Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

A sodium salt on treatment with \(\mathrm{MgCl}_{2}\) gives white precipitate only on heating. The anion of the sodium salt is (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{CO}_{3}^{2-}\) (c) \(\mathrm{NO}_{3}^{-}\) (d) \(\mathrm{SO}_{4}^{2-}\)

Step-by-Step Solution

Answer

The anion is \(\mathrm{HCO}_{3}^{-}\) (option a).

Step 1: Understand the Problem

We need to identify the anion of a sodium salt that gives a white precipitate upon treatment with \(\mathrm{MgCl}_{2}\) only after heating.

Step 2: Consider the Reaction Conditions

The formation of a white precipitate only after heating suggests a thermal decomposition or change that results in an insoluble compound.

Step 3: Analyze Reaction with Each Anion

- (a) \(\mathrm{HCO}_{3}^{-}\): When heated, \(\mathrm{HCO}_{3}^{-}\) decomposes to \(\mathrm{CO}_{3}^{2-}\), which can form an insoluble \(\mathrm{MgCO}_{3}\) precipitate with \(\mathrm{MgCl}_{2}\).- (b) \(\mathrm{CO}_{3}^{2-}\): Forms \(\mathrm{MgCO}_{3}\) with \(\mathrm{MgCl}_{2}\) directly, not requiring heat for precipitation.- (c) \(\mathrm{NO}_{3}^{-}\): \(\mathrm{Mg(NO}_{3})_{2}\) is soluble in water, no precipitate forms.- (d) \(\mathrm{SO}_{4}^{2-}\): \(\mathrm{MgSO}_{4}\) is soluble in water, no precipitate forms.

Step 4: Identify Correct Anion

Only \(\mathrm{HCO}_{3}^{-}\) forms \(\mathrm{MgCO}_{3}\) on heating, which precipitates.

Key Concepts

Thermal Decomposition

Thermal decomposition is an important chemical process that involves the breakdown of a compound into two or more simpler substances when heat is applied. This process is crucial in identifying certain anions where heat triggers specific reactions. In our example, we have the anion \(\mathrm{HCO}_{3}^{-}\) (bicarbonate ion), which undergoes thermal decomposition. When heated, this anion decomposes into \(\mathrm{CO}_{3}^{2-}\) (carbonate ion). Therefore, understanding thermal decomposition helps identify substances based on how they react to heat.

• Thermal stability decides whether a compound will decompose upon heating.
• Different compounds decompose into various products.
• For instance, \(\mathrm{HCO}_{3}^{-}\) becomes \(\mathrm{CO}_{3}^{2-}\) when heated, forming a new compound.

Understanding which compounds decompose can help us design experiments or predict the behavior of chemicals under thermal conditions.

White Precipitate Formation

White precipitate formation is a visual indicator of a chemical reaction where two solutions react to form an insoluble solid. This solid is called a precipitate. In the problem with the sodium salt and \(\mathrm{MgCl}_{2}\), the heating process leads to the formation of a white precipitate of \(\mathrm{MgCO}_{3}\).

• A precipitate forms when the product of a reaction is insoluble in water.
• Visual cues like color change can help in identifying the products and their transformations.
• The white precipitate indicates the reaction has occurred and the specific insoluble product, \(\mathrm{MgCO}_{3}\), has been formed.

White precipitate formation provides practical evidence of reaction pathways and can be used in experiments to identify unknown substances or confirm reaction conditions, making it a key tool in analytical chemistry.

Chemical Reaction Analysis

Chemical reaction analysis involves breaking down and understanding the sequence of events in a reaction to determine the final products and their properties. By analyzing the reaction between sodium bicarbonate and \(\mathrm{MgCl}_{2}\), we identify that heating induces the reaction to form \(\mathrm{MgCO}_{3}\) precipitate.

• The given reaction starts with identifying the reactants involved.
• Analyzing how the reaction conditions (like heating) influence the pathway.
• Examining the formation of products to understand which reactions occur.

Analyzing chemical reactions helps to predict outcomes, troubleshoot potential issues, and optimize the reaction conditions for desired products. In this case, understanding that only \(\mathrm{HCO}_{3}^{-}\) reacts with heat to form \(\mathrm{MgCO}_{3}\) highlights the importance of heat in facilitating the proper chemical pathway.