A) \[KCN\]
B) \[BaC{{l}_{2}}\]
C) \[NaCl\]
D) \[M{{g}_{3}}{{(P{{O}_{4}})}_{2}}\]
Correct Answer: D
Solution :
Key Idea: According to Hardy Schuize rule coagulating power of ions is\[+\alpha \]harge on ion \[\because \]\[Fe{{(OH)}_{3}}\]is positively charged colloid. \[\therefore \]It will be coagulated by anion. (a) \[KCN-{{K}^{+}}\]and \[C{{N}^{-}}\] (b)\[BaC{{l}_{2}}-B{{a}^{2+}}\]and \[C{{l}^{-}}\] (c)\[NaCl-N{{a}^{+}}\]and \[C{{l}^{-}}\] (d)\[M{{g}_{3}}{{(P{{O}_{4}})}_{2}}-M{{g}^{2+}}\]and \[PO_{4}^{3-}\] \[\because \]\[PO_{4}^{3-}\]has highest charge in it among anions. \[\therefore \]\[M{{g}_{3}}{{(P{{O}_{4}})}_{2}}\]is the most effective in coagulation of\[Fe{{(OH)}_{3}}\]
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