Understanding Chlorine Disproportionation in Hot NaOH

The question asks about the changes in the oxidation state of chlorine when it undergoes a disproportionation reaction in a hot solution of sodium hydroxide (NaOH). A disproportionation reaction is a type of redox reaction where an element in a specific oxidation state is simultaneously oxidized and reduced to higher and lower oxidation states, respectively.

In this reaction, chlorine gas ($\text{Cl}_2$) is the reactant, where the oxidation state of chlorine is 0.

When chlorine gas is passed through a hot concentrated solution of NaOH, the following reaction occurs:

\begin{equation*} 3\text{Cl}_2 + 6\text{NaOH (hot)} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O} \end{equation*}

Let's determine the oxidation states of chlorine in the products:

• In $\text{NaCl}$: Sodium (Na) is in group 1, so its oxidation state is +1. Chloride (Cl) is bonded to Na, so its oxidation state is -1. Thus, chlorine changes from 0 to -1.
• In $\text{NaClO}_3$: Sodium (Na) is +1. Oxygen (O) usually has an oxidation state of -2. Let the oxidation state of chlorine be \(x\). The sum of oxidation states in a neutral compound is 0. So, we have:

\begin{equation*} (+1) + (x) + 3(-2) = 0 \end{equation*}

\begin{equation*} 1 + x - 6 = 0 \end{equation*}

\begin{equation*} x - 5 = 0 \end{equation*}

\begin{equation*} x = +5 \end{equation*}

So, in $\text{NaClO}_3$, the oxidation state of chlorine is +5. Thus, chlorine changes from 0 to +5.

In the disproportionation reaction with hot NaOH, chlorine's oxidation state changes from 0 to -1 (reduction) and from 0 to +5 (oxidation). This confirms that it is indeed a disproportionation reaction.

Analyzing the Oxidation State Changes and Options

We found that the zero-oxidation state of chlorine changes to -1 and +5.

Let's look at the given options for the changes in oxidation state:

• (A) 0 to +5
• (B) 0 to -1
• (C) 0 to +3
• (D) 0 to +1

Our analysis shows that the correct changes are 0 to +5 and 0 to -1, which correspond to options (A) and (B).

Now, let's evaluate the multiple-choice selections based on these findings:

• Option 1: (A), (C) only - This suggests changes to +5 and +3. Incorrect, as we found +5 and -1.
• Option 2: (B), (C) only - This suggests changes to -1 and +3. Incorrect, as we found +5 and -1.
• Option 3: (A), (B) only - This suggests changes to +5 and -1. Correct, as our analysis showed these changes.
• Option 4: (A), (D) only - This suggests changes to +5 and +1. Incorrect, as we found +5 and -1 (the change to +1 occurs with cold NaOH, not hot).

Therefore, the correct combination of oxidation state changes for chlorine in hot NaOH disproportionation is 0 to +5 and 0 to -1, corresponding to options (A) and (B).

Revision Table: Chlorine Disproportionation

Additional Information: Chlorine Disproportionation with NaOH

The disproportionation reaction of chlorine with sodium hydroxide is a classic example of how reaction conditions (specifically temperature) can influence the products formed and thus the oxidation states involved. While both cold and hot NaOH solutions lead to disproportionation, the higher temperature favors the formation of the more oxygenated species, chlorate (\(\text{ClO}_3^-\)), where chlorine has a higher oxidation state (+5), instead of hypochlorite (\(\text{ClO}^-\)), where chlorine has a +1 oxidation state.

The reaction with cold dilute NaOH is:

\begin{equation*} \text{Cl}_2 + 2\text{NaOH (cold)} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \end{equation*}

In this case, the oxidation states of chlorine in the products are -1 (in NaCl) and +1 (in NaClO). Here, the changes are from 0 to -1 and 0 to +1. This highlights why specifying 'hot' NaOH in the question is crucial.

Disproportionation reactions are important in chemistry as they demonstrate how a single substance can act as both an oxidizing agent and a reducing agent.