What are group reagents?
Answer
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Hint: The Group reagent is a cocktail of reagents that serves as a prima facie indication of a particular cation, characterized by the formation of a precipitate during a positive test/result and no precipitation in the event of a negative test. Each group has a common reagent which can be used to separate them from the solution.
Complete answer:
The reagent which is employed in identifying a group of radicals is known as group reagent . Ammonium sulphide is used for testing 5th group cations. Sodium hydroxide is the group reagent for the zero group.
Group reagent for \[II,III,IV\] are:
In group $II$,cations are $P{b^{2 + }},C{u^{2 + }},A{s^{2 + }}$ and group reagent is ${H_2}S$ in the presence of dil.$HCl$.
In group $III$ cations are $F{e^{3 + }},A{l^{3 + }}$ and group reagent is $N{H_4}OH$ in the presence of dil $HCl$
In group $IV$, cations are $N{i^{2 + }},C{o^{2 + }},M{n^{2 + }},Z{n^{2 + }}$ and group reagent is ${H_2}S$ in the presence of $N{H_4}OH$. The reagent for the determination of metals should, first of all, contain a proper functional analytical group that enables its interaction with the determined element and subsequent observation of the respective analytical signal. Each group has a common reagent which can be used to separate them from the solution. To obtain meaningful results, the separation must be done in the sequence specified below, as some ions of an earlier group may also react with the reagent of a later group, causing ambiguity as to which ions are present.
Note:
Ammonium ions belong to the zero group as ammonium is a non-metallic cation and it has different reagents for testing. Hence, $N{H_4}^ + $ is tested first, keeping it in the zero group. The group reagent is ammonium hydroxide in the presence of ammonium chloride. $N{H_4}Cl$ is added to suppress the ionization of $N{H_4}OH$ so that only the third group radicals are precipitated as their insoluble hydroxide and not the fourth and fifth group radicals as the solubility products of their hydroxides are much higher.
Complete answer:
The reagent which is employed in identifying a group of radicals is known as group reagent . Ammonium sulphide is used for testing 5th group cations. Sodium hydroxide is the group reagent for the zero group.
Group reagent for \[II,III,IV\] are:
In group $II$,cations are $P{b^{2 + }},C{u^{2 + }},A{s^{2 + }}$ and group reagent is ${H_2}S$ in the presence of dil.$HCl$.
In group $III$ cations are $F{e^{3 + }},A{l^{3 + }}$ and group reagent is $N{H_4}OH$ in the presence of dil $HCl$
In group $IV$, cations are $N{i^{2 + }},C{o^{2 + }},M{n^{2 + }},Z{n^{2 + }}$ and group reagent is ${H_2}S$ in the presence of $N{H_4}OH$. The reagent for the determination of metals should, first of all, contain a proper functional analytical group that enables its interaction with the determined element and subsequent observation of the respective analytical signal. Each group has a common reagent which can be used to separate them from the solution. To obtain meaningful results, the separation must be done in the sequence specified below, as some ions of an earlier group may also react with the reagent of a later group, causing ambiguity as to which ions are present.
Note:
Ammonium ions belong to the zero group as ammonium is a non-metallic cation and it has different reagents for testing. Hence, $N{H_4}^ + $ is tested first, keeping it in the zero group. The group reagent is ammonium hydroxide in the presence of ammonium chloride. $N{H_4}Cl$ is added to suppress the ionization of $N{H_4}OH$ so that only the third group radicals are precipitated as their insoluble hydroxide and not the fourth and fifth group radicals as the solubility products of their hydroxides are much higher.
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