CrCl₃ · xNH₃ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558° C. Assuming 100 % ionisation of this complex and coordination number of Cr is 6 , the complex will be (Given K_f=1.86 K kg mol⁻¹ )

Solution

Chemistry ma’am · 3 steps

Get the van't Hoff factor from the freezing point, then pick the complex that breaks into that many ions.

1. Find i. Depression in freezing point depends on the total number of particles, so use Δ T_f = i K_f m and solve for the van't Hoff factor i.

   i = (Δ T_f)/(K_f m) = 0.558/(1.86 × 0.1) = 0.558/0.186 = 3

2. What i means. With 100% ionisation, i = 3 means one formula unit gives 3 ions in water. Only the Cl⁻ outside the square bracket comes off as free ions. The ligands inside the bracket stay bound to Cr.

   Complex	Ions in water	i	Δ T_f
   [Cr(NH₃)₅Cl]Cl₂	[Cr(NH₃)₅Cl]²⁺ + 2Cl⁻	3	0.558 °C
   [Cr(NH₃)₄Cl₂]Cl	[Cr(NH₃)₄Cl₂]⁺ + Cl⁻	2	0.372 °C
   [Cr(NH₃)₆]Cl₃	[Cr(NH₃)₆]³⁺ + 3Cl⁻	4	0.744 °C
   [Cr(NH₃)₃Cl₃]	does not ionise	1	0.186 °C

3. So. The complex is [Cr(NH₃)₅Cl]Cl₂, option (a). Five NH₃ and one Cl fill the coordination number of 6, and two Cl⁻ sit outside as counter ions.

The idea to keep. For a Werner complex, i = 1 (the complex ion) + the number of counter ions outside the bracket. Get i from the colligative data, then count.

Answer from the NTA final answer key. Solution by PrepWiser.