Statement I:  The dipole moment of CH₃F is greater than CH₃Cl

Statement II:  The c–F bond is more polar than C–Cl bond

Statement I is false but statement II is correc

The answer is 4. Statement I is false but statement II is correct.

Statement I is false because the dipole moment of CH₃F is less than CH₃Cl. This is because the electronegativity of chlorine (3.0) is greater than the electronegativity of fluorine (4.0). This means that the electrons in the C–Cl bond are more polarized towards chlorine, resulting in a greater dipole moment.

Statement II is correct because the C–Cl bond is more polar than the C–F bond. This is for the same reason as stated above. The greater the electronegativity difference between two atoms, the more polar the bond.

Therefore, statement I is false but statement II is correct.

Here is a table that summarizes the electronegativity of the atoms involved:

The dipole moment of a molecule is a measure of the separation of charge in the molecule. A molecule with a higher dipole moment has a greater separation of charge and is, therefore, more polar.

Extra Reading:

Statement I: \(CH_3F\) has a greater dipole moment than \(CH_3Cl\). This is false. The dipole moment of \(CH_3Cl\) (1.94 D) is actually higher than that of \(CH_3F\) (1.82 D).

Statement II: The \(C–F\) bond is more polar than the \(C–Cl\) bond. This is true. Fluorine's higher electronegativity leads to a greater pull of electron density towards itself, creating a more polar bond.

Therefore, the correct answer is 4. Statement I is false but statement II is correct.