Assertion : HC ≡ C– is more stable than H2C = CH–Reason : HC ≡ C– has more s-character than H2C = CH–

The assertion is correct. The HC≡C– ion is more stable than the H2C=CH– ion. This is due to the fact that the HC≡C– ion has a higher s-character in its hybridization.

Here's the step-by-step explanation:

1. The hybridization of the carbon atoms in HC≡C– is sp, which means it has 50% s-character and 50% p-character.

2. On the other hand, the hybridization of the carbon atoms in H2C=CH– is sp2, which means it has 33.33% s-character and 66.67% p-character.

3. The s-orbital is closer to the nucleus and thus more stable than the p-orbital. Therefore, the more s-character a molecule has, the more stable it is.

4. Therefore, since HC≡C– has more s-character than H2C=CH–, it is more stable.

So, the reason is correct and it is the correct explanation of the assertion.

The assertion and reason are both correct and they are related.

Step 1: Understanding the structures HC ≡ C– is an alkyne with a triple bond between the two carbon atoms. H2C = CH– is an alkene with a double bond between the two carbon atoms.

Step 2: Understanding s-character In hybridization, s-character refers to the percentage of an orbital's character that is derived from the s orbital. The more s-character an orbital has, the closer the electrons are to the nucleus, and the lower the energy and more stable the molecule.

Step 3: Applying s-character to the structures In HC ≡ C–, the carbon atoms are sp hybridized, meaning they have 50% s-character. In H2C = CH–, the carbon atoms are sp2 hybridized, meaning they have 33% s-character.

Step 4: Conclusion Since HC ≡ C– has more s-character than H2C = CH–, it is more stable. Therefore, the assertion is correct, and the reason correctly explains why.