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 > F₂ – bond energy

Choose the correct answer from the options given below:

To identify the correct statements, let us analyze each of the given orders one by one:

Analysis of Order A (Dipole Moment):
The given order is H₂O > NH₃ > CHCl₃.
- Water (H₂O) has a bent shape with two highly electronegative O-H bonds and two lone pairs on the oxygen atom, resulting in a very high dipole moment (about 1.85 D).
- Ammonia (NH₃) has a trigonal pyramidal shape. The dipole moments of the three N-H bonds and the lone pair reinforce each other, giving a high dipole moment (about 1.47 D).
- Chloroform (CHCl₃) has a tetrahedral geometry. The C-Cl dipoles are opposed by the C-H dipole, resulting in a lower net dipole moment (about 1.04 D).
Thus, the dipole moment order is H₂O > NH₃ > CHCl₃. Therefore, statement A is correct.

Analysis of Order B (Number of lone pairs on the central atom):
The given order is XeF₄ > XeO₃ > XeF₂.
Let us calculate the number of lone pairs on the central xenon (Xe) atom in each molecule (Xe has 8 valence electrons):
- In XeF₄: Xenon forms 4 single bonds with fluorine, using 4 electrons. The remaining 4 electrons form 2 lone pairs.
- In XeO₃: Xenon forms 3 double bonds with oxygen, using 6 electrons. The remaining 2 electrons form 1 lone pair.
- In XeF₂: Xenon forms 2 single bonds with fluorine, using 2 electrons. The remaining 6 electrons form 3 lone pairs.
Thus, the correct order for the number of lone pairs on the central atom is XeF₂ (3) > XeF₄ (2) > XeO₃ (1). Therefore, statement B is incorrect.

Analysis of Order C (Bond length):
The given order is O–H > C–H > N–O.
Bond length depends heavily on the atomic radii of the bonded atoms. The atomic size order is H < O < N < C. Comparing the bonds:
- Both O-H and C-H involve the extremely small hydrogen atom, but oxygen is smaller than carbon, so the O-H bond length (approx. 0.96 Å) is shorter than the C-H bond length (approx. 1.09 Å).
- The N-O bond involves two second-period atoms, which are significantly larger than hydrogen, making the N-O bond length (approx. 1.40 Å or shorter depending on bond order) much longer than both O-H and C-H.
Thus, the correct bond length order is N–O > C–H > O–H. Therefore, statement C is incorrect.

Analysis of Order D (Bond energy):
The given order is N₂ > O₂ > F₂.
Bond energy is directly proportional to the bond order (number of shared electron pairs) between the atoms:
- Nitrogen (N₂) has a triple bond (N≡N) with a very high bond energy (approx. 945 kJ/mol).
- Oxygen (O₂) has a double bond (O=O) with a moderate bond energy (approx. 498 kJ/mol).
- Fluorine (F₂) has a single bond (F-F) with a low bond energy (approx. 159 kJ/mol), which is further weakened by strong inter-electronic repulsion between the non-bonding lone pairs on adjacent small fluorine atoms.
Thus, the bond energy order is N₂ > O₂ > F₂. Therefore, statement D is correct.

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