Exams › JEE Main › Chemistry

For OF2 molecule consider the following A. Number of lone pairs on oxygen is 2. B. OFF angle is less than 104.5°. C. Oxidation state of O is -2. D. Molecule is bent 'V' shaped. E. Molecular geometry is linear. Correct options are :

1. C, D, E only
2. B, E, A only
3. A, C, D only
4. A, B, D only

Correct answer: A, B, D only

Solution

Options A, B, and D are correct because oxygen in OF2 has two lone pairs, the OFF bond angle is indeed less than 104.5° due to the repulsion from these lone pairs, and the molecular shape is bent or 'V' shaped, consistent with the presence of lone pairs.

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 ______.
• Using molecular orbital theory, which sequence arranges the nitrogen species in order of increasing bond order?
• 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, \(\mathrm{PO_4^{3-}}\), what is the formal charge on an oxygen atom that is singly bonded to phosphorus?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →