Question

The products obtained when chlorine gas reacts with cold and dilute aqueous $NaOH$ are:

• (A) $C{l}^{-}$ and $Cl{O}^{-}$
• (B) $C{l}^{-}$ and $Cl{O}_2^{-}$
• (C) $C{l}^{-}$ and $Cl{O}_3^{-}$
• (D) $C{l}^{-}$ and $Cl{O}_4^{-}$

Answer 1

The Correct Option is A

To determine the products of the reaction between chlorine gas and cold, dilute aqueous $NaOH$, we begin by considering the chemical reaction that occurs under these conditions.

When chlorine gas ($\mathrm{Cl}{2}_{}$) reacts with cold and dilute $NaOH$, a disproportionation reaction occurs. In a disproportionation reaction, a single substance is simultaneously reduced and oxidized to form two different products. For chlorine reacting with sodium hydroxide, the balanced chemical equation is:

$2\mathrm{Na}\mathrm{OH}+{\mathrm{Cl}}_2\to \mathrm{NaCl}+\mathrm{NaClO}+\mathrm{H₂O}$

In this reaction:

• $\mathrm{Cl}-$ (chloride ion) is formed from the reduction of chlorine.
• $\mathrm{ClO}-$ (hypochlorite ion) is formed from the oxidation of chlorine.

This means the products of the reaction are sodium chloride ($\mathrm{NaCl}$) and sodium hypochlorite ($\mathrm{NaClO}$), which contain the chloride ion and hypochlorite ion, respectively.

Now, let us analyze the given options:

• Option (A): $C{l}^{-}$ and $Cl{O}^{-}$ — This option is correct as it states the chloride ion and hypochlorite ion, which are indeed the products of this reaction.
• Option (B) - (D): These options propose different ions or incorrect ions such as $Cl{O}_2^{-}$, $Cl{O}_3^{-}$, or $Cl{O}_4^{-}$, which are incorrect under these reaction conditions.

Therefore, the correct answer is Option (A): $C{l}^{-}$ and $Cl{O}^{-}$.