The hybrid state of S in `SO_2`, is similar to that of

Options

A. C in `C_(2)H_(2)`
B. C in `C_(2) H_(4)`
C. C in `CH_(4)`
D. C in `CO_(2)`

Explanation

To determine the hybrid state of sulfur in \( SO_2 \), we will follow these steps:

### Step 1: Determine the Valence Electrons of Sulfur
Sulfur (S) is in Group 16 of the periodic table and has 6 valence electrons.

### Step 2: Analyze the Structure of \( SO_2 \)
In \( SO_2 \), sulfur forms two double bonds with two oxygen atoms. Each double bond consists of one sigma bond and one pi bond.

### Step 3: Count the Sigma Bonds and Lone Pairs
- **Sigma Bonds**: There are 2 sigma bonds (one from each double bond).
- **Lone Pairs**: Sulfur has 6 valence electrons and uses 4 electrons to form the two double bonds (2 electrons per bond). Therefore, it has 2 electrons left, which means there is 1 lone pair.

### Step 4: Calculate the Steric Number
The steric number is calculated as follows:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} \]
\[ \text{Steric Number} = 2 + 1 = 3 \]

### Step 5: Determine the Hybridization
According to the steric number:
- A steric number of 3 corresponds to \( sp^2 \) hybridization.

### Step 6: Identify Similar Hybridization States
Now, we need to find a molecule with sulfur that also exhibits \( sp^2 \) hybridization.

### Conclusion
The hybrid state of sulfur in \( SO_2 \) is \( sp^2 \).