CrCl₃.xNH₃ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558ºC. 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)

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$[Cr {({NH}_{3})}_{5} Cl] {Cl}_{2}$

$[Cr {({NH}_{3})}_{3} {Cl}_{3}]$

$[Cr {({NH}_{3})}_{6}] {Cl}_{3}$

$[Cr {({NH}_{3})}_{4} {Cl}_{2}] Cl$

answer is C.

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Detailed Solution

$\begin{aligned} Given : {ΔT}_{f} = {0.558}^{\circ} C \\ \begin{aligned} k_{f} = 1.86 \frac{K \times kg}{mol} \\ 0.1 m aq. sol . \\ \Rightarrow & {ΔT}_{f} = i \times k_{f} \times m \\ \Rightarrow & 0.558 = i \times 1.86 \times 0.1 \\ \Rightarrow & i = 3 \end{aligned} \end{aligned}$