Choose the CORRECT statement regarding bond angle :

Options

A. `hat(FCF)` in `F_(2)CO lt hat(HCH)` in `H_(2)CO`
B. `hat(BrPBr)` in `PBr_(3) lt hat (FPF) ` in `PF_(3)`
C. `hat(FSF) (eq) gt hat (FSF) (ax) ` in `SF_(4)`
D. All `hat(FIF)` angles in `IF_(5)` are identical

Answer

To solve the question regarding bond angles, let's analyze the statements provided and determine which one is correct step by step.

### Step 1: Analyze the first statement regarding F2CO and H2CO
- **F2CO (Formaldehyde with fluorine)** has a bond angle that is less than that of **H2CO (Formaldehyde with hydrogen)**.
- In F2CO, the two fluorine atoms are highly electronegative, which pulls the electron density towards themselves. This results in a reduction of electron pair repulsion between the bonding pairs, leading to a smaller bond angle.
- In H2CO, the hydrogen atoms are less electronegative than fluorine, allowing for greater electron pair repulsion and thus a larger bond angle.
- **Conclusion**: The first statement is correct.

### Step 2: Analyze the second statement regarding PBr3 and PF3
- **PBr3 (Phosphorus tribromide)** and **PF3 (Phosphorus trifluoride)** are being compared.
- Bromine is larger than fluorine, which means that the P-Br bonds will have a larger bond length compared to P-F bonds.
- However, fluorine is more electronegative than bromine, which means that in PF3, the electron density is pulled more towards the fluorine atoms, increasing the repulsion and thus the bond angle.
- **Conclusion**: The second statement is incorrect.

### Step 3: Analyze the third statement regarding axial and equatorial bonds in SF6
- In **SF6 (Sulfur hexafluoride)**, the axial bonds (the bonds that are perpendicular to the equatorial plane) experience more repulsion due to the lone pairs being positioned in the equatorial plane.
- The axial bonds are generally longer than the equatorial bonds due to this increased repulsion.
- **Conclusion**: The third statement is incorrect.

### Step 4: Analyze the fourth statement regarding bond lengths in SF6
- Similar to the previous analysis, in SF6, the axial bonds are longer than the equatorial bonds because of the greater repulsion experienced by the axial bonds.
- **Conclusion**: The fourth statement is also incorrect.

### Final Conclusion
After analyzing all the statements, we find that only the first statement regarding the bond angles in F2CO and H2CO is correct.