Assertion `BF_(3)` has greater dipole momnet than `H_(2)S`
Reasoning Fluorine is more electronegative than sulphur .

Options

A. If both `(A)` and `(R )` are correct and `(R )` is the correct
explanation of `(A)`
B. If both `(A)` and `(R )` are correct and `(R )` is the correct
explanation of `(A)`
C. If `(A)` is correct but `(R )` is incorrect
D. If `(A)` is incorrect but `(R )` is correct

Explanation

To solve the assertion and reasoning question regarding the dipole moments of BF₃ and H₂S, we will analyze both the assertion and the reasoning step by step.

### Step 1: Analyze the Assertion
The assertion states that "BF₃ has a greater dipole moment than H₂S."

- **BF₃ (Boron Trifluoride)**:
- BF₃ has a trigonal planar structure due to sp² hybridization.
- The bond angles are 120 degrees, and the molecule is symmetrical.
- Although fluorine is highly electronegative, the symmetrical arrangement of the three fluorine atoms around the boron atom causes the dipole moments to cancel each other out.
- Therefore, the overall dipole moment of BF₃ is **zero**.

- **H₂S (Hydrogen Sulfide)**:
- H₂S has a bent molecular geometry due to sp³ hybridization.
- The presence of lone pairs on the sulfur atom contributes to an asymmetrical distribution of charge.
- As a result, H₂S has a net dipole moment that is **not zero**.

Thus, the assertion that "BF₃ has a greater dipole moment than H₂S" is **incorrect**.

### Step 2: Analyze the Reasoning
The reasoning states that "Fluorine is more electronegative than sulfur."

- **Electronegativity Comparison**:
- Fluorine (F) is indeed more electronegative than sulfur (S).
- Electronegativity decreases as we move down the periodic table due to increasing atomic size and shielding effect.
- Fluorine is in the second period, while sulfur is in the third period, confirming that fluorine is more electronegative.

Thus, the reasoning that "Fluorine is more electronegative than sulfur" is **correct**.

### Conclusion
- The assertion is **incorrect** because BF₃ has a dipole moment of zero, while H₂S has a non-zero dipole moment.
- The reasoning is **correct** because fluorine is indeed more electronegative than sulfur.

### Final Answer
The correct conclusion is that the assertion is incorrect and the reasoning is correct.

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