Which of the following species has the same hybridization as C in `CH_(2)=CH_(2)`?

Options

A. `CO_(3)^(2-)`
B. `XeF_(4)`
C. `PCl_(3)`
D. `I_(3)^(-)`

Answer

To determine which of the following species has the same hybridization as carbon in ethylene (CH₂=CH₂), we need to first establish the hybridization of carbon in ethylene.

### Step 1: Determine the hybridization of carbon in CH₂=CH₂
1. **Identify the central atom**: In CH₂=CH₂, the central atom is carbon (C).
2. **Count the valence electrons**: Carbon has 4 valence electrons.
3. **Count the number of monovalent atoms**: Each hydrogen (H) is monovalent, and there are 2 hydrogens attached to each carbon. Therefore, for one carbon, we have 2 monovalent atoms.
4. **Calculate the steric number**: The formula for steric number is:
\[ \text{Steric Number} = \text{Valence Electrons of Central Atom} + \text{Number of Monovalent Atoms} + \text{Charge} \]
Since there is no charge on CH₂=CH₂, we can simplify:
\[ \text{Steric Number} = 4 + 2 + 0 = 6 \]
5. **Divide by 2**:
\[ \text{Steric Number} = \frac{6}{2} = 3 \]
6. **Determine hybridization**: A steric number of 3 corresponds to sp² hybridization.

### Step 2: Analyze the options provided
Now, we will analyze each option to find which one has the same hybridization (sp²).

1. **Option 1: CO₃²⁻ (Carbonate ion)**
- **Valence electrons of carbon**: 4
- **Number of monovalent atoms**: 3 oxygens (each oxygen forms a double bond or single bond with carbon).
- **Charge**: -2 (which adds 2 to the count).
- **Steric number calculation**:
\[ \text{Steric Number} = 4 + 3 + 2 = 9 \quad \text{(divide by 2)} \quad \frac{9}{2} = 4.5 \quad \text{(not applicable)} \]
- **Correct calculation**: The correct steric number is 3 (due to resonance structures), leading to sp² hybridization.

2. **Option 2: XeF₄ (Xenon tetrafluoride)**
- **Valence electrons of xenon**: 8
- **Number of monovalent atoms**: 4 fluorines.
- **Charge**: 0.
- **Steric number calculation**:
\[ \text{Steric Number} = 8 + 4 + 0 = 12 \quad \text{(divide by 2)} \quad \frac{12}{2} = 6 \]
- **Hybridization**: sp³d² (not matching).

3. **Option 3: PCl₃ (Phosphorus trichloride)**
- **Valence electrons of phosphorus**: 5
- **Number of monovalent atoms**: 3 chlorines.
- **Charge**: 0.
- **Steric number calculation**:
\[ \text{Steric Number} = 5 + 3 + 0 = 8 \quad \text{(divide by 2)} \quad \frac{8}{2} = 4 \]
- **Hybridization**: sp³ (not matching).

4. **Option 4: I₃⁻ (Triiodide ion)**
- **Valence electrons of iodine**: 7
- **Number of monovalent atoms**: 2 (since it forms bonds with 2 iodines).
- **Charge**: -1 (which adds 1).
- **Steric number calculation**:
\[ \text{Steric Number} = 7 + 2 + 1 = 10 \quad \text{(divide by 2)} \quad \frac{10}{2} = 5 \]
- **Hybridization**: sp³d (not matching).

### Conclusion
The only species that has the same hybridization (sp²) as carbon in CH₂=CH₂ is **Option 1: CO₃²⁻**.