Question

Hybridization of positively charged and negatively charged carbons of the following respectively are

• A. \( sp^2 \), \( sp \)
• B. \( sp^2 \), \( sp^2 \)
• C. \( sp^3 \), \( sp^3 \)
• D. \( sp^3 \), \( sp^2 \)

Hint:

To determine the hybridization of a carbon atom, count the number of sigma bonds and lone pairs around it. A positively charged carbon with three sigma bonds and no lone pairs is \( sp^2 \) hybridized. A negatively charged carbon in a triple bond with one sigma bond and one lone pair is \( sp \) hybridized. Remember that pi bonds do not contribute to hybridization.

Answer 1

The Correct Option is A

To determine the hybridization of a carbon atom, we count the number of sigma (\( \sigma \)) bonds and lone pairs around it.
The steric number (number of sigma bonds + number of lone pairs) determines the hybridization: - Steric number = 2: \( sp \) hybridization - Steric number = 3: \( sp^2 \) hybridization - Steric number = 4: \( sp^3 \) hybridization Let's analyze the positively charged carbon in \( (CH_3)_2\overset{+}{C}H \): The positively charged carbon is bonded to two methyl groups (\( -CH_3 \)) and one hydrogen atom.
There are three sigma bonds and no lone pairs.
Thus, the steric number is 3.
Therefore, the hybridization of the positively charged carbon is \( sp^2 \).
Now, let's analyze the negatively charged carbon in \( CH_3C \equiv C^{\ominus} \): The negatively charged carbon is bonded to one methyl group (\( -CH_3 \)) via a sigma bond and to another carbon via one sigma bond and two pi (\( \pi \)) bonds (in the triple bond).
The negative charge indicates the presence of a lone pair of electrons on this carbon.
Thus, there are two sigma bonds and one lone pair.
The steric number is 2 + 1 = 3.
Wait, this is incorrect.
Let's re-evaluate the negatively charged carbon in \( CH_3C \equiv C^{\ominus} \).
The negatively charged carbon is part of a triple bond.
In a triple bond (\( C \equiv C \)), one bond is a sigma (\( \sigma \)) bond, and two are pi (\( \pi \)) bonds.
The negatively charged carbon is bonded to one carbon via a sigma bond and has one lone pair of electrons (responsible for the negative charge).
Thus, there is one sigma bond to another carbon.
Considering the overall molecule, the terminal carbon with the negative charge is bonded to one carbon via a triple bond (one sigma, two pi) and has a lone pair.
For hybridization, we consider sigma bonds and lone pairs.
The negatively charged carbon forms one sigma bond to the adjacent carbon and has one lone pair.
Steric number = 1 (sigma bond) + 1 (lone pair) = 2.
Therefore, the hybridization of the negatively charged carbon is \( sp \).
The hybridization of the positively charged carbon is \( sp^2 \), and the hybridization of the negatively charged carbon is \( sp \).