Problem 154 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(\mathrm{Cl}_{2}\) is passed through hot conc. \(\mathrm{NaOH}\), Salts \(\mathrm{A}\) and \(\mathrm{B}\) are formed which are separated by fractional crystallization. When \(\mathrm{B}\) is treated with a calculated amount of \(\mathrm{KCl}\), a substance \(\mathrm{C}\) is produced. When \(\mathrm{C}\) is gently heated, it disproportionates to give \(\mathrm{D}\) and \(\mathrm{E}\). When \(\mathrm{E}\) is heated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\), the strongest bronsted acid \(\mathrm{F}\) is formed which is isoelectronic with \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Identify \(\mathrm{F}\). (a) \(\mathrm{HClO}_{2}\) (b) \(\mathrm{HClO}_{3}\) (c) \(\mathrm{HClO}_{4}\) (d) \(\mathrm{HCl}\)

Step-by-Step Solution

Answer

The compound \( \mathrm{F} \) is \( \mathrm{HClO}_4 \) (perchloric acid).

Step 1: Reaction of Chlorine with Hot Concentrated NaOH

The reaction of chlorine gas \( \mathrm{Cl}_2 \) with hot concentrated \( \mathrm{NaOH} \) leads to the formation of two salts, \( \mathrm{NaCl} \) and \( \mathrm{NaClO}_3 \). This can be represented by the equation: \[ 3 \mathrm{Cl}_2 + 6 \mathrm{NaOH} \rightarrow 5 \mathrm{NaCl} + \mathrm{NaClO}_3 + 3 \mathrm{H}_2\mathrm{O} \]. Here, \( \mathrm{A} \) is \( \mathrm{NaCl} \) and \( \mathrm{B} \) is \( \mathrm{NaClO}_3 \).

Step 2: Reaction of B with KCl to Form C

Salt \( \mathrm{B} \), which is \( \mathrm{NaClO}_3 \), reacts with a calculated amount of \( \mathrm{KCl} \) to form \( \mathrm{KClO}_3 \). This double displacement reaction can be represented as \[ \mathrm{NaClO}_3 + \mathrm{KCl} \rightarrow \mathrm{KClO}_3 + \mathrm{NaCl} \].

Step 3: Disproportionation of C

When \( \mathrm{C} \) (\( \mathrm{KClO}_3 \)) is gently heated, it undergoes disproportionation to form \( \mathrm{KCl} \) and \( \mathrm{KClO}_4 \). The thermal decomposition can be represented as: \[ 4 \mathrm{KClO}_3 \rightarrow 3 \mathrm{KClO}_4 + \mathrm{KCl} \]. Here, \( \mathrm{D} \) is \( \mathrm{KCl} \) and \( \mathrm{E} \) is \( \mathrm{KClO}_4 \).

Step 4: Reaction of E with Concentrated H2SO4

When \( \mathrm{E} \) (\( \mathrm{KClO}_4 \)) is treated with concentrated \( \mathrm{H}_2 \mathrm{SO}_4 \), it reacts to form \( \mathrm{HClO}_4 \), which is the strongest Bronsted acid that is isoelectronic with \( \mathrm{H}_2 \mathrm{SO}_4 \). The reaction can be summarized as: \[ \mathrm{KClO}_4 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{HClO}_4 + \mathrm{KHSO}_4 \]. Thus, \( \mathrm{F} \) is \( \mathrm{HClO}_4 \).

Key Concepts

Disproportionation Reaction

Disproportionation reactions are unique as a single substance is both oxidized and reduced simultaneously, forming two distinct products. It occurs when an element in one oxidation state transforms into both a higher and a lower oxidation state. Suppose you have a compound, like potassium chlorate (\( \text{KClO}_3 \)), and you gently heat it. It breaks down into potassium chloride (\( \text{KCl} \)) and potassium perchlorate (\( \text{KClO}_4 \)). Here, chlorine is undergoing disproportionation:

• One part of the chlorine in \( \text{KClO}_3 \) is reduced to \( \text{Cl}^- \) in \( \text{KCl} \).
• Another part is oxidized to \( \text{Cl}^{+7} \) in \( \text{KClO}_4 \).

This type of reaction is significant in inorganic chemistry for the synthesis of various compounds and understanding redox principles.

Fractional Crystallization

Fractional crystallization, also known as selective crystallization, is a technique to separate compounds based on their solubility differences. It is a valuable method when a mixture contains multiple salts or substances differing slightly in solubility in a particular solvent. Consider a mixture of \( \text{NaCl} \) and \( \text{NaClO}_3 \) produced from a reaction involving chlorine and hot \( \text{NaOH} \). Using water, you can cause one salt to crystallize first upon cooling. By selectively separating the newly formed crystals, you can isolate the desired compound efficiently:

• \( \text{NaCl} \) has lower solubility compared to \( \text{NaClO}_3 \).
• Upon cooling, \( \text{NaCl} \) crystallizes out first.
• \( \text{NaClO}_3 \) remains in solution and can be crystallized out at a different cooling step or using a different solvent.

Fractional crystallization is widely used in laboratories for purifying inorganic salts and even in industries for manufacturing precise chemical products.

Bronsted Acid

The concept of a Brønsted Acid is pivotal in understanding acid-base reactions in inorganic chemistry. A Brønsted Acid, according to the theory proposed by Johannes Brønsted, is any substance capable of donating a proton (\( \text{H}^+ \)). In the reaction involving \( \text{KClO}_4 \) and concentrated \( \text{H}_2\text{SO}_4 \), \( \text{HClO}_4 \) (perchloric acid) is formed as the strongest Brønsted acid.

• \( \text{HClO}_4 \) donates a \( \text{H}^+ \), which categorizes it as a Brønsted acid.
• It is known for its stability and strength, making it an excellent acid in various reactions.
• \( \text{HClO}_4 \) is isoelectronic with \( \text{H}_2\text{SO}_4 \), meaning they share the same number of electrons and similar structural framework.

Understanding Brønsted Acids helps in predicting reaction pathways and constructing mechanisms for more complex chemical transformations.