Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

Find \([\mathrm{A}],[\mathrm{B}]\) and \([\mathrm{C}]\) respectively in the reaction given below: \([\mathrm{A}]+\mathrm{Na}\left[\mathrm{Cr}(\mathrm{OH})_{4}\right] \stackrel{\mathrm{H}_{2} \mathrm{O}_{2}, \text { boil }}{\longrightarrow}[\mathrm{B}] \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}[\mathrm{C}]\) Yellow Orange colour compound (a) \(\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{Na}_{2} \mathrm{CrO}_{4}\) (b) \(\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (c) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{NaOH}, \mathrm{K}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

Step-by-Step Solution

Answer

The answer is (b): \(\mathrm{NaOH}, \mathrm{Na}_2 \mathrm{CrO}_4, \mathrm{Na}_2 \mathrm{Cr}_2\mathrm{O}_7\).

Step 1: Identify initial reactant

In the given reaction, the starting material is \([\mathrm{A}]\) which is sodium hydroxide, \(\mathrm{NaOH}\). \([\mathrm{A}] = \mathrm{NaOH}\). This is consistent with both the options (a), (b), and (d). The \(\mathrm{NaOH}\) is needed to provide the alkaline medium for the reaction to occur.

Step 2: Determine intermediate product

The reaction described is a chromate preparation reaction. \([\mathrm{B}]\) is formed when \(\mathrm{Na[Cr(OH)_4]}\) reacts with \(\mathrm{NaOH}\) in the presence of \(\mathrm{H}_2\mathrm{O}_2\) and upon boiling, creating \(\mathrm{Na}_2\mathrm{CrO}_4\). So, \([\mathrm{B}] = \mathrm{Na}_2\mathrm{CrO}_4\). This matches options (b) and (c). \(\mathrm{Na}_2\mathrm{CrO}_4\) is a yellow compound, which supports our choice so far.

Step 3: Identify final product

\([\mathrm{C}]\) is formed when \(\mathrm{Na}_2\mathrm{CrO}_4\) reacts with \(\mathrm{H}_2\mathrm{SO}_4\) to form \(\mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7\), a dichromate with an orange color. The color change from yellow to orange confirms the conversion from chromate to dichromate. Thus, \([\mathrm{C}] = \mathrm{Na}_2\mathrm{Cr}_2 \mathrm{O}_7\). This matches only option (b).

Key Concepts

Sodium Hydroxide in Reactions

Sodium hydroxide, or \( \text{NaOH} \), is a common chemical widely used in various reactions and processes. In the chromate-dichromate reaction, \( \text{NaOH} \) plays a crucial role by providing an alkaline environment. Additionally, the alkaline nature of \( \text{NaOH} \) helps in breaking down complex metal hydroxides into simpler forms. As the starting reactant \([\text{A}]\) in this specific reaction, \( \text{NaOH} \), helps facilitate the conversion of chromium compounds.
Sodium hydroxide is often favored for its high solubility and reactive nature. In this reaction, it sets up conditions that allow other reactants to transform efficiently. It's important in maintaining the right pH level, which optimizes the reaction conditions for converting chromium hydroxide complexes into chromates.

• Provides an alkaline medium.
• Maintains the correct pH level.
• Enhances reaction speed by creating optimal conditions.

Chromium Compounds

Chromium compounds play a significant role in chemistry due to their vibrant colors and versatility in reactions. In this problem, the transformation of chromium compounds is central to identifying the correct sequence of products and reactants. Initially, \( \text{Na[Cr(OH)]}_4 \) undergoes a series of changes due to interactions with \( \text{NaOH} \) and \( \text{H}_2\text{O}_2 \), leading to the formation of \( \text{Na}_2\text{CrO}_4 \).
Chromates \([\text{B}]\) and dichromates \([\text{C}]\) are two primary forms of chromium compounds that can be interconverted depending on the reaction conditions. The chromate-dichromate equilibrium is particularly fascinating due to its dependence on the pH level:

• Chromates (\( \text{CrO}_4^{2-} \)): Yellow compounds.
• Dichromates (\( \text{Cr}_2\text{O}_7^{2-} \)): Orange-red compounds.

These transformations often involve hydrogen peroxide (\( \text{H}_2\text{O}_2 \)) or sulfuric acid (\( \text{H}_2\text{SO}_4 \)), contributing to their conversion by altering pH.

Reaction Intermediates

Intermediates are the middle-men of a reaction. They don't appear in the initial or final equation but are crucial for the reaction's progression. In this task, \( \text{Na}_2\text{CrO}_4 \) acts as such an intermediate \([\text{B}]\). It forms when \( \text{Na[Cr(OH)]}_4 \) reacts in the presence of sodium hydroxide and hydrogen peroxide while boiling. This yellow compound then moves forward in the reaction scheme.
The role of intermediates is crucial as they bridge the gap between reactants and products. They often determine the pathway and speed of the reaction. In the chromate-dichromate transition, the intermediate confirms the direction of the reaction shift crucial for identifying the subsequent product, \( \text{Na}_2\text{Cr}_2\text{O}_7 \). Understanding the behavior of intermediates helps predict and control the conditions necessary for desired outputs.

Color Change in Chemistry

In chemistry, color change often signifies a reaction's progression. In the chromate-dichromate reaction, it is particularly noticeable due to the vivid colors of the products. \( \text{Na}_2\text{CrO}_4 \), a yellow compound, serves as an intermediate, and when reacted with sulfuric acid, it transitions to \( \text{Na}_2\text{Cr}_2\text{O}_7 \), which has an intense orange color.
This yellow to orange transition not only confirms the progression to the final product but also helps anticipate the correct reaction pathway. The specific color change is a distinct marker in the study of chromium compounds for educational and practical applications globally.

• Yellow: Chromates typically present in alkaline solutions.
• Orange: Indicates the presence of dichromates in acidic conditions.

Understanding these color shifts is vital for students in learning reaction pathways and identifying reaction completion.

Acid-Base Reactions in Chemistry

Acid-base reactions are fundamental in chemistry, involving the transfer of protons between reactant species. In the chromate-dichromate reaction, the pH plays a prominent role in determining the equilibrium position. Sodium hydroxide serves as the base, while sulfuric acid serves as the acid, driving the transformation of chromates to dichromates.
Through this acid-base dynamic, students can appreciate how pH alters reaction pathways and outcomes. When \( \text{H}_2\text{SO}_4 \) is added to \( \text{Na}_2\text{CrO}_4 \), it shifts the equilibrium toward forming \( \text{Na}_2\text{Cr}_2\text{O}_7 \), showcasing the powerful influence of acid inputs.
In such reactions:

• Bases adjust conditions, often promoting different pathways and products.
• Acids can steer the reactions toward specific products by altering the environment.

Grasping these interactions aids learners in predicting and manipulating reaction outcomes in laboratory settings.