Which of the following molecule has two lone pairs and bond angle `lt 109^(@)28.` :-

Options

A. `SF_(2)`
B. `KrF_(4)`
C. `I Cl_(4)^(-)`
D. All of these

Answer

To determine which of the given molecules has two lone pairs and a bond angle less than 109.28°, we will analyze each molecule step by step.

### Step 1: Analyze SF2
1. **Valence Electrons**: Sulfur (S) has 6 valence electrons, and each fluorine (F) has 7 valence electrons. Therefore, SF2 has:
\[ 6 + 2 \times 7 = 20 \text{ valence electrons} \]
2. **Bond Formation**: Sulfur forms two bonds with fluorine, using 2 of its 6 valence electrons. This leaves:
\[ 6 - 2 = 4 \text{ electrons} \]
These 4 electrons form 2 lone pairs on sulfur.
3. **Geometry**: The molecular geometry is tetrahedral due to sp³ hybridization. However, the presence of lone pairs alters the bond angles.
4. **Bond Angle**: The ideal tetrahedral bond angle is 109.28°, but due to lone pair-lone pair repulsion, the bond angle in SF2 is less than 109.28°.

### Step 2: Analyze KrF4
1. **Valence Electrons**: Krypton (Kr) has 8 valence electrons, and each fluorine (F) has 7 valence electrons. Therefore, KrF4 has:
\[ 8 + 4 \times 7 = 36 \text{ valence electrons} \]
2. **Bond Formation**: Krypton forms four bonds with fluorine, using 4 of its 8 valence electrons. This leaves:
\[ 8 - 4 = 4 \text{ electrons} \]
These 4 electrons form 2 lone pairs on krypton.
3. **Geometry**: The molecular geometry is octahedral due to sp³d² hybridization.
4. **Bond Angle**: In an octahedral geometry, the bond angles are typically 90°. Thus, the bond angles in KrF4 are less than 109.28°.

### Step 3: Analyze ICl4⁻
1. **Valence Electrons**: Iodine (I) has 7 valence electrons, and each chlorine (Cl) has 7 valence electrons. The negative charge adds one more electron. Therefore, ICl4⁻ has:
\[ 7 + 4 \times 7 + 1 = 30 \text{ valence electrons} \]
2. **Bond Formation**: Iodine forms four bonds with chlorine, using 4 of its 7 valence electrons. This leaves:
\[ 7 - 4 + 1 = 4 \text{ electrons} \]
These 4 electrons form 2 lone pairs on iodine.
3. **Geometry**: The molecular geometry is also octahedral due to sp³d² hybridization.
4. **Bond Angle**: Similar to KrF4, the bond angles in ICl4⁻ are less than 109.28°.

### Conclusion
All three molecules (SF2, KrF4, and ICl4⁻) have two lone pairs and bond angles less than 109.28°. Therefore, the answer is that all of these molecules satisfy the given conditions.