Question

1×10-3 m solution of Pt(NH3)4Cl4 in H2O shows depression in freezing point by 0.0054°C .The structure of the compound will be(given Kf= 1,860 km-1)​

Answer 1

Depression in freezing point = Kf*m*i

i = vant hoff constant, m = molality

0.0054 = 1.86* 10^-3* i

i = 3 = no. of dissociable ions

So the structure is [Pt(NH3)2(Cl)2](NH3)2(Cl)2

Answer 2

• Depression in freezing point (ΔTf) = i*Kf*m

• Substituting the given values in the above equation gives, i = 3

• This means that when the coordination complex dissociates, it gives 3 ionic species.

• Since NH₃ is a neutral molecule, it will not bear any charge on dissociation. Therefore, all ammonia molecules should be placed within the coordination sphere.

• This leaves only one appropriate complex, i.e. [Pt(NH₃)₄C₂l]Cl₂.