Question

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

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

Answer 1

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

Option (A) is correct.

$B{i}^{3+}$, $S{n}^{4+}$ ions cannot be separated by $H_{2}S$ in $diluteHCl$.

In the presence of $HCl$, both the ions are precipitated by $H_{2}S$.

$\begin{array}{ccccc}B{i}^{3+}& +& H_{2}S& \stackrel{HCl}{\to }& B{i}_2{S}_3\downarrow \\ & & & & Brown\end{array}$

$\begin{array}{ccccc}S{n}^{4+}& +& H_{2}S& \stackrel{HCl}{\to }& Sn{S}_2\downarrow \\ & & & & Yellow\end{array}$