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, and any net formal charge located on E. I is neutral, has four bonded neighbours and one central lone pair. II has unit negative charge, five bonded neighbours and one central lone pair. Both contain O and F. Determine the absolute difference in TOTAL lone-pair counts. Each ligand O has two lone pairs and F three. Infer both formulas from formal charge and geometry; these are specified Lewis models, not an assertion that every ion is isolable.
Answer:
Solution:
Let o, f be ligand counts and L central lone pairs. Central formal charge q=8−2L−(2o+f). I has o+f=4, L=1, q=0, giving o=2, f=2. Total LP=1+4+6=11. II has o+f=5, L=1, q=−1, giving 2o+f=7 and thus o=2, f=3. Total LP=1+4+9=14. Difference3. The charge changes the allowed bond-order sum; treating II as neutral would give the wrong formula.
