When salt BA is treated with concentrated H_2 | vedclass

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(D) Step $1$: Reaction with concentrated $H_2SO_4$: Bromide salts $(Br^-)$ react with concentrated $H_2SO_4$ to produce hydrogen bromide gas, which is

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$Br^-$

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(D) Step $1$: Reaction with concentrated $H_2SO_4$: Bromide salts $(Br^-)$ react with concentrated $H_2SO_4$ to produce hydrogen bromide gas, which is further oxidized to reddish-brown bromine $(Br_2)$ gas.Step $2$: Reaction with $AgNO_3$: Bromide ions $(Br^-)$ react with silver nitrate $(AgNO_3)$ to form silver bromide $(AgBr)$, which appears as a pale yellow precipitate.Step $3$: Conclusion: Since both observations match the characteristics of the bromide ion, the anion $A^-$ is $Br^-$.

When salt $BA$ is treated with concentrated $H_2SO_4$, a reddish-brown gas is liberated. The aqueous solution of $BA$ gives a pale yellow precipitate with $AgNO_3$ solution. Which of the following anions $(A^-)$ is present in the salt $BA$?

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