On passing $H_2S$ gas,a black precipitate of $II$ group is obtained. The mixture may not contain:
- A$Pb^{2+}$
- B$Cd^{2+}$
- C$Hg^{2+}$
- D$Cu^{2+}$
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Similar Questions
Which of the following compounds can produce Rinmann's green with $Co(NO_3)_2$ solution?
Medium
View SolutionConsider the sulphides $HgS$,$PbS$,$CuS$,$Sb_{2}S_{3}$,$As_{2}S_{3}$ and $CdS$. Number of these sulphides soluble in $50\% \ HNO_{3}$ is .... .
Which one among the following pairs of ions cannot be separated by $H_2S$ in dilute hydrochloric acid?
DifficultIIT 1986
View SolutionIn a solution containing $Hg^{2+}, Cd^{2+}, Sr^{2+}, Fe^{2+},$ and $Cu^{2+}$ ions in $0.3 \, M \, HCl$,which ions will precipitate upon the addition of $H_2S$?
Medium
View SolutionMatch the List-$I$ with List-$II$ :
Cations Group reaction $P \rightarrow Pb^{2+}, Cu^{2+}$ $(i)$ $H_2S$ gas in presence of dilute $HCl$ $Q \rightarrow Al^{3+}, Fe^{3+}$ $(ii)$ $(NH_4)_2CO_3$ in presence of $NH_4OH$ $R \rightarrow Co^{2+}, Ni^{2+}$ $(iii)$ $NH_4OH$ in presence of $NH_4Cl$ $S \rightarrow Ba^{2+}, Ca^{2+}$ $(iv)$ $H_2S$ in presence of $NH_4OH$
| Cations | Group reaction |
| $P \rightarrow Pb^{2+}, Cu^{2+}$ | $(i)$ $H_2S$ gas in presence of dilute $HCl$ |
| $Q \rightarrow Al^{3+}, Fe^{3+}$ | $(ii)$ $(NH_4)_2CO_3$ in presence of $NH_4OH$ |
| $R \rightarrow Co^{2+}, Ni^{2+}$ | $(iii)$ $NH_4OH$ in presence of $NH_4Cl$ |
| $S \rightarrow Ba^{2+}, Ca^{2+}$ | $(iv)$ $H_2S$ in presence of $NH_4OH$ |
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