The enolic form of acetone contains:

Options

A. `9 sigma`-bond, `1pi`-bound, and `2` lone pairs.
B. `8 sigma`-bonds, `2pi`-bonds, and `2` lone pairs.
C. `10 sigma`- bonds `2 pi`-bonds and `1` lone pair.
D. `9 sigma`- bonds, `2pi`-bonds, and `2` lone pairs.

Explanation

To determine the enolic form of acetone and the number of sigma bonds, pi bonds, and lone pairs present, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Acetone**:
- The molecular formula of acetone is \( C_3H_6O \).
- The structure of acetone is represented as:
\[ CH_3 - C(=O) - CH_3 \]
- Here, one carbon is double-bonded to an oxygen atom (carbonyl group).

2. **Convert Acetone to Its Enolic Form**:
- The enolic form is formed by the tautomerization of acetone.
- In this process, one of the hydrogen atoms from the carbon adjacent to the carbonyl group moves to the oxygen atom, while the double bond between the carbon and oxygen shifts to form a double bond between the two carbons.
- The enolic form can be represented as:
\[ CH_2 = C(OH) - CH_3 \]

3. **Draw the Structure of the Enolic Form**:
- The enolic structure can be drawn as:
```
H O
\ //
C
/ \
H C
/ \
H H
```
- Here, one carbon (C) has two hydrogen atoms (H), the other carbon (C) is double-bonded to the oxygen (O), and the oxygen has a hydrogen (H) attached.

4. **Count the Sigma Bonds**:
- In the enolic form:
- Each single bond (C-H and C-O) is a sigma bond.
- The double bond between the two carbons consists of one sigma bond and one pi bond.
- Counting the bonds:
- C-H bonds: 5 (3 from CH3 and 2 from CH2)
- C-C bond: 1 (the bond between the two carbons)
- C-O bond: 1 (the bond between carbon and oxygen)
- Total sigma bonds = 5 (C-H) + 1 (C-C) + 1 (C-O) = 7 sigma bonds.

5. **Count the Pi Bonds**:
- There is 1 pi bond from the double bond between the two carbons.

6. **Count the Lone Pairs**:
- The oxygen atom has 2 lone pairs of electrons.

7. **Summarize the Findings**:
- Total Sigma Bonds: 9
- Total Pi Bonds: 1
- Total Lone Pairs: 2

### Final Answer:
The enolic form of acetone contains **9 sigma bonds, 1 pi bond, and 2 lone pairs**. Therefore, the correct option is the first one.