Identify the correct orders against the property mentioned

A. H₂O > NH₃ > CHCl₃ – dipole moment
B. XeF₄ > XeO₃ > XeF₂ – number of lone pairs on central atom
C. O–H > C–H > N–O – bond length
D. N₂ > O₂ > H₂ – bond enthalpy

Choose the correct answer from the options given below:

To identify the correct orders against the given properties, let us analyze each statement one by one:

Statement A: Dipole moment order: H₂O > NH₃ > CHCl₃
- Water (H₂O) has two lone pairs and two polar O-H bonds. The bond dipoles and lone pair dipoles reinforce each other, resulting in a very high dipole moment (approximately 1.85 D).
- Ammonia (NH₃) has one lone pair and three polar N-H bonds. The bond dipoles and lone pair dipole also reinforce each other, giving a high dipole moment (approximately 1.47 D), but less than H₂O due to fewer lone pairs and lower electronegativity of nitrogen compared to oxygen.
- Chloroform (CHCl₃) is a tetrahedral molecule with three polar C-Cl bonds and one C-H bond. While it is polar, the net dipole moment (approximately 1.04 D) is significantly lower than that of H₂O and NH₃.
Therefore, the order H₂O > NH₃ > CHCl₃ is correct.

Statement B: Number of lone pairs on central atom order: XeF₄ > XeO₃ > XeF₂
- In XeF₄, Xenon has 8 valence electrons, shares 4 with fluorine atoms, leaving 4 non-bonding electrons, which corresponds to 2 lone pairs.
- In XeO₃, Xenon forms 3 double bonds with oxygen atoms using 6 valence electrons, leaving 2 non-bonding electrons, which corresponds to 1 lone pair.
- In XeF₂, Xenon shares 2 electrons with fluorine atoms, leaving 6 non-bonding electrons, which corresponds to 3 lone pairs.
The correct order for the number of lone pairs on the central atom is XeF₂ (3) > XeF₄ (2) > XeO₃ (1). Thus, statement B is incorrect.

Statement C: Bond length order: O–H > C–H > N–O
- The bond length typically increases with the size of the bonding atoms. Nitrogen and Oxygen atoms are larger than Hydrogen atoms.
- Therefore, a bond between two non-hydrogen atoms like N–O (where both atoms are from the second period) is longer than bonds involving hydrogen (like O–H and C–H).
- Thus, the bond length order O–H > C–H > N–O is incorrect because N–O has the longest bond length among them.

Statement D: Bond enthalpy order: N₂ > O₂ > H₂
- Nitrogen (N₂) contains a triple bond ($N\equiv N$), which is extremely strong and requires a very high amount of energy to break (bond enthalpy is approximately 945 kJ/mol).
- Oxygen (O₂) contains a double bond ($O=O$), which is also strong but weaker than the triple bond in N₂ (bond enthalpy is approximately 498 kJ/mol).
- Hydrogen (H₂) contains a single bond ($H-H$), which has a lower bond enthalpy compared to N₂ and O₂ (approximately 436 kJ/mol).
Therefore, the order of bond enthalpy is N₂ > O₂ > H₂, making statement D correct.

Since only statements A and D are correct, the correct option is A, D only.