(B) The correct answer is $B$. Sometimes,yellow turbidity is observed upon passing $H_2S$ gas even when second group radicals are absent. This occurs

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The fourth group radicals are precipitated as sulphides

(B) The correct answer is $B$. Sometimes,yellow turbidity is observed upon passing $H_2S$ gas even when second group radicals are absent. This occurs because certain fourth group radicals (such as $Zn^{2+}$,$Mn^{2+}$,$Ni^{2+}$,$Co^{2+}$) may precipitate as sulphides if the concentration of $S^{2-}$ ions in the solution is sufficiently high,leading to the formation of a colloidal suspension or turbidity.

Class 12 Chemistry Principles Related to Practical Chemistry Wet Test for Basic Radical

Medium

Sometimes yellow turbidity appears on passing $H_2S$ gas even in the absence of the second group radicals. This happens because

AIIMS 1982 All AIIMS

A
Sulphur is present in the mixture as an impurity
B
The fourth group radicals are precipitated as sulphides
C
The $H_2S$ is oxidized by some acid radicals
D
The third group radicals are precipitated

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Wet Test for Basic Radical

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AIIMS 1982 Paper All AIIMS years →

On addition of aqueous $NaOH$ to a salt solution,a white gelatinous precipitate is formed,which dissolves in excess alkali. The salt solution contains

Medium

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Identify ions $A$ and $B$,respectively,based on the following reaction scheme:

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Which ion will not be precipitated by aqueous $(NH_4)_2S$?

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$A$ precipitate of ....... would be obtained on adding $HCl$ to a solution of $(Sb_2S_3)$ in yellow ammonium sulphide.

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The only cations present in a slightly acidic solution are $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$. The reagent that when added in excess to this solution would identify and separate $Fe^{3+}$ in one step is

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Sometimes yellow turbidity appears on passing $H_2S$ gas even in the absence of the second group radicals. This happens because

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On addition of aqueous $NaOH$ to a salt solution,a white gelatinous precipitate is formed,which dissolves in excess alkali. The salt solution contains

Identify ions $A$ and $B$,respectively,based on the following reaction scheme:

Which ion will not be precipitated by aqueous $(NH_4)_2S$?

$A$ precipitate of ....... would be obtained on adding $HCl$ to a solution of $(Sb_2S_3)$ in yellow ammonium sulphide.

The only cations present in a slightly acidic solution are $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$. The reagent that when added in excess to this solution would identify and separate $Fe^{3+}$ in one step is

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