Question

Dehydration of an organic acid X with concentrated \(\text{H}_2\text{SO}_4\) at 373K gives \(\text{H}_2\text{O}\) and gas Y. The hybridisation of the carbon in Y and nature of Y are respectively

• A. \(sp^2\), Neutral
• B. \(sp\), Neutral
• C. \(sp^2\), acidic
• D. \(sp\), acidic

Hint:

Common Neutral Oxides: \(\text{CO}, \text{NO}, \text{N}_2\text{O}\). All other non-metal oxides are generally acidic.

Answer 1

The Correct Option is B

Step 1: Identify X and Y:
The dehydration of carboxylic acids with conc. \(\text{H}_2\text{SO}_4\) is a specific reaction. Formic acid (\(\text{HCOOH}\)) undergoes dehydration to give Carbon Monoxide (\(\text{CO}\)) and water. Reaction: \(\text{HCOOH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \Delta} \text{CO} \uparrow + \text{H}_2\text{O}\). So, Acid X = Formic Acid, Gas Y = Carbon Monoxide (\(\text{CO}\)).
Step 2: Determine Hybridisation of C in CO:
The Lewis structure of CO is \(:\text{C} \equiv \text{O}:\). The carbon atom forms a triple bond (1 sigma + 2 pi) and has one lone pair. Steric number = (Number of sigma bonds) + (Lone pairs) = \(1 + 1 = 2\). Steric number 2 corresponds to \(sp\) hybridisation.
Step 3: Determine Nature of Y (CO):
Carbon monoxide (\(\text{CO}\)) is a well-known neutral oxide (along with \(\text{N}_2\text{O}\) and \(\text{NO}\)). It does not react with acids or bases to form salts. Conclusion:
Hybridisation is \(sp\) and nature is Neutral. Final Answer:
Option (B).