`BrO_3^(-)` is isostructural with which of the following -

Options

A. `XeO_3`
B. `XeF_3`
C. `XeF_4`
D. `XeO_2`

Explanation

To determine which of the given options is isostructural with \( \text{BrO}_3^{-} \), we need to analyze the structure and hybridization of \( \text{BrO}_3^{-} \) first.

### Step 1: Determine the Valence Electrons
- **Bromine (Br)** is in Group 17 and has **7 valence electrons**.
- The \( \text{BrO}_3^{-} \) ion has an additional electron due to the negative charge, giving a total of \( 7 + 3 \) (for three oxygen atoms) + 1 (for the negative charge) = **11 valence electrons**.

### Step 2: Calculate the Steric Number
The steric number can be calculated using the formula:
\[ \text{Steric Number} = \frac{(\text{Number of valence electrons}) + (\text{Monovalent atoms}) - (\text{Cationic charge}) + (\text{Anionic charge})}{2} \]
- Here, we have:
- Valence electrons = 7 (from Br) + 3 (from 3 O) = 10
- Monovalent atoms = 0 (none in this case)
- Cationic charge = 0
- Anionic charge = 1 (from the negative charge)

Plugging in the values:
\[ \text{Steric Number} = \frac{10 + 0 - 0 + 1}{2} = \frac{11}{2} = 5.5 \]
Since we cannot have a half steric number, we consider the bonding and lone pairs.

### Step 3: Determine Hybridization
The steric number of 4 indicates that the hybridization is \( sp^3 \), which corresponds to a tetrahedral arrangement. However, since we have one lone pair, the geometry will be trigonal pyramidal.

### Step 4: Analyze the Options
Now we need to check which of the given options has a similar trigonal pyramidal structure.

1. **Option A: \( \text{XCO}_3 \)**
- Calculate steric number: \( \text{X} \) (noble gas) has 8 valence electrons, no monovalent atoms, no cationic charge, and no anionic charge.
- Steric number = \( \frac{8 + 0 - 0 + 0}{2} = 4 \) → This gives a tetrahedral geometry, not trigonal pyramidal.

2. **Option B: \( \text{XF}_3 \)**
- This has 7 valence electrons (from F) + 0 (monovalent) = 7.
- Steric number = \( \frac{7 + 0 - 0 + 0}{2} = 3.5 \) → This is not stable and does not have a trigonal pyramidal shape.

3. **Option C: \( \text{XZO}_4 \)**
- This has 8 valence electrons from \( \text{X} \) + 4 (from 4 O) = 12.
- Steric number = \( \frac{12 + 0 - 0 + 0}{2} = 6 \) → This corresponds to an octahedral geometry, not trigonal pyramidal.

4. **Option D: \( \text{XZOF}_2 \)**
- This has 8 (from \( \text{X} \)) + 2 (from O) + 0 (from F) = 10.
- Steric number = \( \frac{10 + 0 - 0 + 0}{2} = 5 \) → This gives a trigonal bipyramidal geometry, not trigonal pyramidal.

### Conclusion
After analyzing all options, **none of the options provided are isostructural with \( \text{BrO}_3^{-} \)**. However, if we consider \( \text{XCO}_3 \) as the closest structure, it is tetrahedral but does not match the trigonal pyramidal shape of \( \text{BrO}_3^{-} \).