Three separate samples of a solution of a single salt gave these test results: one formed a white precipitate with excess ammonia solution,one formed a white precipitate with dilute $NaCl$ solution,and one formed a black precipitate with $H_2S$. The salt could be:
- A$AgNO_3$
- B$Pb(NO_3)_2$
- C$Hg(NO_3)_2$
- D$MnSO_4$
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Similar Questions
Match List-$I$ with List-$II$.
$LIST$-$I$ (Precipitating reagent and conditions) $LIST$-$II$ (Cation) $A$. $NH_4Cl + NH_4OH$ $I$. $Mn^{2+}$ $B$. $NH_4OH + (NH_4)_2CO_3$ $II$. $Pb^{2+}$ $C$. $NH_4OH + NH_4Cl + H_2S$ gas $III$. $Al^{3+}$ $D$. Dilute $HCl$ $IV$. $Sr^{2+}$
Choose the correct answer from the options given below:
| $LIST$-$I$ (Precipitating reagent and conditions) | $LIST$-$II$ (Cation) |
| $A$. $NH_4Cl + NH_4OH$ | $I$. $Mn^{2+}$ |
| $B$. $NH_4OH + (NH_4)_2CO_3$ | $II$. $Pb^{2+}$ |
| $C$. $NH_4OH + NH_4Cl + H_2S$ gas | $III$. $Al^{3+}$ |
| $D$. Dilute $HCl$ | $IV$. $Sr^{2+}$ |
Choose the correct answer from the options given below:
In the wet tests for identification of various cations by precipitation,which transition element cation does not belong to group $IV$ in qualitative inorganic analysis?
$A$ metal salt solution forms a yellow precipitate with potassium chromate in acetic acid,a white precipitate with dilute $H_2SO_4$,but gives no precipitate with $NaCl$. The metal salt solution will consist of:
DifficultAIIMS 2011
View Solution$A$ black sulphide is formed by the action of $H_2S$ on
MediumIIT 1978
View Solution$A$ precipitate of ....... would be obtained on adding $HCl$ to a solution of $(Sb_2S_3)$ in yellow ammonium sulphide.
Difficult
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