The hybridization of sulphur in `SO_2` is

Options

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

Explanation

To determine the hybridization of sulfur in \( SO_2 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
Sulfur (S) is in group 16 of the periodic table and has 6 valence electrons. Each oxygen (O) atom also has 6 valence electrons.

### Step 2: Draw the Lewis Structure
In \( SO_2 \), sulfur forms double bonds with two oxygen atoms. Each double bond consists of one sigma bond and one pi bond. The Lewis structure can be represented as:
```
O
||
S==O
```
Here, sulfur is in the center, bonded to two oxygen atoms, and there is a lone pair of electrons on sulfur.

### Step 3: Count the Steric Number
The steric number is calculated by adding the number of sigma bonds and the number of lone pairs around the central atom (sulfur in this case).

- There are 2 sigma bonds (one with each oxygen).
- There is 1 lone pair on sulfur.

So, the steric number = 2 (sigma bonds) + 1 (lone pair) = 3.

### Step 4: Determine Hybridization
The hybridization can be determined based on the steric number:
- A steric number of 3 corresponds to \( sp^2 \) hybridization.

### Step 5: Conclusion
Thus, the hybridization of sulfur in \( SO_2 \) is \( sp^2 \).

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