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: \([\mathrm{A}]+\mathrm{Na}\left[\mathrm{Cr}(\mathrm{OH})_{4}\right] \frac{\mathrm{H}_{2} \mathrm{O}_{z} \text {, boil }}{\longrightarrow}[\mathrm{B}] \frac{\mathrm{H}_{,} \mathrm{SO}_{4}}{\longrightarrow} \underset{\text { Yellow }}{\text { Orange colour }}[\mathrm{C}]\) 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

Option (b): \(\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\).

Step 1: Analyze the Reaction Part 1

The reaction begins with \( [\mathrm{A}] + \mathrm{Na}[\mathrm{Cr}(\mathrm{OH})_{4}] \). \( \mathrm{Na}[\mathrm{Cr}(\mathrm{OH})_{4}] \) is sodium chromite, which reacts with a base \([\mathrm{A}]\) when boiled. Problem suggests \([\mathrm{A}]\) is a common base. Examine options: (a), (b), and (d) have \(\mathrm{NaOH}\) as \([\mathrm{A}]\). Thus, \([\mathrm{A}] = \mathrm{NaOH}\).

Step 2: Reaction with Sodium Chromite

Upon boiling \( \mathrm{Na}[\mathrm{Cr}(\mathrm{OH})_{4}] \) with \( \mathrm{NaOH} \), it is converted to a yellow compound \([\mathrm{B}]\), which is sodium chromate \(\mathrm{Na}_{2}\mathrm{CrO}_{4}\). Look at options: (b) and (c) suggest this transformation for \([\mathrm{B}]\).

Step 3: Conversion to Final Product

When adding \( \mathrm{H}_{2}\mathrm{SO}_{4} \) to \([\mathrm{B}] \), an orange color compound \([\mathrm{C}]\) forms, which indicates the formation of sodium dichromate \(\mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\). This matches with \([\mathrm{C}]\) as proposed in options (b) and (c).

Step 4: Identify Correct Option

We found \([\mathrm{A}] = \mathrm{NaOH} \), \([\mathrm{B}] = \mathrm{Na}_{2}\mathrm{CrO}_{4} \), and \([\mathrm{C}] = \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\). Option (b) aligns with these values: \(\mathrm{NaOH}, \mathrm{Na}_{2}\mathrm{CrO}_{4}, \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\).

Key Concepts

Inorganic Chemistry Reactions

In chemistry, reactions occurring between inorganic substances are termed inorganic chemistry reactions. These interactions involve compounds that do not necessarily contain carbon as their primary element. In the context of our exercise, we are dealing with a reaction involving sodium chromite and sodium chromate.
Inorganic reactions can exhibit various characteristics such as color changes, precipitate formations, gas evolutions, or change in temperature. Understanding these features is crucial for identifying the progression and outcome of a reaction.
Our given reaction vividly highlights a color change—a transition from a yellow to an orange compound, indicative of the creation of new substances.
This exercise exemplifies the transformation of sodium chromate (\[ Na_2CrO_4 \]) to sodium dichromate (\[ Na_2Cr_2O_7 \]), emphasizing the fascinating interplay of chemical substances that defines inorganic chemistry reactions.

Sodium Chromate and Dichromate Formation

Sodium chromate and sodium dichromate are crucial compounds in various chemical reactions, known for their distinctive colors. Sodium chromate (\[ Na_2CrO_4 \]) is yellow, whereas sodium dichromate (\[ Na_2Cr_2O_7 \]) appears orange.
In our exercise, this transformation from yellow to orange signifies the conversion of chromate to dichromate when acid (usually sulfuric acid) is introduced into the system.
During the formation of sodium chromate, sodium chromite (\[ Na[Cr(OH)_4] \]) reacts with a base like sodium hydroxide. Upon the addition of sulfuric acid, the chromate ions combine to form dichromate ions, leading to this color shift.

• First, sodium chromite is boiled with \[ NaOH \] resulting in sodium chromate, a yellow compound.
• Second, the addition of sulfuric acid converts sodium chromate into sodium dichromate, an orange-colored compound.

This clear sequence provides an easy visualization of the formation and conversion process, essential for comprehending the nature of chromate and dichromate compounds.

Chemical Reaction Analysis

Chemical reaction analysis involves dissecting a reaction process step by step to understand the behavior and transformation of reactants into products.
In the context of our exercise, we begin by identifying the reactants: sodium chromite and a strong base, \[ NaOH \]. On boiling, a significant transformation into sodium chromate occurs, which is evident through the color change to yellow.
Next, with the addition of sulfuric acid, a further conversion is observed, forming sodium dichromate, as indicated by the orange color. Analyzing such transformations helps clarify the role of each reactant and the influencing conditions.
This systematic approach to reaction analysis allows students to:

• Understand reactant-product relationships.
• Monitor changes—in this case, through color.
• Predict the results of similar reactions by understanding the influence of different chemical agents.

By breaking down these reactions, students gain deeper insights into the mechanisms that drive chemical transformation, a foundational skill in chemistry.