`SCl_2` is the best known dihalice of sulphur, hybrid state of sulphur in `SCl_2` is

Options

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

Answer

To determine the hybrid state of sulfur in \( SCl_2 \), we can follow these steps:

### Step 1: Identify the Valence Electrons
Sulfur (S) is in Group 16 of the periodic table and has 6 valence electrons. Each chlorine (Cl) atom contributes 7 valence electrons, but since each Cl forms a single bond with sulfur, we will focus on the valence electrons of sulfur.

### Step 2: Draw the Lewis Structure
In \( SCl_2 \), sulfur is bonded to two chlorine atoms. We can represent this as follows:

```
Cl
|
S
|
Cl
```

### Step 3: Count Lone Pairs and Bond Pairs
In the Lewis structure of \( SCl_2 \):
- There are 2 bond pairs (one for each S-Cl bond).
- Sulfur has 4 remaining electrons, which will form 2 lone pairs.

### Step 4: Determine the Steric Number
The steric number is calculated by adding the number of bond pairs and lone pairs around the central atom (sulfur in this case):
- Bond pairs = 2
- Lone pairs = 2
- Steric number = 2 (bond pairs) + 2 (lone pairs) = 4

### Step 5: Identify the Hybridization
The hybridization corresponding to a steric number of 4 is \( sp^3 \). This means that sulfur undergoes \( sp^3 \) hybridization to form the bonds in \( SCl_2 \).

### Step 6: Determine the Molecular Geometry
Due to the presence of 2 lone pairs, the molecular geometry of \( SCl_2 \) is bent or V-shaped, which is a result of the repulsion between the lone pairs.

### Conclusion
Thus, the hybrid state of sulfur in \( SCl_2 \) is \( sp^3 \).

### Final Answer
The hybrid state of sulfur in \( SCl_2 \) is \( sp^3 \).

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