A) \[KBr\]
B) \[{{K}_{2}}S{{O}_{4}}\]
C) \[{{K}_{2}}Cr{{O}_{4}}\]
D) \[{{K}_{4}}[Fe{{(CN)}_{6}}]\]
Correct Answer: A
Solution :
\[Fe{{(OH)}_{3}}\],is a positively charged sol, thus coagulated by negative ion (anion). Smaller the charge on anion, smaller is its coagulating power or higher is its flocculation value. (a)\[KBr\xrightarrow{{}}{{K}^{+}}+B{{r}^{-}}\] (b)\[{{K}_{2}}S{{O}_{4}}\xrightarrow{{}}2{{K}^{+}}+SO_{4}^{2-}\] (c)\[{{K}_{2}}Cr{{O}_{4}}\xrightarrow{{}}2{{K}^{+}}+CrO_{4}^{2-}\] (d)\[{{K}_{4}}[Fe{{(CN)}_{6}}]\xrightarrow{{}}4{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{4-}}\] \[\because \,\,B{{r}^{-}}\]has smaller charge. \[\therefore \,\,KBr\]is least effective in coagulating\[Fe{{(OH)}_{3}}\]sol.
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