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).

Explanation

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^(+)`