A complex is represented as ${CoCl}_{3} x {NH}_{3^{-}}$ lts 0.1 molal solution in water shows $- {ΔT}_{f} = 0 .558 K . K_{f}$ for $H_{2} O$ is 1.86 K molality $^{- 1}$ . Assuming 100% ionization of complex and co-ordination number of Co is six, calculate formula of complex.

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

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

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

None of these

answer is B.

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Detailed Solution

$- {ΔT}_{f} = K_{f} x molality x i; 0.558 = 1.86 x 0.1 x i or i = 3$ Thus complex is $[{CoClxNH}_{3}] {Cl}_{2}$

since co-ordination number of CO is six, thus complex is $[CO {({NH}_{3})}_{5} Cl] {Cl}_{2}$

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