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Which one among the following pairs of ions cannot be separated by \mathrm{H}_{2} \mathrm{S} in dilute hydrochloric acid

Option: 1

\mathrm{Bi}^{3+}, \mathrm{Sn}^{4+}


Option: 2

\mathrm{Al}^{3+}, \mathrm{Hg}^{2+}


Option: 3

\mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}


Option: 4

\mathrm{Ni}^{2+}, \mathrm{Cu}^{2+}


Answers (1)

best_answer

Separation can be carried out by a reagent when it reacts with only one of the given ions. Since, \text{Bi}^{3+} and \text{Sn}^{4+} both form precipitate with \text{H}_{2}\text{S}\: \: \text{(in HCl)} , these will not be separated.

Hence, the correct answer is Option (1)

Posted by

Rakesh

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