Which of the following species does not have tetrahedral geometry ?

Options

A. `BH_(4)^(-)`
B. `NH_(2)^(-)`
C. `CO_(3)^(2-)`
D. `H_(3)O^(+)`

Explanation

To determine which of the following species does not have tetrahedral geometry, we will calculate the hybridization of each species using the following formula:

**Hybridization Formula:**
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence electrons on central atom} + \text{Number of monovalent atoms} + \text{Charge} \right) \]

### Step 1: Analyze BH₄⁻
- **Valence electrons on Boron (B)**: 3
- **Number of monovalent groups (H)**: 4 (from 4 hydrogen atoms)
- **Charge**: -1 (since it's BH₄⁻, we add 1)

Calculating:
\[ \text{Hybridization} = \frac{1}{2} \left( 3 + 4 + 1 \right) = \frac{1}{2} \times 8 = 4 \]
- **Hybridization**: sp³
- **Geometry**: Tetrahedral

### Step 2: Analyze NH₂⁻
- **Valence electrons on Nitrogen (N)**: 5
- **Number of monovalent groups (H)**: 2 (from 2 hydrogen atoms)
- **Charge**: -1 (since it's NH₂⁻, we add 1)

Calculating:
\[ \text{Hybridization} = \frac{1}{2} \left( 5 + 2 + 1 \right) = \frac{1}{2} \times 8 = 4 \]
- **Hybridization**: sp³
- **Geometry**: Tetrahedral

### Step 3: Analyze CO₃²⁻
- **Valence electrons on Carbon (C)**: 4
- **Number of monovalent groups**: 0 (no hydrogen atoms)
- **Charge**: -2 (since it's CO₃²⁻, we add 2)

Calculating:
\[ \text{Hybridization} = \frac{1}{2} \left( 4 + 0 + 2 \right) = \frac{1}{2} \times 6 = 3 \]
- **Hybridization**: sp²
- **Geometry**: Trigonal planar

### Step 4: Analyze H₃O⁺
- **Valence electrons on Oxygen (O)**: 6
- **Number of monovalent groups (H)**: 3 (from 3 hydrogen atoms)
- **Charge**: +1 (since it's H₃O⁺, we subtract 1)

Calculating:
\[ \text{Hybridization} = \frac{1}{2} \left( 6 + 3 - 1 \right) = \frac{1}{2} \times 8 = 4 \]
- **Hybridization**: sp³
- **Geometry**: Tetrahedral

### Conclusion:
The only species that does not have tetrahedral geometry is **CO₃²⁻** (with sp² hybridization and trigonal planar geometry).