Exams › JEE Main › Chemistry
Which among the following molecules is (a) involved in sp^3d hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom? (1) XeF4 (2) XeF2 (3) PF5 (4) SF4
- (1) XeF4
- (2) XeF2
- (3) PF5
- (4) SF4
Correct answer: (4) SF4
Solution
SF4 exhibits sp^3d hybridization due to the presence of five electron pairs around the sulfur atom, including one lone pair. This lone pair leads to a distorted seesaw molecular geometry, resulting in different bond lengths between the sulfur and fluorine atoms.
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 →