Question
Which one among the following pairs of ions cannot be separated by $${H_2}S$$ in dilute hydrochloric acid?
A.
$$B{i^{3 + }},S{n^{4 + }}$$
B.
$$A{l^{3 + }},H{g^{2 + }}$$
C.
$$Z{n^{2 + }},C{u^{2 + }}$$
D.
$$N{i^{2 + }},C{u^{2 + }}$$
Answer :
$$B{i^{3 + }},S{n^{4 + }}$$
Solution :
NOTE : The ions of group $$II$$ of salt analysis are
precipitated by $$HCl\,{\text{and}}\,{H_2}S$$ whereas members of group $$IV$$ are precipitated by $${H_2}S$$ in alkaline medium.
$$\because B{i^{3 + }}\,{\text{and}}\,S{n^{4 + }}$$ both belong to group $$II$$
∴ They will be precipitated by $$HCl$$ in presence of $${H_2}S.$$
Both $$B{i^{3 + }}\,{\text{and}}\,S{n^{4 + }}$$ belong to group $$II$$ of qualitative inorganic analysis and will get precipitated by $${H_2}S.$$