Consider the following reactions:(A) (text{Io | vedclass

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(D) The reaction of ionic bromides (like $NaBr$) with $MnO_2$ and conc. $H_2SO_4$ produces $Br_2$ (red-orange gas),which is $(C)$.The reaction of ioni

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$(2)$ and $(3)$ both

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(D) The reaction of ionic bromides (like $NaBr$) with $MnO_2$ and conc. $H_2SO_4$ produces $Br_2$ (red-orange gas),which is $(C)$.The reaction of ionic iodides (like $KI$) with $MnO_2$ and conc. $H_2SO_4$ produces $I_2$ (violet gas),but the question specifies $(E)$ has the same colour as $(C)$,implying $(E)$ is also $Br_2$ or a similar red-orange species.However,in standard qualitative analysis,$Br_2$ and $I_2$ are distinct. If $(C)$ is $Br_2$ and $(E)$ is $Br_2$,they are identical. If the question implies $(C)$ is $Br_2$ and $(E)$ is $I_2$,they differ in colour.Assuming the standard context where $(C)$ is $Br_2$ and $(E)$ is $I_2$,they differ in colour,atomicity (both are diatomic $X_2$),and their reaction with $NaOH$. Since $Br_2$ and $I_2$ both react with $NaOH$,the resulting solutions have different colours ($Br_2$ gives a colorless solution of $NaBr/NaBrO_3$,while $I_2$ gives a colorless solution of $NaI/NaIO_3$).Given the options,$(2)$ and $(3)$ are the most likely differences if $(C)$ and $(E)$ are different halogens.

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