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The gas liberated on heating a mixture o...

The gas liberated on heating a mixture of two salts with `NaOH`, gives a reddish brown precipitate with an alkaline solution of `K_(2)Hgl_(4)`. The aqueous solution of the mixture on treatment with `BaCl_(2)` gives a white precipitate which is sparingly soluble in conc. HCl. On heating the mixture of `K_(2)Cr_(2)O_(7)` and conc `H_(2)SO_(4)` red vapours (A) are produced. The aqueous solution of the mixture gives a deep blue colouration (B) with potassium ferricyanide solution. Idnetify the radicals in the given mixture and write the balanced equations for the formation of (A) and (B).

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Let us summarize given facts of question.
`"Red vapours (A)"overset(K_(2)Cr_(2)O_(7))underset("Conc. "H_(2)SO_(4)" heat")larr" Mixture of two salts " overset(NaOH)underset(Delta)rarr "Gas" overset(Alk. K_(2)Hgl_(4))rarr " Reddish brown ppt."`
`"Deep blue colour (B)" overset(K_(3)[Fe(CN)_(6))larr "Ag. solution of the mixture "overset(BaCl_(2))rarr" white ppt."`
The given reaction leads to the following conclusions.
(i) Heating of mixture with NaOH to give `NH_(3)` gas (indicated by reddish brwon ppt with alkaline solution of `K_(2)Hgl_(4)`) indicates the presence of `NH_(4)^(+)` ion in the mixture.
(ii) Heating of mixture with `K_(2)Cr_(2)O_(7)` and conc. `H_(2)SO_(4)` to give red vapours (of chromyl chloride) indicates the presence of `Cl^(-)` ion in the mixture.
(iii) Reaction of aqueous solution of the mixture barium chloride solution to give white ppt. (of `BaSO_(4)`) sparingly soluble in conc. HCl indicates the presence of `SO_(4)^(2-)` ions in the mixture.
(iv) Reaction of aqueous solution of the mixture with potassium ferricyanide solution to give deep blue colour indicates presence of `Fe^(2+)` ions in the mixture.
Hence the mixture contains following four ions `NH_(4)^(+),Fe^(2+),SO_(4)^(2-)` and `Cl^(-)`.
Equations for the formation of A and B.
`4NaCl+K_(2)Cr_(2)O_(7)+3H_(2)SO_(4)overset("Heat")rarr K_(2)SO_(4)+2Na_(2)SO_(4)+underset("(Orange)(A)")underset("Chromyl chloride")(2CrO_(2)Cl_(2))+3H_(2)O`
`3Fe^(2+)+2K_(3)[Fe(CN)_(6)] rarr underset("blue ppt (B)")(Fe_(3)[Fe(CN)_(6)]_(2))+6K^(+)`
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