Problem 51 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following pairs of reactant does not form oxygen when they react with each other? (a) \(\mathrm{Cl}_{2}, \mathrm{NaOH}\) solution (cold, dilute) (b) \(\mathrm{F}_{2}, \mathrm{NaOH}\) solution (hot, concentrated) (c) \(\mathrm{F}_{2}, \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{CaOCl}_{2}, \mathrm{H}_{2} \mathrm{SO}_{s}\) (dilute)

Step-by-Step Solution

Answer

The pair \( \text{Cl}_{2}, \text{NaOH} \) does not form oxygen when reacting.

Step 1: Analyzing Reaction A

In the reaction between \( \text{Cl}_{2} \) and cold, dilute \( \text{NaOH} \), chlorine disproportionates to form hypochlorite and chloride ions. This reaction does not produce oxygen as a by-product. The reaction can be written as: \[ 2 \text{Cl}_2 + 2 \text{NaOH} \rightarrow \text{NaClO} + \text{NaCl} + \text{H}_2\text{O} \]

Step 2: Analyzing Reaction B

In the reaction between \( \text{F}_{2} \) and hot, concentrated \( \text{NaOH} \), fluorine produces oxygen along with other products such as sodium fluoride and ozone (\( \text{O}_3 \)). The reaction may be complex, resulting in the production of various oxides and fluorides.

Step 3: Analyzing Reaction C

When \( \text{F}_{2} \) reacts with \( \text{H}_{2} \text{O} \), it oxidizes water to produce oxygen along with hydrogen fluoride. The reaction is: \[ 2 \text{F}_2 + 2 \text{H}_2\text{O} \rightarrow 4 \text{HF} + \text{O}_2 \] This clearly results in the formation of oxygen gas.

Step 4: Analyzing Reaction D

In the reaction between \( \text{CaOCl}_{2} \) and dilute \( \text{H}_{2} \text{SO}_{4} \), hypochlorous acid and calcium sulfate are produced, and oxygen gas is released as a by-product. The simplified reaction can be:\[ \text{Ca(OCl)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} + \text{O}_2 \]

Step 5: Conclusion

Upon examining all reactions, the pair that does not produce oxygen gas is option (a) \( \text{Cl}_{2} \) and \( \text{NaOH} \) solution (cold, dilute). All other options lead to the formation of oxygen.

Key Concepts

Disproportionation Reaction

In a disproportionation reaction, a single substance is simultaneously oxidized and reduced, resulting in the formation of two different products. This type of reaction is common in chemistry and is a fascinating example of redox processes.
In the case of chlorine reacting with sodium hydroxide, the reaction is:

• Chlorine (Cl₂) undergoes disproportionation.
• It splits into two different oxidation states, forming hypochlorite (NaClO) and chloride ions (NaCl).

As observed here, the chlorine is both reduced to form chloride ( Cl^{-} ) and oxidized to form hypochlorite ( ClO^{-} ). This type of reaction highlights the dual nature where the same element undergoes both reduction and oxidation in a single chemical process.
Disproportionation reactions showcase the dynamic behavior of elements and their ability to change oxidation states, creating various products from the same set of reactants. This is crucial in understanding the nature of complex chemical transformations.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are fundamental chemical processes where electrons are transferred between substances. They are central to many biological and chemical systems and are essential for the operation of mechanisms like batteries and metabolic pathways.
In a redox reaction:

• Oxidation: A substance loses electrons.
• Reduction: A substance gains electrons.

The reaction between fluorine and water is a classic example. Here, water is oxidized to release oxygen, while fluorine is reduced to produce hydrogen fluoride. This can be represented by the equation:\[2\,\text{F}_2 + 2\,\text{H}_2\text{O} \rightarrow 4\,\text{HF} + \text{O}_2\]
In this case, water serves as the reducing agent, losing electrons to form oxygen, and fluorine acts as the oxidizing agent, gaining those electrons.
Understanding redox reactions not only helps in solving chemical equations but also in grasping the fundamental principles of electron transfer in various scientific and industrial applications.

Reaction Mechanisms

Chemical reactions are often complex, involving multiple steps and intermediates. A reaction mechanism is the step-by-step sequence of elementary reactions by which overall chemical change occurs. It provides a detailed picture of the pathway from reactants to products, highlighting the formation and breaking of bonds.
When analyzing the reaction mechanisms for the exercise given, each reaction has its unique pathway. For instance, when calcium hypochlorite reacts with sulphuric acid, oxygen production occurs through a series of transformations involving the breaking and formation of chemical bonds.
Different chemical species, such as intermediates and activated complexes, that are formed transiently during the reactions are key to understanding these pathways.
Mechanisms may vary in complexity, from straightforward single-step reactions to multi-step processes involving several intermediates. By studying reaction mechanisms, chemists can comprehend not only how a reaction proceeds but also the energies involved at each stage. This knowledge extends to practical applications like designing more efficient industrial processes and creating new materials.