For 1 molal solution of each compound minimum freezing point will be assuming complete ionisation in each case:

[F e ((H_{2} {O)}_{6}] {C l}_{3}

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F e (H_{2} {O)}_{5} C l] {C l}_{2} . H_{2} O

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F e (H_{2} {O)}_{4} C l_{2}] C l .2 H_{2} O

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F e (H_{2} {O)}_{3} C l_{3}] . H_{2} O

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Solution

The correct option is D $F e (H_{2} {O)}_{3} C l_{3}] . H_{2} O$
$∵$ , Maximum freezing point $\propto \frac{1}{N o . o f d i s s o c i a t e d p a r t i c l e s}$
So, $[F e (H_{2} O)_{6}] C l_{3}, [F e (H_{2} O)_{5} C l] C l_{2} . H_{2} O$ and $[F e (H_{2} O)_{4} C l_{2}] C l .2 H_{2} O$ has $C l^{-}$ ion which can be dissociated.
But $[F e (H_{2} O)_{3} C l_{3}] .3 H_{2} O$ does not have any $C l^{-}$ ion to get dissociated thereby having maximum freezing point.
$∴$ Answer is option D $[F e (H_{2} O)_{3} C l_{3}] . H_{2} O$

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