Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following halogen liberates oxygen, when passed through hot concentrated KOH solution? (a) \(\mathrm{Cl}_{2}\) (b) \(\mathrm{F}_{2}\) (c) \(\mathrm{I}_{2}\) (d) \(\mathrm{Br}_{2}\)

Step-by-Step Solution

Answer

Fluorine (F_2) liberates oxygen when passed through hot concentrated KOH solution.

Step 1: Understanding the Reaction with KOH

When a halogen is passed through hot, concentrated KOH solution, it generally participates in a disproportionation reaction. In such reactions, part of the halogen is reduced and part is oxidized.

Step 2: Recognizing Halogen Behavior

The halogens are characterized by their ability to undergo disproportionation reactions. However, not all halogens will produce oxygen when mixed with KOH. We need to assess which halogen reacts in such a way as to liberate oxygen.

Step 3: Focusing on Fluorine

Fluorine (F_2) is unique among the halogens because it is the strongest oxidizing agent. Its reaction with hot concentrated KOH is special in that it doesn't just undergo disproportionation; instead, it oxidizes water from the solution to form oxygen.

Step 4: Writing the Chemical Reaction for F_2

When F_2 is passed through hot concentrated KOH, it reacts vigorously, and one of the products is oxygen gas (O_2). The balanced chemical equation is:\[ \mathrm{2F_2 + 2KOH \rightarrow 2KF + OF_2 + H_2O} \]Note that oxygen is liberated from this reaction as a result of the strong oxidizing property of fluorine.

Key Concepts

Disproportionation Reaction

A disproportionation reaction is a fascinating type of chemical reaction where a single substance undergoes simultaneous oxidation and reduction, forming two different products. In the context of halogens reacting with potassium hydroxide (KOH), the halogen acts both as an oxidizing and a reducing agent. Here’s how it works:

• One part of the halogen molecule gains electrons (is reduced).
• The other part loses electrons (is oxidized).

This dual nature of the reaction is especially noteworthy because not all elements exhibit such behavior. Disproportionation reactions are common in chemistry because they reflect the intermediate states that many elements can achieve. For halogens, this type of reaction often results in one product being a halide ion and another more oxidized form, showcasing the versatility of these elements.

Halogen Oxidation

Halogen oxidation is a crucial concept in understanding the reactivity of these elements. Oxidation refers to the process where an element loses electrons, increasing its oxidation state. In the case of halogens:

• They typically start at an oxidation state of 0 when in their elemental form.
• When they oxidize something, their oxidation state increases as they take on a more positive charge.

This electron transfer is fundamental to many chemical reactions involving halogens. For instance, when interacting with KOH, the most reactive halogen, fluorine, effectively removes electrons from water molecules (a component of the solution), demonstrating a shift in its own oxidation state. Understanding halogen oxidation helps us predict how these elements will behave in chemical reactions, making it a crucial point in the study of chemistry.

Oxidizing Agents

Oxidizing agents play a vital role in chemical reactions as they are the substances that undergo reduction by accepting electrons from another species. Halogens are prime examples of powerful oxidizing agents, and among them, fluorine stands at the top of the list. Here's why:

• Fluorine has the highest electronegativity of all elements, making it extremely effective in pulling electrons away from other substances.
• It can oxidize water itself to release oxygen, something other halogens cannot achieve.

Being a potent oxidizing agent makes fluorine highly reactive and somewhat dangerous. Its ability to oxidize water directly is an unusual trait that sets it apart, allowing it to liberate oxygen when reacting with compounds like hot concentrated KOH. This property is key to understanding the behavior and applications of fluorine in both laboratory and industrial settings.

Chemical Reactions of Fluorine

The chemical reactions of fluorine are unique due to its high reactivity and strong oxidizing power. When fluorine interacts with various substances, like KOH, it showcases unique properties distinct from other halogens.

• Fluorine's reaction with KOH leads to the liberation of oxygen, a reaction notable for both its vigor and its products.
• This reaction is represented by the equation: \[ \mathrm{2F_2 + 2KOH \rightarrow 2KF + OF_2 + H_2O} \]
• Here, fluorine not only forms potassium fluoride (KF) but also generates oxygen gas.

Understanding this reaction helps illustrate why fluorine is considered such a powerful and reactive element. Its ability to oxidize compounds, including water, makes it extraordinarily effective yet requires careful handling. This reaction with KOH is an excellent demonstration of fluorine's potential and challenges in chemical processes.