Structure of `S_(2)Cl_(2)` is analogous to :

Options

A. `SOCl_(2)`
B. `CO_(2)`
C. `H_(2)S`
D. `H_(2)O_(2)`

Answer

To determine the structure of \( S_2Cl_2 \) and find its analogous structure, we can follow these steps:

### Step 1: Determine the Valence Electrons
Sulfur (S) has 6 valence electrons, and chlorine (Cl) has 7 valence electrons. For \( S_2Cl_2 \), we have:
- 2 Sulfur atoms: \( 2 \times 6 = 12 \) electrons
- 2 Chlorine atoms: \( 2 \times 7 = 14 \) electrons
- Total valence electrons = \( 12 + 14 = 26 \) electrons

### Step 2: Draw the Lewis Structure
1. Place the two sulfur atoms in the center and the two chlorine atoms on the outside.
2. Connect each sulfur atom to a chlorine atom with a single bond.
3. This uses 4 electrons (2 bonds).
4. The remaining electrons are \( 26 - 4 = 22 \) electrons.
5. Distribute the remaining electrons as lone pairs on the chlorine atoms first. Each chlorine will get 3 lone pairs (6 electrons total).
6. Now we have used \( 4 + 6 = 10 \) electrons, leaving us with \( 26 - 10 = 16 \) electrons.
7. Place lone pairs on the sulfur atoms. Each sulfur will get 3 lone pairs (6 electrons total), using \( 10 + 6 = 16 \) electrons.
8. The final structure shows each sulfur atom with 2 bond pairs (to chlorine) and 2 lone pairs.

### Step 3: Determine Hybridization
- Each sulfur atom has 2 bond pairs and 2 lone pairs.
- Hybridization can be calculated as:
\[ \text{Hybridization} = \text{Number of bond pairs} + \text{Number of lone pairs} = 2 + 2 = 4 \quad \Rightarrow \quad sp^3 \]

### Step 4: Compare with Other Structures
Now we need to compare \( S_2Cl_2 \) with the given options to find an analogous structure:
1. **SOCl2**: Has 3 bond pairs and 1 lone pair (not analogous).
2. **CO2**: Has 2 double bonds (not analogous).
3. **H2S**: Has 2 bond pairs and 2 lone pairs (analogous).
4. **H2O2**: Has a different structure.

### Conclusion
The structure of \( S_2Cl_2 \) is analogous to **H2S** because both have the same hybridization \( sp^3 \), with 2 bond pairs and 2 lone pairs.

---