Which of the following solutions will yield a black precipitate when $H_2S$ gas is passed through it?

$M^{n+}(aq.) + KI \to X\downarrow \text{ (ppt.)}$ $\xrightarrow{Excess KI} \text{ppt. remains insoluble in excess } KI \text{ solution. Then cation } M^{n+}(aq.) \text{ can be}$

An inorganic compound $X$ on treatment with concentrated $H_{2}SO_{4}$ produces brown fumes and gives a dark brown ring with $FeSO_{4}$ in the presence of concentrated $H_{2}SO_{4}$. Also,compound $X$ gives a precipitate $Y$ when its solution in dilute $HCl$ is treated with $H_{2}S$ gas. The precipitate $Y$ on treatment with concentrated $HNO_{3}$ followed by excess of $NH_{4}OH$ further gives a deep blue coloured solution. Compound $X$ is:

$A$ mixture of two salts is used to prepare a solution $S$,which gives the following results:
White precipitate$(s)$ $\xleftarrow{\text{Dilute } NaOH(aq.) \text{ at Room temperature}}$ $S$ (aq. solution of the salts) $\xrightarrow{\text{Dilute } HCl(aq.) \text{ at Room temperature}}$ White precipitate$(s)$ only
The correct option$(s)$ for the salt mixture is(are):
$A$. $Pb(NO_3)_2$ and $Zn(NO_3)_2$
$B$. $Pb(NO_3)_2$ and $Bi(NO_3)_3$
$C$. $AgNO_3$ and $Bi(NO_3)_3$
$D$. $Pb(NO_3)_2$ and $Hg(NO_3)_2$

The ion that cannot be precipitated by both $HCl$ and $H_2S$ is

The only cations present in a slightly acidic solution are $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$. The reagent that when added in excess to this solution would identify and separate $Fe^{3+}$ in one step is