Which of the following statements is correct?

Options

A. `ClF_(3)` molecule is bent `T` shape
B. In `SF_(4)` molecule, `F-S-F` equatorial bond angle is `103^(@)` due to lp-lp repulsion
C. In `[Cl_(4)]^(-)` molecular ion, `Cl-I-Cl` bond angle is `90^(@)`
D. In `OBr_(2)`, the bond angle is less than `OCl_(2)`

Answer

To determine which of the provided statements is correct, let's analyze each statement step by step.

### Step 1: Analyze Statement A
**Statement A**: ClF3 molecule has a bent or T shape.

1. **Calculate Steric Number**: The steric number is calculated using the formula:
\[ \text{Steric Number} = \text{Number of Valence Electrons} + \text{Number of Monovalent Atoms} - \text{Charge} / 2 \]
For ClF3:
- Chlorine (Cl) has 7 valence electrons.
- There are 3 fluorine (F) atoms (monovalent).
- There is no charge.
\[ \text{Steric Number} = 7 + 3 - 0 / 2 = 5 \]

2. **Determine Hybridization**: A steric number of 5 indicates sp3d hybridization.

3. **Shape of the Molecule**: With 3 bonded pairs and 2 lone pairs, the molecular geometry is T-shaped.

**Conclusion for Statement A**: Correct.

### Step 2: Analyze Statement B
**Statement B**: In SF4 molecule, the F-S-F equatorial bond angle is 103 degrees due to lone pair-lone pair repulsion.

1. **Calculate Steric Number**: For SF4:
- Sulfur (S) has 6 valence electrons.
- There are 4 fluorine atoms.
\[ \text{Steric Number} = 6 + 4 - 0 / 2 = 5 \]

2. **Determine Hybridization**: A steric number of 5 indicates sp3d hybridization.

3. **Shape of the Molecule**: The presence of one lone pair results in a seesaw shape. The equatorial bond angles are indeed affected by lone pair repulsion, reducing the bond angle from the ideal value.

**Conclusion for Statement B**: Correct.

### Step 3: Analyze Statement C
**Statement C**: In ICl4- molecule, the Cl-I-Cl bond angle is 90 degrees.

1. **Calculate Steric Number**: For ICl4-:
- Iodine (I) has 7 valence electrons.
- There are 4 chlorine atoms.
- The negative charge adds 1 electron.
\[ \text{Steric Number} = 7 + 4 + 1 / 2 = 6 \]

2. **Determine Hybridization**: A steric number of 6 indicates sp3d2 hybridization.

3. **Shape of the Molecule**: The geometry is square planar, and the Cl-I-Cl bond angle is indeed 90 degrees.

**Conclusion for Statement C**: Correct.

### Step 4: Analyze Statement D
**Statement D**: In OBr2, the bond angle is less than in OCl2.

1. **Calculate Hybridization**: For both OBr2 and OCl2:
- Oxygen (O) has 6 valence electrons.
- There are 2 halogens (Br or Cl).
\[ \text{Steric Number} = 6 + 2 / 2 = 4 \]

2. **Determine Hybridization**: Both have sp3 hybridization.

3. **Shape of the Molecule**: Both OBr2 and OCl2 have a bent shape due to 2 lone pairs on oxygen.

4. **Bond Angle Comparison**: The bond angle in OBr2 is expected to be larger than in OCl2 because bromine is larger than chlorine, which causes less repulsion and a larger angle.

**Conclusion for Statement D**: Incorrect.

### Final Conclusion
The correct statements are A, B, and C.