Problem 151 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\) solution, oxidation number of chlorine changes from (a) 0 to \(-1\) (b) 0 to \(+5\) (c) 0 to \(+7\) (d) \(-1\) to 0

Step-by-Step Solution

Answer

The oxidation number of chlorine changes from 0 to -1 and from 0 to +5. (Options a and b)

Step 1: Understanding the Reaction

The reaction between chlorine gas \(\mathrm{Cl}_{2}\) and hot sodium hydroxide \(\mathrm{NaOH}\) is a disproportionation reaction. In this type of reaction, one element is simultaneously oxidized and reduced. The chemical equation for the reaction is: \[ \mathrm{Cl}_{2} + 6 \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{NaClO_{3}} + 3 \mathrm{H_{2}O} \] Here, chlorine goes from an oxidation state of 0 to -1 in \(\mathrm{NaCl}\) and from 0 to +5 in \(\mathrm{NaClO_{3}}\).

Step 2: Identifying the Oxidation States

Initially, chlorine atoms in \(\mathrm{Cl}_{2}\) have an oxidation number of 0. After the reaction: - In \(\mathrm{NaCl}\), chlorine is reduced to its -1 oxidation state. - In \(\mathrm{NaClO_{3}}\), chlorine is oxidized to a +5 oxidation state. Thus, chlorine undergoes both reduction and oxidation in this reaction.

Step 3: Conclusion Based on Oxidation Number Change

The reaction results in chlorine changing its oxidation state both from 0 to -1 and from 0 to +5. This demonstrates the disproportionation reaction where both reduction and oxidation of the chlorine take place. Therefore, the options correctly describing this change are (a) and (b).

Key Concepts

Oxidation States

Understanding oxidation states is key in chemistry as it helps us track electron transfer in reactions. An oxidation state, also known as an oxidation number, indicates the degree of oxidation of an atom in a chemical compound. It's useful to think of it as a charge on the atom if we consider the compound to be ionic.

• Oxidation states are assigned within molecules using specific rules.
• The sum of the oxidation states in a neutral compound is zero.
• In ions, the sum equals the charge of the ion.

For example, in the molecule \[\mathrm{Cl}_{2}\], each chlorine atom has an oxidation state of 0 because they are in their elemental form. The step-by-step tutorial about our reaction shows chlorine's oxidation state changing from 0 to -1 and from 0 to +5 after reacting with \(\mathrm{NaOH}\). This involves a transfer of electrons, which is fundamental in understanding redox reactions.

Chlorine Reactions

Chlorine is a highly reactive halogen known for its ability to participate in various chemical reactions. One interesting reaction involving chlorine is its reaction with sodium hydroxide \(\mathrm{NaOH}\), especially in a hot solution, which is a classic example of a disproportionation reaction.

• In a disproportionation reaction, a single element undergoes simultaneous oxidation and reduction.
• Chlorine reacts with hot \(\mathrm{NaOH}\) to form \(\mathrm{NaCl}\) and \(\mathrm{NaClO_{3}}\), illustrating its versatile chemistry.

This reaction is important for students to understand because it challenges the idea that a single element cannot be both oxidized and reduced in the same reaction. In this case, chlorine achieves this by moving to both a lower oxidation state in \(\mathrm{NaCl}\) and a higher one in \(\mathrm{NaClO_{3}}\).

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are key to understanding chemical processes. They involve the transfer of electrons between species, leading to the change in oxidation states.

• Reduction involves the gain of electrons, leading to a decrease in oxidation state.
• Oxidation involves the loss of electrons, resulting in an increase in oxidation state.

In our example, chlorine gas reacts with sodium hydroxide in a redox reaction. During this reaction, chlorine is both reduced and oxidized, a hallmark of a disproportionation reaction. In \(\mathrm{NaCl}\), chlorine is reduced as it gains electrons; in \(\mathrm{NaClO_{3}}\), it is oxidized since it loses electrons. Understanding these electron transfers helps clarify why the net number of electrons conserved in any redox reaction remains balanced.