The molecule which contain `sigma` bonds, `pi` bonds and lone pairs in 1 : 1 : 1 ratio :-

Options

A. `C_(2)N_(2)` (Cyanogen)
B. `C_(6)H_(6)` (Benzene)
C. `C(CN)_(4)` (Tetracyano methane)
D. `C_(3)O_(2)` (Carbon suboxide)

Explanation

To solve the problem of finding a molecule that contains sigma bonds, pi bonds, and lone pairs in a 1:1:1 ratio, we will analyze each molecule provided in the question step by step.

### Step 1: Identify the Molecules
We need to analyze the following molecules:
1. Cyanogen (C2N2)
2. Benzene (C6H6)
3. Tetracyanomethane (C(CN)4)
4. Carbon suboxide (C3O2)

### Step 2: Analyze Cyanogen (C2N2)
- **Structure**: The structure of cyanogen consists of a carbon-carbon triple bond and two carbon-nitrogen triple bonds.
- **Sigma Bonds**:
- 1 sigma bond from C≡C
- 2 sigma bonds from C≡N (one from each C to N)
- Total = 3 sigma bonds
- **Pi Bonds**:
- 2 pi bonds from C≡C
- 2 pi bonds from each C≡N
- Total = 4 pi bonds
- **Lone Pairs**: Each nitrogen has 1 lone pair.
- Total = 2 lone pairs
- **Ratio**: 3 sigma : 4 pi : 2 lone pairs (not equal)

### Step 3: Analyze Benzene (C6H6)
- **Structure**: Benzene has a hexagonal ring with alternating double bonds.
- **Sigma Bonds**:
- Each C-H bond is a sigma bond (6 C-H)
- Each C-C bond is a sigma bond (6 C-C)
- Total = 12 sigma bonds
- **Pi Bonds**:
- There are 3 double bonds, each contributing 1 pi bond.
- Total = 3 pi bonds
- **Lone Pairs**: There are no lone pairs in benzene.
- Total = 0 lone pairs
- **Ratio**: 12 sigma : 3 pi : 0 lone pairs (not equal)

### Step 4: Analyze Tetracyanomethane (C(CN)4)
- **Structure**: Tetracyanomethane has a central carbon atom bonded to four cyanide groups.
- **Sigma Bonds**:
- 1 sigma bond from C to C
- 4 sigma bonds from C to N
- Total = 5 sigma bonds
- **Pi Bonds**:
- Each C≡N bond has 2 pi bonds.
- Total = 8 pi bonds
- **Lone Pairs**: Each nitrogen has 1 lone pair.
- Total = 4 lone pairs
- **Ratio**: 5 sigma : 8 pi : 4 lone pairs (not equal)

### Step 5: Analyze Carbon Suboxide (C3O2)
- **Structure**: Carbon suboxide has a structure with alternating double bonds.
- **Sigma Bonds**:
- 1 sigma bond from C=C
- 2 sigma bonds from C=O
- Total = 4 sigma bonds
- **Pi Bonds**:
- 1 pi bond from C=C
- 3 pi bonds from C=O
- Total = 4 pi bonds
- **Lone Pairs**: Each oxygen has 2 lone pairs.
- Total = 4 lone pairs
- **Ratio**: 4 sigma : 4 pi : 4 lone pairs (equal)

### Conclusion
The molecule that contains sigma bonds, pi bonds, and lone pairs in a 1:1:1 ratio is **Carbon Suboxide (C3O2)**.

### Final Answer
**Carbon Suboxide (C3O2)** is the correct answer.

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