This is a qualitative analysis problem. We need to identify the ions present in the mixture based on a series of chemical tests.

Step 1: Analysis of the reaction with dilute HCl

An aqueous solution of the mixture of two inorganic salts is treated with dilute HCl. A precipitate (P) is formed. This indicates the presence of a Group I cation, as they precipitate as chlorides. The Group I cations are $Pb^{2+}$, $Ag^+$, and $Hg_2^{2+}$. The problem states that the precipitate (P) was found to dissolve in hot water. This is a characteristic property of lead(II) chloride ($PbCl_2$). Silver chloride ($AgCl$) and mercury(I) chloride ($Hg_2Cl_2$) are insoluble in hot water. $Pb^{2+}(aq) + 2Cl^-(aq) \xrightarrow{cold} PbCl_2(s) \ (White \ ppt \ P)$ $PbCl_2(s) \xrightarrow{hot \ water} Pb^{2+}(aq) + 2Cl^-(aq) \ (Dissolves)$ So, one of the salts contains the $Pb^{2+}$ cation. The other salt's cation is in the filtrate (Q).

Step 2: Analysis of filtrate (Q) with $H_2S$ in acidic medium

The filtrate (Q) is treated with $H_2S$ in a dilute mineral acid medium, and no precipitate is formed. This is the test for Group II cations ($Cu^{2+}$, $Pb^{2+}$, $Bi^{3+}$, $Cd^{2+}$, $As^{3+}$, etc.). Since no precipitate is formed, Group II cations are absent from the filtrate (Q).

Step 3: Analysis of filtrate (Q) with $H_2S$ in ammoniacal medium

The filtrate (Q) is then treated with $H_2S$ in an ammoniacal medium ($NH_4OH + H_2S$), which gives a precipitate (R). This is the test for Group III and Group IV cations.

• Group III cations ($Al^{3+}$, $Fe^{3+}$, $Cr^{3+}$) precipitate as hydroxides in the presence of $NH_4OH$. The $OH^-$ concentration from the buffer is sufficient to precipitate them.
• Group IV cations ($Zn^{2+}$, $Ni^{2+}$, $Co^{2+}$, $Mn^{2+}$) precipitate as sulfides in this medium. So, the cation in filtrate Q belongs to either Group III or Group IV.

Step 4: Analysis of precipitate (R) with $H_2O_2$ and $NaOH$

The precipitate (R) is treated with $H_2O_2$ in an aqueous $NaOH$ medium, which gives a coloured solution (S). This is a characteristic test for the chromium(III) ion, $Cr^{3+}$.

1. The cation in filtrate Q is $Cr^{3+}$. In an ammoniacal medium, it precipitates as chromium(III) hydroxide, $Cr(OH)_3$, which is a grey-green precipitate (R). $Cr^{3+}(aq) + 3NH_4OH(aq) \rightarrow Cr(OH)_3(s) + 3NH_4^+(aq)$
2. Chromium(III) hydroxide is amphoteric and reacts with a strong base like $NaOH$. However, the key reaction is the oxidation by hydrogen peroxide ($H_2O_2$) in a basic medium. $Cr^{3+}$ (oxidation state +3) is oxidized to chromate ion, $CrO_4^{2-}$ (oxidation state +6). $2Cr(OH)_3(s) + 4NaOH(aq) + 3H_2O_2(aq) \rightarrow 2Na_2CrO_4(aq) + 8H_2O(l)$
3. The resulting solution (S) contains sodium chromate ($Na_2CrO_4$). The chromate ion ($CrO_4^{2-}$) imparts a distinct yellow colour to the solution.

Let's check why other possibilities are incorrect:

• If R were $Al(OH)_3$ (white), it would dissolve in $NaOH$ to form a colourless solution of sodium tetrahydroxoaluminate, $Na[Al(OH)_4]$.
• If R were $Fe(OH)_3$ (reddish-brown), it would be insoluble in $NaOH$.
• If R were $ZnS$ (white), it would react with $NaOH$ to form $Zn(OH)_2$, which dissolves in excess $NaOH$ to give a colourless solution of sodium tetrahydroxozincate(II), $Na_2[Zn(OH)_4]$.
• Precipitates of Ni, Co, Mn would not form a coloured solution under these conditions.

Step 5: Conclusion

The coloured solution (S) contains sodium chromate, $Na_2CrO_4$. Comparing this with the given options: A: $F{e_2}{(S{O_4})_3}$ - Incorrect. B: $CuS{O_4}$ - Incorrect. C: $ZnS{O_4}$ - Incorrect. D: $N{a_2}Cr{O_4}$ - Correct.