Question

Which 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

Answer: (A)

(a) $B{i}^{3+}$ and $S{n}^{4+}$ cannot be separated $by{H}_{2}S$ in dilute $HCl$, because both are present in second group and both are precipitated by $H_{2}S$ in dil. $HCl$ as $B{i}_2{S}_3$ and $Sn{S}_2$.

(b) $A{l}^{3+}$ is not precipitated by $H_{2}S$ in dil $HCl$ whereas $H{g}^{2+}$ is precipitated as $HgS$.

(c) $Z{n}^{2+}$ is not precipitated, but $C{u}^{2+}$ is precipitated as CuS.

(d) $N{i}^{2+}$ is not precipitated, but $C{u}^{2+}$ is precipitated as mention earlier.