Question

Assertion (A): Chlorine gas disproportionate in hot and conc. \[\text{ NaOH}\] solution.
Reason (R): \[\text{ NaCI}\] and \[\text{ NaOCI}\] are formed in the above reaction.
A) Both (A) and (R) are true and (R) is the correct explanation of (A).
B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
C) (A) is true but (R) is false.
D) (A) is false and (R) is true.

Answer 1

Hint: The sodium hydroxide reacts with the chlorine at various temperatures and concentration and gives the different products. Sodium hydroxide and chlorine is an oxidation-reduction (redox) reaction. Here the chlorine exhibits the variable oxidation state, such that one of the oxidation states is higher and the other is lower than the intermediate or initial oxidation state of chlorine. This is a special type of reaction as they undergo the disproportion.

Complete step by step answer:
In redox chemistry, the reaction in which the one compound having the intermediate oxidation state converted itself into two compounds, such that one acquires the higher and another one acquires the lower oxidation state is called as the disproportion reaction. The general reaction can be depicted as follows:
$\text{ 2A}\to {{\text{A}}^{\text{ }\!\!'\!\!\text{ }}}\text{+}{{\text{A}}^{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}\text{ }$
Here, the oxidation state of A changes to ${{\text{A}}^{\text{ }\!\!'\!\!\text{ }}}$ and${{\text{A}}^{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}$.

Let us consider the reaction of chlorine $\text{ C}{{\text{l}}_{\text{2}}}$ with hot and concentrated sodium hydroxide$\text{ NaOH}$. The hot and concentrated sodium hydroxide and chlorine gas reaction is a disproportion reaction. In hot condition the chlorine gas reacts with the sodium hydroxide as follows:
\[\begin{matrix}
   \text{3C}{{\text{l}}_{\text{2}}} & \text{+} & \text{6NaOH} & \to & \text{5NaCl} & \text{+} & \text{NaCl}{{\text{O}}_{\text{3}}} & \text{+} & \text{3}{{\text{H}}_{\text{2}}}\text{O} \\
   \text{(Chlorine gas)} & {} & \text{(Sodium hydroxide)} & {} & {} & {} & \text{(Sodium Chlorate)} & {} & {} \\
\end{matrix}\]

Here, we observed that chlorine $\text{ C}{{\text{l}}_{\text{2}}}$ has an oxidation state of zero. It is oxidized to $\text{ ClO}_{\text{3}}^{-}$ and reduced to the $\text{ C}{{\text{l}}^{-}}$ion.
Let's calculate the change oxidation in the oxidation state to chlorine. The $\text{ C}{{\text{l}}_{\text{2}}}$is initially has the zero oxidation state. It gains the 2 electrons and is reduced to the $\text{ C}{{\text{l}}^{-}}$ion. The oxidation state for the chloride ion would be,
$\text{Oxidation state C}{{\text{l}}^{\text{-}}}\text{= -1}$

Now, the chlorine is oxidized to form the chlorate ion. The oxidation state of chorine is chlorate is
$\begin{align}
  & \text{Total charge on ion = O}\text{.S}\text{. of Cl + 3 (O}\text{.S}\text{. of O)} \\
 & \Rightarrow \text{ -1 = x + 3(-2)} \\
 & \Rightarrow \text{ x = -1 + 6} \\
 & \therefore \text{ x = +5} \\
\end{align}$

Therefore the oxidation state of chlorine in chlorate $\text{ ClO}_{\text{3}}^{-}$ is$+5$. Chlorine is oxidized into the chlorate ion. The chlorine exhibit the variable oxidation state $(-1\text{ , +5 )}$ in presence of sodium hydroxide. Therefore the reaction of chlorine with sodium hydroxide is a disproportion reaction.
Therefore the given assertion is correct.

However, the chlorine does not produce hypochlorite $\text{OC}{{\text{l}}^{\text{-}}}$ ion in hot conditions. This is because the hypochlorite ion $\text{OC}{{\text{l}}^{\text{-}}}$ exists only in cold states. The $\text{OC}{{\text{l}}^{\text{-}}}$ion is unstable in cold conditions. In the hot conditions, it disproportionate into chloride ion and chlorate ion.
$\text{ 3OC}{{\text{l}}^{\text{-}}}\to \text{ ClO}_{\text{3}}^{\text{-}}\text{+2C}{{\text{l}}^{\text{-}}}$
Thus the given reason does not justify the assertion, therefore, the reason is false.

Hence, (C) is the correct option.

Note: As a chlorine gas, other halogens also react with the sodium hydroxide to give similar disproportionate products. This is not applicable for fluorine. It cannot exhibit the multiple oxidation states, it only exists in $-1$ the oxidation state due to the absence of p and d orbital.