Identify the correct order of flocculation value for $F e (O H)_{3}$ sol:

K B r < K_{2} C_{2} O_{4} < K_{3} [F e (C N)_{6}] < K_{4} [F e (C N)_{6}]

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K B r = K_{2} C_{2} O_{4} = K_{3} [F e (C N)_{6}] = K_{4} [F e (C N)_{6}]

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K B r < K_{4} [F e (C N)_{6}] < K_{2} C_{2} O_{4} < K_{3} [F e (C N)_{6}]

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K_{4} [F e (C N)_{6}] < K_{3} [F e (C N)_{6}] < K_{2} C_{2} O_{4} < K B r

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Solution

The correct option is D $K_{4} [F e (C N)_{6}] < K_{3} [F e (C N)_{6}] < K_{2} C_{2} O_{4} < K B r$
$F e (O H)_{3}$ sol is positively charged sol. So, coagulation (flocculation) is due to anions.
Flocculation value is defined as the minimum concentration of an electrolyte required to cause the coagulation of the sol.

Coagulating power $α$ $\frac{1}{F l o c c u l a t i o n v a l u e}$

With respect to anions $B r^{-}, C_{2} O_{4}^{- 2}, [F e (C N)_{6}]^{3 -}, [F e (C N)_{6}]^{4 -}$ ,
the increasing order of coagulation value is
$K_{4} [F e (C N)_{6}] < K_{3} [F e (C N)_{6}] < K_{2} C_{2} O_{4} < K B r$

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