Between ionic and covalent bonds, there are large majority of bonds, in which the bonding electrons are shared unequally between two atoms but are not completely transferred.Such bonds are said to be polar covalent bonds and the bond polarity is due to difference in electronegativity, the ability of an atom in a molecule to attract the shared electrons in a covalent bond.
The measure of net polarity is a quantity called the dipole moment, `mu` , which is defined as the magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges : `mu=Qxxr`.Molecular polarities give rise to some of the forces that occur between molecules and these forces are of several different types including dipole-dipole forces, London dispersion forces, and hydrogen bonds. All these intermolecular forces are electrical in origin and result from the mutual attraction of unlike charges or the mutual repulsion of like charges.
Which of the following statements is incorrect ?

Options

A. Out of trimethylamine and trimethylphsophine , trimethylamine has higher dipole moment
B. Out of `(SiH_3)_2O` and `(CH_3)_2O, (SIH_3)_2O` is more basic
C. The critical temperature of water is higher than that of `O_2` because the `H_2O` molecule has dipole moment
D. Intermolecular hydrogen bonding increases the enthalpy of vaporization of a liquid due to the increase in the attraction between molecules .

Explanation

To determine which statement is incorrect among the given options regarding dipole moments and molecular properties, we will analyze each statement step by step.

### Step 1: Analyze Option A
**Statement A:** "Out of trimethyl amine and trimethyl phosphine, trimethyl amine has a higher dipole moment."

- **Structure of Trimethyl Amine (TMA):** TMA has a nitrogen atom bonded to three methyl groups (CH₃). Nitrogen has a higher electronegativity (3.0) compared to carbon (2.5), which means it will attract the bonding electrons more strongly, resulting in a net dipole moment directed towards the nitrogen atom.

- **Structure of Trimethyl Phosphine (TMP):** TMP has a phosphorus atom bonded to three methyl groups. Phosphorus has a lower electronegativity (2.1) than nitrogen, meaning it will not attract the bonding electrons as strongly as nitrogen does. The dipole moments from the methyl groups will cancel each other out, resulting in a net dipole moment of zero.

**Conclusion for Option A:** Trimethyl amine has a higher dipole moment than trimethyl phosphine. Thus, this statement is **correct**.

### Step 2: Analyze Option B
**Statement B:** "Out of SiH₃O and CH₃O, SiH₃O is more basic."

- **Structure of SiH₃O:** In SiH₃O, silicon can participate in back bonding due to its vacant d-orbitals, which can stabilize the lone pair of electrons on oxygen, reducing its availability for proton abstraction.

- **Structure of CH₃O:** In CH₃O, there is no back bonding because carbon does not have vacant d-orbitals. Therefore, the lone pair on oxygen is more available for proton abstraction, making it a stronger base.

**Conclusion for Option B:** SiH₃O is less basic than CH₃O, making this statement **incorrect**.

### Step 3: Analyze Option C
**Statement C:** "The critical temperature of water is higher than O₂ because H₂O has a higher dipole moment."

- **Dipole Moment of H₂O:** Water is a polar molecule with a significant dipole moment due to the electronegativity difference between hydrogen and oxygen, leading to strong hydrogen bonding.

- **Dipole Moment of O₂:** Oxygen is a nonpolar molecule with no dipole moment, resulting in weaker London dispersion forces.

**Conclusion for Option C:** The higher dipole moment of water leads to stronger intermolecular forces, resulting in a higher critical temperature compared to O₂. Thus, this statement is **correct**.

### Step 4: Analyze Option D
**Statement D:** "Intermolecular hydrogen bonding increases the enthalpy of vaporization of a liquid due to increased attraction of molecules."

- **Hydrogen Bonding:** Hydrogen bonds are strong intermolecular forces that require more energy to break, thus increasing the enthalpy of vaporization.

**Conclusion for Option D:** This statement is **correct** as stronger intermolecular forces lead to higher enthalpy of vaporization.

### Final Conclusion
After analyzing all the options, we find that the **incorrect statement is Option B**.