Exams › JEE Main › Chemistry
Using a simple molecular orbital treatment, the bond order of the hypothetical molecule OF would be:
- 2
- 1.5
- 1.0
- 0.5
Correct answer: 1.5
Solution
OF has 13 valence electrons; filling gives 8 bonding and 5 antibonding electrons, so bond order = (8-5)/2 = 1.5.
Related JEE Main Chemistry questions
- Although the electronegativity gap between N and F is larger than that between N and H, ammonia has a dipole moment of 1.5 D while nitrogen trifluoride has only 0.2 D. The reason is that
- When N₂ is converted into N₂⁺, the dissociation energy of the N–N bond ______, and when O₂ is converted into O₂⁺, the dissociation energy of the O–O bond ______.
- From the ions listed below, which pair has geometries that can be accounted for by the same type of orbital hybridization? NO₂⁻, NO₃⁻, NH₂⁻, NH₄⁺, SCN⁻
- Atoms A and B have electronegativities of 1.20 and 4.0, respectively. What is the percentage ionic character of the A–B bond?
- In the phosphate ion, PO₄³⁻, what is the formal charge on an oxygen atom that is singly bonded to phosphorus in a P–O bond?
- Among the following species, which one has no unpaired electrons?
⚔️ Practice JEE Main Chemistry free + battle 1v1 →