{3}] C l_{3}$

$[Pt (N H_{3})_{2} C l_{2}] C l_{2}$

$[Pt (N H_{3}) C l_{3}] Cl$

$Δ T_{f} = i K_{f} m$
or $i = \frac{Δ T _{f}}{K _{f} \times m} = \frac{0.0054}{1.86 \times 1 \times 1 0 ^{- 3}} = 2.9$
$= i \approx 3$ Thus, each molecule of the complex dissociates to form three ions.
This can be so only if the formula of the complex is $[Pt (N H_{3})_{2} C l_{2}] C l_{2}$ .

Q. 1×10−3m solution of Pt(NH3​)4​Cl4​ in H2​O shows depression in freezing point by 0.0054∘C. The structure of the compound will be [Given, Kf​(H2​O)=1.860km−1 ]

[Pt(NH3​)4​]Cl4​

[Pt(NH3​)3​]Cl3​

[Pt(NH3​)2​Cl2​]Cl2​

[Pt(NH3​)Cl3​]Cl