$$CrCl_{3}\cdot xNH_{3}$$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of $$0.558^{\circ}C$$. Assuming 100% ionisation of this complex and coordination number of Cr is 6 , the complex will be (Given $$K_{f}$$ = 1.86 K kg $$mol^{-1}$$)

We use the colligative property of freezing point depression to determine the van’t Hoff factor $$i$$.

The formula for depression in freezing point is $$\Delta T_f = i \cdot K_f \cdot m$$.

Given: $$\Delta T_f = 0.558°C$$, $$K_f = 1.86$$ K kg mol^{-1}, $$m = 0.1$$ molal.

Substituting these values yields $$0.558 = i \times 1.86 \times 0.1$$ so that $$i = \frac{0.558}{0.186} = 3$$.

A van’t Hoff factor of $$3$$ indicates that the complex dissociates into three ions in solution.

Complex $$[Cr(NH_3)_5Cl]Cl_2$$ dissociates into $$[Cr(NH_3)_5Cl]^{2+}$$ and $$2Cl^-$$, giving three ions (i = 3).

Complex $$[Cr(NH_3)_6Cl]Cl_3$$ would involve a coordination number of seven (which is not possible for Cr since CN = 6).

Complex $$[Cr(NH_3)_3Cl_3]$$ is a neutral species and does not produce free ions (i = 1).

Complex $$[Cr(NH_3)_4Cl_2]Cl$$ dissociates into two ions (i = 2).

Since chromium must have coordination number six and the van’t Hoff factor is three, the only viable structure is $$[Cr(NH_3)_5Cl]Cl_2$$.

The correct answer is Option 1: $$[Cr(NH_3)_5Cl]Cl_2$$.