Question

When $C{l_2}$ is passed through cold dilute $NaOH$, the products are
(A) $NaCl$, $NaOCl$ and ${H_2}O$
(B) $NaCl$, $NaC{l_2}$ and ${H_2}O$
(C) $NaCl$, $NaC{l_2}$ and ${H_2}O$
(D) $NaCl$, $NaC{l_4}$ and ${H_2}O$

Answer 1

Hint: As cold and dilute $NaOH$solution is used, the addition of $C{l_2}$ gas will be comparatively less prominent.So, we can easily rule out the options in which there are more chlorine atoms Chlorine reacts violently with a hot and concentrated solution of $NaOH$. The reaction is so fast that it forms $NaC{l_2}$ , $NaC{l_2}$ , $NaC{l_4}$ .

Complete step by step answer:
The reaction between cold and dilute $NaOH$ and $C{l_2}$ gas is given as

$2{NaOH} + C{l_2} \to NaCl + NaOCl + {H_2}O$
$Cold\; and\; dil.$

This reaction is a disproportionation reaction,$C{l_2}$ gets converted to $NaCl$, this is a reduction reactionAlso, $C{l_2}$ gets converted to $NaOCl$, this is an oxidation reaction.Now, the reaction between hot and concentrated $NaOH$ and $C{l_2}$ gas is given as

$6NaOH + 3C{l_2} \to 5NaCl + NaCl{O_3} + 3{H_2}O$

This reaction is also disproportionation reaction,$C{l_2}$ gets converted to $NaCl$, this is a reduction reactionAlso, $C{l_2}$ gets converted to $NaCl{O_3}$, this is an oxidation reaction.
Disproportionation reaction involves simultaneous oxidation and reduction of the same reactant in the same reaction.The reason behind this is, the hypochlorite ion i.e.$OC{l^ - }$ ion is unstable and is stable at lower temperatures.With an increase in temperature, the $OC{l^ - }$ion disproportionates to chloride and chlorate ion.

$3OC{l^ - } \to 2C{l^ - } + ClO_3^ - $

As the temperature increases, the oxidation reaction is more favored.Therefore, the products should be $NaCl$, $NaOCl$ and ${H_2}O$.

So, the correct answer is Option A .

Note: The solution of $NaCl$ and $NaOCl$ is called the Milton solution. A concentrated solution of $NaOH$ is corrosive. Chlorine, a halogen, is a highly electronegative element. Its affinity is so strong that the reaction proceeds violently.