Upon treatment with ammoniacal H_2S,the metal | vedclass

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(D) Ammoniacal $H_2S$ is the group reagent for the fourth group of cationic radicals in qualitative analysis.$Fe^{3+}$ and $Al^{3+}$ belong to the thi

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$Zn(II)$

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(D) Ammoniacal $H_2S$ is the group reagent for the fourth group of cationic radicals in qualitative analysis.$Fe^{3+}$ and $Al^{3+}$ belong to the third group and precipitate as hydroxides,$Fe(OH)_3$ and $Al(OH)_3$,in the presence of $NH_4OH$ and $NH_4Cl$.$Mg^{2+}$ does not precipitate with $H_2S$ in the presence of $NH_4OH$.$Zn^{2+}$ belongs to the fourth group and reacts with $H_2S$ in the presence of $NH_4OH$ to form a white precipitate of $ZnS$.

Upon treatment with ammoniacal $H_2S$,the metal ion that precipitates as a sulfide is

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Which one among the following pairs of ions cannot be separated by $H_2S$ in dilute hydrochloric acid?

Statement-$1$: Few drops of $HNO_3$ are added before proceeding to $III^{rd}$ group.
Statement-$2$: For the oxidation of $Fe^{+2} \rightarrow Fe^{+3}$.

Sodium nitroprusside when reacts with a solution of $Na_2S$,produces:

$Al^{3+}$,$Fe^{3+}$,$Zn^{2+}$,and $Ni^{2+}$ ions are present in an acidic solution. Excess of ammonium chloride $(NH_4Cl)$ solution is added followed by the addition of ammonium hydroxide $(NH_4OH)$ solution. The resulting precipitate will contain:

Which ion is not precipitated by both $HCl$ and $H_2S$?

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