Question

In qualitative analysis, ${III}^{rd}$ group includes $F{e}^{3+},A{l}^{3+}$ and $C{r}^{3+}$. The group reagent is ammonium hydroxide in the presence of ammonium chloride. $N{H}_{4}CI$ is added to suppress the ionisation of $N{H}_{4}OH$ so that only the $II{I}^{rd}$ group radicals are precipitated as their insoluble hydroxide and not the $I{V}^{th}$ and $V^{th}$ group radicals as the solubility products of their hydroxides are much higher. Before adding group reagent to the filtrate of IInd group, one drop of concentrated $HN{O}_3$ (oxidising agent) is added and the content is boiled for 2-3 minutes and then cool because the solubility product of $Fe(OH{)}_2$ is very high as compared to $Fe(OH{)}_3$. When a light bluish-green crystalline double salt-containing cations of III${}^{rd}$ group and zero groups are analysed. it responds to the following tests/reactions.
(a) Its aqueous solution gives a reddish-brown precipitate with an alkaline solution of potassium tetraiodomercurate (II)
(b) Its aqueous solution after boiling with conc. $HN{O}_3$ gives a reddish-brown precipitate with sodium hydroxide.
(c) The solution of reddish-brown precipitate in oil.HCI gives blood red colouration with ammonium sulphocyanide and Prussian blue with potassium ferrocyanide.
(d) It's a very dilute solution gives a white precipitate with $(C{H}_{3}COO{)}_{2}Pb$ solution and a white precipitate are only soluble in hot ammonium acetate.

Why one mL conc. $HN{O}_3$ is added to $I{I}^{nd}$ group filtrate before proceeding for $II{I}^{rd}$ group radicals?

• A. Because if oxidises the dissolved $H_{2}S$ to colloidal sulphur
• B. Because it oxidises $F{e}^{2+}$ to $F{e}^{3+}$ can be completely precipitated as its hydroxide
• C. Because precipitation of cations belonging to $II{I}^{rd}$ group by group reagent takes place only in presence of conc. $HN{O}_3$
• D. None of these

Answer 1

Answer: (B)

Because it oxidises $F{e}^{2+}$ to $F{e}^{3+}$ can be completely precipitated as its hydroxide

As complete precipitation of $Fe(OH{)}_2$ ($K_{sp}$ very high) does not take place in presence of $N{H}_{4}Cl$ so it is oxidised to $F{e}^{3+}$ by $HN{O}_3$(acts as oxidising agent) and $Fe(OH{)}_3$ has lower $K_{sp}$ which is completely separated as reddish brown precipitate.
$F{e}^{2+}$ reacts with $H^{+}$(of nitric acid), acts as an oxidising agent,converts ferrous to ferric ion.This ferric ion reacts with group reagents to form red precipitate.

Hence, the correct option is $B$