Question

The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions is:

List-I	List-II
(P) Passing H2S in the presence of NH4OH	(1) Cu2+
(Q) (NH4)2CO3 in the presence of NH4OH	(2) Al3+
(R) NH4OH in the presence of NH4Cl	(3) Mn2+
(S) Passing H2S in the presence of dilute HCl	(4) Ba2+ (5) Mg2+

• 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

Hint:

Group analysis of cations in qualitative inorganic analysis follows a defined order based on solubility of precipitates in acidic/basic media. Know the solubility rules and group reagents.

Answer 1

The Correct Option is A

(P) H\(_2\)S + NH\(_4\)OH → Precipitates Group IV cations 
Group IV cations include Mn\(^{2+}\), Zn\(^{2+}\), Co\(^{2+}\), etc., which precipitate as sulfides in basic medium. \[ \Rightarrow \text{P} \rightarrow \boxed{3} \quad \text{(Mn}^{2+}\text{)} \] (Q) (NH\(_4\))\(_2\)CO\(_3\) + NH\(_4\)OH → Precipitates Group V cations 
This group includes alkaline earth metals like Ba\(^{2+}\), Ca\(^{2+}\), Sr\(^{2+}\), which form carbonates. \[ \Rightarrow \text{Q} \rightarrow \boxed{4} \quad \text{(Ba}^{2+}\text{)} \] (R) NH\(_4\)OH + NH\(_4\)Cl → Precipitates Group III cations 
Group III cations like Al\(^{3+}\), Cr\(^{3+}\), Fe\(^{3+}\) form hydroxides in weakly basic conditions. \[ \Rightarrow \text{R} \rightarrow \boxed{2} \quad \text{(Al}^{3+}\text{)} \] (S) H\(_2\)S + dilute HCl → Precipitates Group II cations 
Group II includes Cu\(^{2+}\), Pb\(^{2+}\), Hg\(^{2+}\) which form sulfides in acidic medium. \[ \Rightarrow \text{S} \rightarrow \boxed{1} \quad \text{(Cu}^{2+}\text{)} \] 

• P → 3
• Q → 4
• R → 2
• S → 1

Final Answer: \( \boxed{\text{A}} \)