A sulphate of a metal (A) on heating evolves two gases (B) and (C) and an oxide (D). Gas (B) turns K — Qualitative and Quantitative Analysis Chemistry Question
Question
A sulphate of a metal (A) on heating evolves two gases (B) and (C) and an oxide (D). Gas (B) turns K2Cr2O7 paper green while gas (C) forms a trimer in which there is no S–S bond. Compound (D) with conc. HCl forms a Lewis acid (E) which exists in a dimer. Compounds (A), (B), (C), (D) and (E) are respectively:
💡 Solution & Explanation
Step 1: AgCl and AgI have significantly different solubility products (Ksp of AgCl ~ 1.8 x 10^-10, Ksp of AgI ~ 8.3 x 10^-17). Step 2: Ammonia (NH3) forms a soluble diamminesilver(I) complex, [Ag(NH3)2]+, with silver ions. This complexation is strong enough to dissolve AgCl but is unable to dissolve the far more insoluble AgI. Step 3: Thus, treating the mixture with dilute ammonia (NH3) selectively dissolves AgCl while leaving AgI as an insoluble yellow residue, enabling separation. This corresponds to option (d).