JEE Advanced 2025 Paper 1 · Q14 · Qualitative Analysis

Question

Match each group reagent in List-I with the metal ion in List-II that it precipitates. List-I: (P) Passing $H_2S$ in the presence of $NH_4OH$; (Q) $(NH_4)_2CO_3$ in the presence of $NH_4OH$; (R) $NH_4OH$ in the presence of $NH_4Cl$; (S) Passing $H_2S$ in the presence of dilute $HCl$. List-II: (1) $Cu^{2+}$; (2) $Al^{3+}$; (3) $Mn^{2+}$; (4) $Ba^{2+}$; (5) $Mg^{2+}$. Choose: (A) P->3,Q->4,R->2,S->1; (B) P->4,Q->2,R->3,S->1; (C) P->3,Q->4,R->1,S->5; (D) P->5,Q->3,R->2,S->4.

SolveKar AI · Accepted Answer

Group analysis (standard scheme): Group II (acid radical) — $H_2S$ in dilute $HCl$ precipitates $Cu^{2+}$ as $CuS$. So (S) -> $Cu^{2+}$ = (1). Group III — $NH_4OH + NH_4Cl$ precipitates $Al^{3+}$ (and $Fe^{3+}, Cr^{3+}$) as hydroxide. So (R) -> $Al^{3+}$ = (2). Group IV — $H_2S$ in $NH_4OH$ (basic) precipitates $Mn^{2+}, Zn^{2+}, Ni^{2+}, Co^{2+}$ as sulphides (here $Mn^{2+}$). So (P) -> $Mn^{2+}$ = (3). Group V — $(NH_4)_2CO_3$ with $NH_4OH$ precipitates $Ba^{2+}, Sr^{2+}, Ca^{2+}$ as carbonates. So (Q) -> $Ba^{2+}$ = (4). Match: P->3, Q->4, R->2, S->1. Answer: (A).

JEE Advanced 2025 Paper 1 · Q14 · Qualitative Analysis

Solution