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 `[IC l_(4)]^(-)` molecular ion, Cl-I-Cl bond angle is `90^(@)`
D. In `OB r_(2)`, the bond angle is less than `OC l_(2)`

Answer

To solve the question regarding which of the following statements is correct, we will analyze each statement one by one.

### Step 1: Analyze Statement 1
**Statement:** ClF3 molecule is bent T shape.

- **Valence Electrons:** Chlorine (Cl) has 7 valence electrons. In ClF3, chlorine forms 3 bonds with fluorine (F), using 3 of its valence electrons.
- **Lone Pairs:** This leaves 4 valence electrons, which form 2 lone pairs.
- **Steric Number:** The steric number is calculated as the number of bonded atoms plus lone pairs. Here, it is 3 (bonded Cl-F) + 2 (lone pairs) = 5.
- **Hybridization:** With a steric number of 5, the hybridization is sp3d.
- **Geometry:** The molecular geometry is trigonal bipyramidal. However, with 2 lone pairs, the shape is T-shaped, confirming that this statement is **correct**.

### Step 2: Analyze Statement 2
**Statement:** SF4 molecule's equatorial bond angle is 103 degrees due to lone pair-lone pair repulsion.

- **Valence Electrons:** Sulfur (S) has 6 valence electrons. In SF4, sulfur forms 4 bonds with fluorine, using 4 of its valence electrons.
- **Lone Pairs:** This leaves 2 valence electrons, which form 1 lone pair.
- **Steric Number:** The steric number is 4 (bonded F) + 1 (lone pair) = 5.
- **Hybridization:** The hybridization is sp3d.
- **Geometry:** The geometry is trigonal bipyramidal. The ideal bond angles are 90 degrees (axial) and 120 degrees (equatorial). The presence of a lone pair decreases the equatorial angle, but it is not 103 degrees; it is closer to 118 degrees. Therefore, this statement is **incorrect**.

### Step 3: Analyze Statement 3
**Statement:** ICl4^- molecule Cl-I-Cl bond angle is 90 degrees.

- **Valence Electrons:** Iodine (I) has 7 valence electrons. With a negative charge, it has 8 electrons. In ICl4^-, iodine forms 4 bonds with chlorine, using 4 of its valence electrons.
- **Lone Pairs:** This leaves 4 electrons, which form 2 lone pairs.
- **Steric Number:** The steric number is 4 (bonded Cl) + 2 (lone pairs) = 6.
- **Hybridization:** The hybridization is sp3d2.
- **Geometry:** The geometry is octahedral. In an octahedral arrangement, the bond angles between the Cl-I-Cl are indeed 90 degrees. Thus, this statement is **correct**.

### Step 4: Analyze Statement 4
**Statement:** OBr2 bond angle is less than OCl2.

- **Valence Electrons:** Oxygen (O) has 6 valence electrons. In both OBr2 and OCl2, oxygen forms 2 bonds with bromine and chlorine, respectively.
- **Shape:** Both OBr2 and OCl2 have a bent shape due to 2 lone pairs on oxygen.
- **Size Effect:** However, bromine is larger than chlorine, which leads to greater repulsion in OBr2 compared to OCl2. This results in a larger bond angle in OBr2 compared to OCl2.
- **Conclusion:** Therefore, this statement is **incorrect**.

### Final Conclusion
The correct statements are:
- Statement 1: ClF3 molecule is bent T shape (Correct)
- Statement 3: ICl4^- molecule Cl-I-Cl bond angle is 90 degrees (Correct)