Compound A and B respectively are :-Cl_2 + 2N | vedclass

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(D) The reaction of chlorine with sodium hydroxide depends on the concentration and temperature of the $NaOH$ solution.$1$. With cold and dilute $NaOH

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$NaOCl, NaClO_3$

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(D) The reaction of chlorine with sodium hydroxide depends on the concentration and temperature of the $NaOH$ solution.$1$. With cold and dilute $NaOH$:$Cl_2 + 2NaOH (cold \& dil.) \rightarrow NaCl + NaOCl + H_2O$Here,$A$ is $NaOCl$ (sodium hypochlorite).$2$. With hot and concentrated $NaOH$:$3Cl_2 + 6NaOH (hot \& conc.) \rightarrow 5NaCl + NaClO_3 + 3H_2O$Here,$B$ is $NaClO_3$ (sodium chlorate).Therefore,$A$ is $NaOCl$ and $B$ is $NaClO_3$.

Compound $A$ and $B$ respectively are :-
$Cl_2 + 2NaOH (cold \& dil.) \rightarrow A + NaCl + H_2O$
$3Cl_2 + 6NaOH (hot \& conc.) \rightarrow B + 5NaCl + 3H_2O$

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Compound $A$ and $B$ respectively are :-$Cl_2 + 2NaOH (cold \& dil.) \rightarrow A + NaCl + H_2O$$3Cl_2 + 6NaOH (hot \& conc.) \rightarrow B + 5NaCl + 3H_2O$