Chemistry - CHEMICAL BONDING Question with Solution | TestHub
Consider formal oxyfluoride models of a valence-eight centre E, with terminal E=O and E–F single bonds; all net formal charge lies on E. I has charge+1, four bonded neighbours and one central lone pair. II is neutral, has six bonded neighbours, no central lone pair, and an ideal geometry in which equivalent opposite bond dipoles cancel. Both contain O and F. Infer formulas and find the absolute difference in TOTAL lone-pair counts. Each O has two lone pairs, F three. Treat these as stipulated Lewis/domain models, not an existence claim.
Answer:
Solution:
q=8−2L−(2o+f). For I, o+f=4, L=1, q=+1, so2o+f=5, implying o=1, f=3, total LP=1+2+9=12. For II, o+f=6, L=0, q=0, so2o+f=8, implying o=2, f=4, total LP=4+12=16. The two O occupy opposite octahedral sites and remaining F form opposite pairs, consistent with zero bond-dipole sum. Difference16−12=4.
