Question

Which one of the following pairs of ions cannot be separated by $H_{2}S$ in dilute $HCl$?

• A. $B{i}^{3+},S{n}^{4+}$
• B. $A{l}^{3+},H{g}^{2+}$
• C. $C{u}^{2+},Z{n}^{2+}$
• D. $N{i}^{2+},C{u}^{2+}$

Answer 1

The correct option is A $B{i}^{3+},S{n}^{4+}$

(i) Both $B{i}^{3+}$ and $S{n}^{4+}$ (belong to II group) are precipitated by $H_{2}S$ in presence of $dil.HCl$.

(ii) $H{g}^{2+}$ is precipitated with $H_{2}S$ in presence of $dil.HCl$, while $A{l}^{3+}$ does not.

(iii) $C{u}^{2+}$ precipitated with $H_{2}S$ in presence of $dil.HCI$ while $Z{n}^{2+}$ does not.

(iv) Similarly $N{i}^{2+}$ does not precipitate with $H_{2}S$ in presence of $dil.H_{2}S$, while $C{u}^{2+}$ precipitated.

hence option (A) is correct.