The enolic form of acetone `(CH_(3)-overset(OH)overset(|)C=CH_2)` contains

Options

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

Answer

To determine the number of sigma bonds, pi bonds, and lone pairs in the enolic form of acetone, which is represented as CH₃-C(OH)=CH₂, we can follow these steps:

### Step 1: Identify the Structure
The enolic form of acetone can be represented as:
```
H H
\ /
C
/ \
H C
|
O
|
H
```
This structure shows that we have two carbon atoms (C) connected by a double bond, and one of the carbons is also connected to a hydroxyl group (OH).

### Step 2: Count Sigma Bonds
1. **C-H Bonds**: Each carbon (C) in the structure is bonded to hydrogen (H) atoms.
- The first carbon (C1) is connected to three hydrogen atoms (C-H bonds).
- The second carbon (C2) is connected to two hydrogen atoms (C-H bonds).

Total C-H bonds = 3 (from C1) + 2 (from C2) = 5 sigma bonds.

2. **C-C Bond**: The two carbon atoms are connected by a single bond (C-C).
- This is 1 sigma bond.

3. **C-O Bond**: The carbon (C2) is connected to the oxygen (O) in the hydroxyl group (OH).
- This is also a sigma bond.

Adding these together:
- Total sigma bonds = 5 (C-H) + 1 (C-C) + 1 (C-O) = **7 sigma bonds**.

### Step 3: Count Pi Bonds
- The double bond between the two carbon atoms (C1 and C2) consists of one sigma bond and one pi bond.
- Therefore, there is **1 pi bond**.

### Step 4: Count Lone Pairs
- The oxygen atom in the hydroxyl group (OH) has two lone pairs of electrons.
- The carbon atoms do not have any lone pairs.

Thus, the total number of lone pairs = **2 lone pairs** (from the oxygen atom).

### Final Summary
- **Sigma bonds**: 7
- **Pi bonds**: 1
- **Lone pairs**: 2