A) \[H{{g}^{2+}}\]
B) \[S{{b}^{3+}}\]
C) \[A{{g}^{3+}}\]
D) \[S{{b}^{3+}}\] or \[B{{i}^{3+}}\] or both
Correct Answer: D
Solution :
As the ionic product of\[S{{b}^{3+}}\] and \[B{{i}^{3+}}\] is very low and \[C{{l}^{-}}\]is present in high concentration, therefore Sb and Bi get precipitated, as \[S{{b}^{3+}}+3C{{l}^{-}}\to SbC{{l}_{3}}\]You need to login to perform this action.
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