Passage
Organic compounds mainly consist of covalent bonds. The electron pair in these covalent bonds may undergo displacement either of their own or under the influence of other species. The cleavage of covalent bond between two atoms takes place in homolytic or heterolytic fashion. The homolytic fission results into free radicals while heterolytic fission results into carbocations and carbanions. These are also called reaction intermediates and are attacked by electrophiles and nucleophiles. The electrophiles seek electron rich sites while nucleophiles seek electron deficient sites.
Answer the followings Questions :
The hybridisation state of carbon in the carbocation `C_6 H_5 CH_(2)^(+)` is

Options

A. `sp`
B. `sp^2`
C. `sp^3`
D. `sp` or `sp^2`

Answer

To determine the hybridization state of carbon in the carbocation \( C_6H_5CH_2^+ \), we can follow these steps:

### Step 1: Identify the Structure of the Carbocation
The given carbocation is \( C_6H_5CH_2^+ \). This indicates that we have a phenyl group (\( C_6H_5 \)) attached to a \( CH_2^+ \) group. The structure can be represented as follows:

```
H
|
H - C - C6H5
|
H
+
```

### Step 2: Count the Sigma Bonds
To determine the hybridization, we need to count the number of sigma bonds formed by the carbon atom in the \( CH_2^+ \) group.

- The carbon in \( CH_2^+ \) is bonded to two hydrogen atoms and one carbon atom from the phenyl group.
- Therefore, it forms 3 sigma bonds in total.

### Step 3: Determine the Hybridization
The hybridization can be determined using the formula:
- **Hybridization = Number of Sigma Bonds + Number of Lone Pairs**

In this case:
- Number of Sigma Bonds = 3 (2 from H atoms and 1 from the phenyl group)
- Number of Lone Pairs = 0 (since it is a carbocation, there are no lone pairs on this carbon)

Thus, the total is:
\[ 3 \text{ (sigma bonds)} + 0 \text{ (lone pairs)} = 3 \]

According to hybridization rules:
- If there are 4 total (3 sigma + 1 lone pair), the hybridization is \( sp^3 \).
- If there are 3 total (3 sigma), the hybridization is \( sp^2 \).

Since we have 3 sigma bonds and no lone pairs, the hybridization state of the carbon in \( C_6H_5CH_2^+ \) is \( sp^2 \).

### Final Answer:
The hybridization state of carbon in the carbocation \( C_6H_5CH_2^+ \) is **\( sp^2 \)**.

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