The ratio of pure and hybrid orbitals
`H_(2)C=CH-CH=CH_(2)`

Options

A. `7 : 12`
B. `14 : 13`
C. `12 : 10`
D. `5 : 6`

Answer

To determine the ratio of pure and hybrid orbitals in the compound H₂C=CH-CH=CH₂, we will follow these steps:

### Step 1: Identify the structure of the compound
The compound H₂C=CH-CH=CH₂ is a linear diene with alternating double bonds. It consists of four carbon atoms and is structured as follows:

```
H H
\ /
C
// \
C C
\ /
C
/ \
H H
```

### Step 2: Count the number of hydrogen atoms and pi bonds
- **Hydrogen atoms (H)**: Count the hydrogen atoms in the structure.
- There are 6 hydrogen atoms.

- **Pi bonds**: Identify the number of pi bonds in the structure.
- There are 2 pi bonds (one between each pair of double-bonded carbons).

### Step 3: Calculate the number of pure orbitals
The formula to calculate the number of pure orbitals is:
\[ \text{Number of pure orbitals} = \text{Number of hydrogen atoms} + 2 \times \text{Number of pi bonds} \]

Substituting the values:
\[ \text{Number of pure orbitals} = 6 + 2 \times 2 = 6 + 4 = 10 \]

### Step 4: Determine the hybridization of each carbon atom
- The carbon atoms involved in double bonds (C=C) are sp² hybridized.
- The carbon atoms involved in single bonds (C-C) are also sp² hybridized.

In total, there are 4 carbon atoms, and since each carbon atom is sp² hybridized:
- Each sp² hybridized carbon contributes 3 hybrid orbitals.

### Step 5: Calculate the number of hybrid orbitals
The total number of hybrid orbitals can be calculated as:
\[ \text{Number of hybrid orbitals} = \text{Number of carbon atoms} \times 3 \]

Substituting the values:
\[ \text{Number of hybrid orbitals} = 4 \times 3 = 12 \]

### Step 6: Determine the ratio of pure to hybrid orbitals
Now that we have the number of pure and hybrid orbitals:
- Pure orbitals = 10
- Hybrid orbitals = 12

The ratio of pure to hybrid orbitals is:
\[ \text{Ratio} = \frac{\text{Number of pure orbitals}}{\text{Number of hybrid orbitals}} = \frac{10}{12} = \frac{5}{6} \]

### Final Answer
The ratio of pure to hybrid orbitals in H₂C=CH-CH=CH₂ is **5:6**.

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