Problem 27 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

Option (c): 0 to +5.

Step 1: Identify the Initial Oxidation State of Cl

The initial oxidation state of chlorine in \(\mathrm{Cl}_2\) is 0, because it is in its elemental form.

Step 2: Consider the Reaction With Hot NaOH

When \(\mathrm{Cl}_2\) is reacted with hot \(\mathrm{NaOH}\), it undergoes a disproportionation reaction where it gets both oxidized and reduced. The products formed are \(\mathrm{NaCl}\) and \(\mathrm{NaClO_3}\).

Step 3: Determine the Oxidation States in the Products

In \(\mathrm{NaCl}\), chlorine has an oxidation state of \(-1\) because \(\mathrm{Na}\) typically has an oxidation state of \(+1\). In \(\mathrm{NaClO_3}\), chlorine has an oxidation state of \(+5\).

Step 4: Identify the Change in Oxidation Number

From step 2, we saw that chlorine changes from an oxidation number of 0 in \(\mathrm{Cl}_2\) to \(+5\) in \(\mathrm{NaClO_3}\).

Step 5: Select the Correct Option

Based on the oxidation changes, the option where chlorine changes from 0 to \(+5\) matches, which is option (c).

Key Concepts

Oxidation Number

Understanding oxidation numbers is crucial for identifying and balancing redox reactions. An oxidation number is a hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. It's a useful tool for keeping track of electrons transferred in reactions.

• In elemental form, such as \ Cl_2 \, the oxidation number is 0.
• For monoatomic ions, the oxidation number is equal to the charge of the ion. For example, the Na in NaCl has an oxidation number of +1.
• Oxygen typically has an oxidation number of \(-2\), with some exceptions like in peroxides.

In compound \ \mathrm{NaCl} \, chlorine takes an oxidation number of \(-1\), while in \ \mathrm{NaClO_3} \, due to its bonding with oxygen, chlorine has an oxidation number of \(+5\). Knowing these changes helps us analyze reactions on a deeper level.

Chlorine Reactions

Chlorine is a reactive element that can form various compounds through chemical reactions. When chlorine reacts with different substances, it can change its oxidation number, as it does in disproportionation reactions.
Reaction with Hot NaOH:

• In the reaction between chlorine \ (\mathrm{Cl}_2)\ and hot sodium hydroxide \ (\mathrm{NaOH}) \, chlorine undergoes a disproportionation reaction.
• During this process, chlorine is simultaneously oxidized and reduced, forming two different products: \ \mathrm{NaCl} \ and \ \mathrm{NaClO_3} \.
• \ \mathrm{Cl}_2 \ \, as an element, has an oxidation state of 0, which changes to \(-1\) in \ \mathrm{NaCl} \ and \(+5\) in \ \mathrm{NaClO_3} \.

The balanced reaction shows us how chlorine's reactivity allows it to form compounds where it takes on different oxidation states.

Redox Reactions

Redox reactions involve the transfer of electrons between substances, leading to a change in oxidation numbers. These reactions are characterized by the reduction and oxidation processes occurring simultaneously.

• Oxidation: This is the process of losing electrons and increasing the oxidation number. In our chlorine reaction, Cl is oxidized from 0 to +5 in \ \mathrm{NaClO_3} \.
• Reduction: This is the process of gaining electrons and decreasing the oxidation number. Chlorine is reduced from 0 to \(-1\) in \ \mathrm{NaCl} \.

During the reaction between \ (\mathrm{Cl}_2)\ and hot \ \mathrm{NaOH} \, both these processes occur, showcasing a disproportionation redox reaction. Understanding these principles can help you solve related problems more effectively and appreciate the transfer of electrons in different chemical contexts.