A complex is represented as COCl3XNH3. Its 0.1 molal solution in water shows ${ΔT}_{f} = 0 .558 K . K_{f}$ for H₂O is 1.86K molality^–1. Assuming 100% ionization of complex and co-ordination number of Co is six, calculate formula of complex:

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[Co(NH₃)₄Cl₂ ]Cl

[Co(NH₃)₆]Cl₃

[CoCl₆]NH₃

[Co(NH₃)₅Cl]Cl₂

answer is B.

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Detailed Solution

$\begin{array}{l} {ΔT}_{f} = K_{f} \times molality \times i \\ 0 .558 = 1 .86 \times 0 .1 \times i or i = 3 \end{array}$
Thus complex is [CoCl.xNH₃].Cl₂
Since co. no. Co is six, thus complex is [CO(NH₃)₅Cl]Cl₂.

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