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(D) Let us analyze each reaction:$1$. $3Br_2   6NaOH (	ext{hot and conc.}) ightarrow 5NaBr   NaBrO_3   3H_2O$. Thus,option $A$ is incorrect as it d

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$Cl_2$ reacts with excess of $NH_3$ to give $N_2$ and $HCl$.

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(D) Let us analyze each reaction:$1$. $3Br_2   6NaOH (	ext{hot and conc.}) ightarrow 5NaBr   NaBrO_3   3H_2O$. Thus,option $A$ is incorrect as it does not mention $NaBrO_3$.$2$. $SO_2   O_3 ightarrow SO_3   O_2$. This is correct.$3$. $Si   2NaOH   H_2O ightarrow Na_2SiO_3   2H_2$. This is correct.$4$. With excess $NH_3$,$Cl_2$ reacts as: $3Cl_2   8NH_3 ightarrow N_2   6NH_4Cl$. The statement in option $D$ claims $HCl$ is formed,but $HCl$ reacts with excess $NH_3$ to form $NH_4Cl$. Therefore,both $A$ and $D$ are technically incorrect based on standard chemical descriptions,but $D$ is the most commonly cited incorrect statement in this context due to the product being $NH_4Cl$ instead of $HCl$.

Identify the incorrect statement among the following.

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