Question

If ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ and ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ both are present in group III of qualitative analysis, then distinction can be made by:
A.Addition of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$ in presence of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ , when only ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ is precipitated.
B.Addition of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$ in presence of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ , when ${\text{Cr}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ and ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ both are precipitated; and on adding ${\text{B}}{{\text{r}}_{\text{2}}}$ water and ${\text{NaOH}}$ , ${\text{Cr}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ dissolves.
C.Precipitates of ${\text{Cr}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ and ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ as obtained in (B) are treated with conc. ${\text{HCl}}$ when only ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ dissolves.
D.Both (B) and (C) are correct.

Answer 1

Hint: The classification of the basic radicals into five different groups depends mainly on the different solubilities of the chlorides, sulphides, hydroxides and carbonates. The precipitation of the metal ions into different groups can be explained by the common ion effect and the solubility product.

Complete step by step answer:
The metal ions of group III A are ferric ion, ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$, ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ and ${\text{A}}{{\text{l}}^{{\text{3 + }}}}$. The hydroxides of these radicals are precipitated by ammonium hydroxide in the presence of ammonium chloride. Ammonium hydroxide alone will precipitate not only ferric hydroxide, i.e., ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ , chromium hydroxide, i.e., ${\text{Cr}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ and aluminium hydroxide, i.e., \[{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}\] , but also the hydroxides of the metals of the later groups.
Ammonium hydroxide is a weak base and it ionizes as:
${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}} \rightleftharpoons {\text{N}}{{\text{H}}_{\text{4}}}^ + {\text{ + O}}{{\text{H}}^ - }$
Addition of ammonium chloride increases the concentration of the common ion ammonium and thus pushes the ionization in the backward direction. This leads to the suppression of the ionization of ammonium hydroxide. Thus, the concentration of the hydroxyl ions decreases and the solubility products of only ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ , ${\text{Cr}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ and \[{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}\] are exceeded.
Hence, if both ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ and ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ both are present, then both their hydroxides will be precipitated by ammonium hydroxide in the presence of ammonium chloride.
For the detection of ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ , the precipitate is treated with ${\text{B}}{{\text{r}}_{\text{2}}}$ water and ${\text{NaOH}}$ which gives yellow coloured sodium chromate confirming its presence.
$2{\text{NaOH + B}}{{\text{r}}_{\text{2}}} \to {\text{NaBrO + NaBr + }}{{\text{H}}_{\text{2}}}{\text{O}}$
${\text{NaBrO}} \to {\text{NaBr + }}\left[ {\text{O}} \right]$
${\text{2Cr}}{\left( {{\text{OH}}} \right)_{\text{3}}}{\text{ + 4NaOH + 3}}\left[ {\text{O}} \right] \to 2{\text{N}}{{\text{a}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}{\text{ + 5}}{{\text{H}}_{\text{2}}}{\text{O}}$
The solution is then acidified and treated with lead acetate solution. Yellow colour of lead chromate precipitate confirms the presence of ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$.
${\text{N}}{{\text{a}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}{\text{ + }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}} \right)_{\text{2}}}{\text{Pb}} \to 2{\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa + PbCr}}{{\text{O}}_{\text{4}}}$
For the detection of ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$, the precipitate is treated with dilute hydrochloric acid in which the hydroxide dissolves.

So, the correct option is B.

Note: If iron is present, it will be present as ferrous ion, ${\text{F}}{{\text{e}}^{{\text{2 + }}}}$ after the passing of hydrogen sulphide through the solution. If ammonium hydroxide is added in a solution having ${\text{F}}{{\text{e}}^{{\text{2 + }}}}$ ion, then ${\text{Fe}}{\left( {{\text{OH}}} \right)_2}$ is not precipitated completely. So, the ${\text{F}}{{\text{e}}^{{\text{2 + }}}}$ ions are oxidized to ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ ions by boiling with concentrated nitric acid before the addition of ammonium hydroxide and ammonium chloride.
${\text{6FeS}}{{\text{O}}_{\text{4}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2HN}}{{\text{O}}_{\text{3}}} \to 3{\text{F}}{{\text{e}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}{\text{ + 2NO + 4}}{{\text{H}}_{\text{2}}}{\text{O}}$