JEE Advanced · Chemistry · 31. Practical Chemistry
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 \(H _2 S\) in the presence of \(NH _4 OH\) | (1) \(Cu ^{2+}\) |
| (Q) \(\left( NH _4\right)_2 CO _3\) in the presence of \(NH _4 OH\) | (2) \(Al ^{3+}\) |
| (R) \(NH _4 OH\) in the presence of \(NH _4 Cl\) | (3) \(Mn ^{2+}\) |
| (S) Passing \(H _2 S\) in the presence of dilute HCl | (4) \(Ba ^{2+}\) |
| (5) \(Mg ^{2+}\) |
- A \(\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 1\)
- B \(\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 1\)
- C \(\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 5\)
- D \(\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4\)
Answer & Solution
Correct Answer
(A) \(\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 1\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Mn}^{+2} \xrightarrow{\mathrm{H}_2 \mathrm{~S}+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { Pink/buff ppt. }}{\mathrm{MnS} \downarrow}\)
\(\mathrm{Ba}^{+2} \xrightarrow{\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { White ppt. }}{\mathrm{BaCO}_3 \downarrow}\)
\(\mathrm{Al}^{+3} \xrightarrow{\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { White ppt. }}{\mathrm{Al}(\mathrm{OH})_3 \downarrow}\)
\(\mathrm{Cu}^{+2} \xrightarrow{\mathrm{H}_2 \mathrm{~S}+\mathrm{HCl} \text { (dil.) }} \underset{\text { Black ppt. }}{\mathrm{CuS} \downarrow}\)
\(\mathrm{Ba}^{+2} \xrightarrow{\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { White ppt. }}{\mathrm{BaCO}_3 \downarrow}\)
\(\mathrm{Al}^{+3} \xrightarrow{\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH}} \underset{\text { White ppt. }}{\mathrm{Al}(\mathrm{OH})_3 \downarrow}\)
\(\mathrm{Cu}^{+2} \xrightarrow{\mathrm{H}_2 \mathrm{~S}+\mathrm{HCl} \text { (dil.) }} \underset{\text { Black ppt. }}{\mathrm{CuS} \downarrow}\)
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