Question

Formic acid on heating with concentrated H\(_2\)SO\(_4\) at 373 K gives X, a colourless substance and Y, a good reducing agent. The number of \(\sigma\) and \(\pi\) bonds in X, Y are respectively:

• A. X = 2, 0; Y = 1, 2
• B. X = 1, 2; Y = 2, 2
• C. X = 2, 1; Y = 1, 1
• D. X = 1, 2; Y = 3, 3 \vspace{0.5cm}

Hint:

Decomposition reactions like this one often break molecules into simpler components, which involves analyzing the bonds in the products to understand the reaction mechanism.

Answer 1

The Correct Option is A

Step 1: Understanding the Reaction Formic acid (HCOOH) when heated with concentrated sulfuric acid at 373 K undergoes a decomposition reaction. The reaction breaks down formic acid into carbon monoxide (X) and water (Y): \[ HCOOH \xrightarrow{H_2SO_4} CO + H_2O \] - X is carbon monoxide (CO), a colourless gas. - Y is water (H\(_2\)O), which acts as a good reducing agent. Step 2: Analyzing the Number of Bonds in X (Carbon Monoxide) Carbon monoxide (CO) consists of a triple bond between carbon and oxygen. The bonding in CO is as follows: - 1 \(\sigma\) bond between C and O from the single overlap of their orbitals. - 2 \(\pi\) bonds from the sideways overlap of p-orbitals between C and O. Thus, for X = CO: - \(\sigma\) bonds = 2 - \(\pi\) bonds = 0 Step 3: Analyzing the Number of Bonds in Y (Water) Water (H\(_2\)O) consists of two O-H bonds. Each O-H bond contains: - 1 \(\sigma\) bond between oxygen and hydrogen from their orbital overlap. Thus, for Y = H\(_2\)O: - \(\sigma\) bonds = 2 - \(\pi\) bonds = 0 Step 4: Matching with the Given Options Now, comparing the results: - X (CO) has 2 \(\sigma\) bonds and 0 \(\pi\) bonds. - Y (H\(_2\)O) has 1 \(\sigma\) bond and 2 \(\pi\) bonds. The correct answer is: \[ \boxed{(1) \ X = 2, 0; \ Y = 1, 2} \] \vspace{0.5cm}