`PCl_(5)` is an example of a molecule having `sp^(3)d`-hybridisation. Three out of the orbitals in `sp_(x)p_(y)`-hybridisation while remaining two have `p_(x)d_(x^(2))`-hybridisation. If P-atom is attached to substitutuents differ in electronegativity, as in `PCl_(x)F_(5-x)`, then it is has been experimently observed that the more electronegativity substituent occupies the axial position of t.b.p geometry.
Q. The highest `H-C-H` bond angle present in :

Answer

To solve the question regarding the highest H-C-H bond angle present in the given compounds, we will analyze the structures and hybridization of each compound step by step.

### Step-by-Step Solution:

1. **Identify the Compounds**: The question implies we need to compare the bond angles in three compounds: CH2F2, CH4, and CH3F.

2. **Determine the Hybridization and Geometry**:
- **For CH2F2**:
- The central atom is Carbon (C).
- Carbon has 4 valence electrons and is bonded to 2 Hydrogen (H) atoms and 2 Fluorine (F) atoms.
- The steric number (SN) = 4 (2 from H + 2 from F).
- This means the hybridization is **sp³** and the geometry is **tetrahedral**.
- **For CH4**:
- The central atom is Carbon (C).
- Carbon is bonded to 4 Hydrogen (H) atoms.
- The steric number (SN) = 4 (4 from H).
- This also means the hybridization is **sp³** and the geometry is **tetrahedral**.
- **For CH3F**:
- The central atom is Carbon (C).
- Carbon is bonded to 3 Hydrogen (H) atoms and 1 Fluorine (F) atom.
- The steric number (SN) = 4 (3 from H + 1 from F).
- This means the hybridization is **sp³** and the geometry is **tetrahedral**.

3. **Analyze the Bond Angles**:
- In a perfect tetrahedral geometry, the bond angles are approximately **109.5°**.
- However, the presence of more electronegative atoms (like F) can affect these angles due to their electron-withdrawing effect.
- In **CH2F2**, there are two F atoms, which are highly electronegative. This will cause the H-C-H bond angle to be less than 109.5° due to the repulsion between the lone pairs on F.
- In **CH4**, there are no electronegative atoms, so the bond angle remains at **109.5°**.
- In **CH3F**, there is one F atom. The presence of one electronegative atom will slightly reduce the H-C-H bond angle compared to CH4 but will still be greater than that in CH2F2.

4. **Conclusion**:
- The highest H-C-H bond angle will be in **CH4** because it has no electronegative atoms to distort the bond angles, maintaining the ideal tetrahedral angle of **109.5°**.
- Therefore, the answer to the question is **CH4**.

### Final Answer:
The highest H-C-H bond angle is present in **CH4**.