The group having triangular planar structures is :

Options

A. `BF_(3), NF_(3), CO_(3)^(2-)`
B. `CO_(3)^(2-), NO_(3)^(-), SO_(3)`
C. `NH_(3), SO_(3), CO_(3)^(2-)`
D. `NCl_(3), BCl_(3), SO_(3)`

Explanation

To determine the group having triangular planar structures, we need to analyze the hybridization and steric numbers of the given compounds. The key to identifying the geometry of a molecule lies in its hybridization, which can be determined by calculating the steric number.

### Step-by-Step Solution:

1. **Understanding Hybridization and Geometry**:
- **sp hybridization** corresponds to a linear geometry (steric number = 2).
- **sp² hybridization** corresponds to a trigonal planar geometry (steric number = 3).
- **sp³ hybridization** corresponds to a tetrahedral geometry (steric number = 4).

2. **Calculating Steric Number**:
- The steric number is calculated as:
\[ \text{Steric Number} = \text{Number of Lone Pairs} + \text{Number of Bond Pairs} \]

3. **Analyzing Each Compound**:
- **Option A: BF₃**:
- Boron has 3 valence electrons and forms 3 bonds with fluorine.
- Steric Number = 0 (lone pairs) + 3 (bond pairs) = 3.
- Hybridization = sp² → Geometry = Trigonal Planar.

- **Option A: NF₃**:
- Nitrogen has 5 valence electrons, forms 3 bonds with fluorine, and has 1 lone pair.
- Steric Number = 1 (lone pair) + 3 (bond pairs) = 4.
- Hybridization = sp³ → Geometry = Tetrahedral (not trigonal planar).

- **Option A: CO₃²⁻**:
- Carbon has 4 valence electrons, forms 3 bonds with oxygen, and has no lone pairs.
- Steric Number = 0 (lone pairs) + 3 (bond pairs) = 3.
- Hybridization = sp² → Geometry = Trigonal Planar.

4. **Continuing with Other Options**:
- **Option B: NO₃⁻**:
- Nitrogen has 5 valence electrons, forms 3 bonds with oxygen, and has no lone pairs.
- Steric Number = 0 (lone pairs) + 3 (bond pairs) = 3.
- Hybridization = sp² → Geometry = Trigonal Planar.

- **Option B: SO₃**:
- Sulfur has 6 valence electrons, forms 3 bonds with oxygen, and has no lone pairs.
- Steric Number = 0 (lone pairs) + 3 (bond pairs) = 3.
- Hybridization = sp² → Geometry = Trigonal Planar.

5. **Final Analysis**:
- From the analysis, BF₃, CO₃²⁻, NO₃⁻, and SO₃ all have a trigonal planar structure.
- NF₃ does not have a trigonal planar structure as it is pyramidal due to the lone pair.

### Conclusion:
The group having triangular planar structures is **Option B: CO₃²⁻, NO₃⁻, and SO₃**.

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