In the given species, carbanion is `sp^3` hybridised?

Answer

To determine whether the carbanion in the given species is sp³ hybridized, we can follow these steps:

### Step 1: Identify the Structure
The structure given is C=O, CH⁻, CH₃, and H. We need to focus on the carbon atom that carries the negative charge (the carbanion).

### Step 2: Count the Sigma Bonds
For hybridization, we need to calculate the steric number. The steric number is determined by counting the number of sigma bonds and lone pairs around the central atom (in this case, the carbanion).

- **Sigma Bonds**:
- The carbanion (CH⁻) is bonded to:
- One carbon atom (C=O) → 1 sigma bond
- One carbon atom (CH₃) → 1 sigma bond
- One hydrogen atom (H) → 1 sigma bond
- Total sigma bonds = 3

### Step 3: Count the Lone Pairs
Next, we need to account for the lone pairs on the carbanion. Since it carries a negative charge, it has one lone pair of electrons.

- **Lone Pairs**: 1 lone pair

### Step 4: Calculate the Steric Number
Now, we can calculate the steric number using the formula:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} \]

Substituting the values we found:
\[ \text{Steric Number} = 3 \text{ (sigma bonds)} + 1 \text{ (lone pair)} = 4 \]

### Step 5: Determine the Hybridization
We know that:
- A steric number of 4 corresponds to sp³ hybridization.

### Conclusion
Thus, we conclude that the carbanion in the given species is sp³ hybridized.

### Summary
The carbanion is sp³ hybridized because it has a steric number of 4, which is derived from 3 sigma bonds and 1 lone pair.

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