Identification of Coloured Solution S Formed from Precipitate R

Based on the systematic qualitative analysis of inorganic salts given in the paragraph, we can deduce the chemical reactions step-by-step:

1. Analysis of Precipitate $P$: Treatment of the aqueous salt solution with dilute $\text{HCl}$ gives a precipitate $P$ which is soluble in hot water. This indicates the presence of $\text{Pb}^{2+}$ ions, as lead(II) chloride ($\text{PbCl}_2$) precipitates in cold water but dissolves upon heating.

2. Analysis of Filtrate $Q$:

   • The filtrate $Q$ does not form a precipitate with $\text{H}_2\text{S}$ in acidic medium, confirming the absence of Group II cations.
   • However, treatment of $Q$ with $\text{H}_2\text{S}$ in an ammoniacal medium ($\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$) yields a precipitate $R$. This indicates the presence of Group III cations such as $\text{Cr}^{3+}$ which precipitate as chromium(III) hydroxide, $\text{Cr(OH)}_3$.

3. Formation of Coloured Solution $S$: When the precipitate $R$ ($\text{Cr(OH)}_3$) is treated with hydrogen peroxide ($\text{H}_2\text{O}_2$) in an aqueous sodium hydroxide ($\text{NaOH}$) medium, $\text{Cr(III)}$ is oxidized to $\text{Cr(VI)}$, forming a soluble yellow chromate complex, sodium chromate ($\text{Na}_2\text{CrO}_4$):

   $2\text{Cr(OH)}_3 + 3\text{H}_2\text{O}_2 + 4\text{NaOH} \rightarrow 2\text{Na}_2\text{CrO}_4 + 8\text{H}_2\text{O}$

Thus, the chemical species present in the yellow coloured solution $S$ is $\text{Na}_2\text{CrO}_4$.

Correct Answer: D