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Group reagent for analytic group IV is [Kurukshetra CET 2002]
In inorganic qualitative analysis, the group reagent for the IV analytical group is a combination of $NH_4Cl$, $NH_4OH$, and $H_2S$. This combination helps in the precipitation of sulfides of Group IV cations.
When $H_2S$ is passed through $Hg_2S$ we get [AIEEE 2002]
$ [X] + H_2SO_4 \rightarrow [Y] $ a colourless gas with irritating smell. $ [Y] + K_2Cr_2O_7 + H_2SO_4 \rightarrow green solution$ [X] and [Y] is [IIT-JEE (Screening) 2003]
Concentrated sodium hydroxide can separate a mixture of [MP PMT 2000]
What product is formed by mixing the solution of $K_4[Fe(CN)_6]$ with the solution of $FeCl_3$ [Roorkee 1989]
When a solution of $K_4[Fe(CN)_6]$ is mixed with a solution of $FeCl_3$, the product formed is ferric-ferrocyanide, which is also known as Prussian blue. The reaction can be represented as: $4FeCl_3 + 3K_4[Fe(CN)_6] ightarrow Fe_4[Fe(CN)_6]_3 + 12KCl$. This results in a blue precipitate of ferric-ferrocyanide.
When $H_2S$ is passed through a mixture containing $Cu^{+2}, Ni^{+2}, Zn^{+2}$ in acidic solution then ion will precipitate [RPMT 2002]
When $H_2S$ is passed through a mixture containing $Cu^{2+}$, $Ni^{2+}$, $Zn^{2+}$ in an acidic solution, $Cu^{2+}$ will precipitate as $CuS$. This is because $CuS$ is insoluble in acidic solutions, whereas $NiS$ and $ZnS$ are soluble in acidic solutions and will not precipitate. Therefore, only $Cu^{2+}$ will precipitate in the given conditions.
A 0.3 M HCl solution contains the following ions $ Hg^ {2+} , Cd ^{2+} , Sr ^{2+} , Fe^{2+} , Cu ^{2+} $ . The addition of $ H_2S$ to above solution will precipitate [CPMT 1973]
When $H_2S$ is added to a solution of 0.3 M HCl containing ions such as $Hg^{2+}$, $Cd^{2+}$, $Sr^{2+}$, $Fe^{2+}$, and $Cu^{2+}$, the sulfide ions ($S^{2-}$) from $H_2S$ will react with the metal cations to form metal sulfides. Among the given ions, $Hg^{2+}$, $Cd^{2+}$, and $Cu^{2+}$ form insoluble sulfides ($HgS$, $CdS$, and $CuS$) in an acidic medium, leading to their precipitation. The other metal ions do not form precipitates under these conditions. Hence, the correct answer is o1: Cd, Cu, and Hg.
Which of the following gives a ppt. with $Pb(NO_3)_2$ but not with $Ba(NO_3)_2$ [CPMT 1979; MP PET 1997]
In the group III radicals, in place of $NH_4Cl$ which of the following can be used [AIIMS 1980, 82; MP PMT 1985]
Which compound does not dissolve in hot dilute $HNO_3$ [IIT 1996]
HgS (Mercury(II) sulfide) is insoluble in hot dilute $HNO_3$ (Nitric acid). This is because HgS forms a very stable and insoluble compound due to the strong Hg-S bond, which does not break down easily in nitric acid.
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