As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order :
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- \(\mathrm{AlCl_3} \gt \mathrm{K_3[Fe(CN)_6]} \gt \mathrm{K_2CrO_4} \gt \mathrm{KBr} = \mathrm{KNO_3}\)
- \(\mathrm{K_3[Fe(CN)_6]} \lt \mathrm{K_2CrO_4} \lt \mathrm{KBr} = \mathrm{KNO_3} = \mathrm{AlCl_3}\)
- \(\mathrm{K_3[Fe(CN)_6]} \lt \mathrm{K_2CrO_4} \lt \mathrm{AlCl_3} \lt \mathrm{KBr} \lt \mathrm{KNO_3}\)
- \(\mathrm{K_3[Fe(CN)_6]} \gt \mathrm{AlCl_3} \gt \mathrm{K_2CrO_4} \gt \mathrm{KBr} \gt \mathrm{KNO_3}\)
The Correct Option is B
Solution and Explanation
Coagulating power rises with the charge of the anion: [Fe(CN)6]3- > CrO42- > Br- = NO3- = Cl-.
Flocculation value is inversely proportional to coagulating power.
AlCl3 furnishes the monovalent anion Cl-, so it is equal in effect to KBr and KNO3 (Al3+ has the same sign as the sol and is irrelevant).
Hence flocculation values: K3[Fe(CN)6] < K2CrO4 < KBr = KNO3 = AlCl3.
Answer: (2) K3[Fe(CN)6] < K2CrO4 < KBr = KNO3 = AlCl3
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