Which one of the following molecule will have all equal `X-F` bonds length? (where X=Central atom)

Options

A. `SOCl_(2)F_(2)`
B. `SeF_(4)`
C. `PBr_(2)F_(3)`
D. `IF_(7)`

Explanation

To determine which molecule has all equal `X-F` bond lengths, we need to analyze the given options one by one. The central atom `X` in each molecule will dictate the hybridization and geometry, which in turn affects the bond lengths.

### Step-by-Step Solution:

1. **Identify the Central Atom and Hybridization:**
- For each molecule, identify the central atom and calculate its hybridization.
- The formula for hybridization is given by:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of single bonded atoms} \right) \]

2. **Analyze Each Option:**
- **Option A: SOCl2F2**
- Central atom: Sulfur (S)
- Valence electrons: 6
- Single bonded atoms: 4 (2 Cl and 2 F)
- Hybridization:
\[ \frac{1}{2} (6 + 4) = 5 \quad \text{(SP}^3\text{D)} \]
- Geometry: Trigonal bipyramidal. According to Bent's rule, the two fluorine atoms will be in axial positions, and the bond lengths of SF will be equal.

- **Option B: SEF4**
- Central atom: Selenium (Se)
- Valence electrons: 6
- Single bonded atoms: 4 (4 F)
- Hybridization:
\[ \frac{1}{2} (6 + 4) = 5 \quad \text{(SP}^3\text{D)} \]
- Geometry: Trigonal bipyramidal. There are two types of bonds (axial and equatorial), so the bond lengths will not be equal.

- **Option C: PBr2F3**
- Central atom: Phosphorus (P)
- Valence electrons: 5
- Single bonded atoms: 5 (2 Br and 3 F)
- Hybridization:
\[ \frac{1}{2} (5 + 5) = 5 \quad \text{(SP}^3\text{D)} \]
- Geometry: Trigonal bipyramidal. Similar to option B, the bond lengths will not be equal due to different axial and equatorial positions.

- **Option D: IF7**
- Central atom: Iodine (I)
- Valence electrons: 7
- Single bonded atoms: 7 (7 F)
- Hybridization:
\[ \frac{1}{2} (7 + 7) = 7 \quad \text{(SP}^3\text{D}^3) \]
- Geometry: Pentagonal bipyramidal. There are different types of bonds (axial and equatorial), so the bond lengths will not be equal.

3. **Conclusion:**
- The only molecule with equal `X-F` bond lengths is **SOCl2F2** (Option A) because it has two identical SF bonds in axial positions.

### Final Answer:
The molecule with all equal `X-F` bond lengths is **SOCl2F2**.