As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:

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$K_{3} [Fe {(CN)}_{6}] < K_{2} {CrO}_{4} < {AlCl}_{3} < KBr < {KNO}_{3}$

$K_{3} [Fe {(CN)}_{6}] > {AlCl}_{3} > K_{2} {CrO}_{4} > KBr > {KNO}_{3}$

${AlCl}_{3} > K_{3} [Fe {(CN)}_{6}] > K_{2} {CrO}_{4} > KBr = {KNO}_{3}$

$K_{3} [Fe {(CN)}_{6}] < K_{2} {CrO}_{4} < KBr = {KNO}_{3} = {AlCl}_{3}$

answer is D.

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Detailed Solution

Ferric Hydroxide is positive sol. Therefore negative ion is effective for coagulation. More is the –ve charge, greater is the flocculation power and lower is the flocculation value.

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