Question

1. How many electrons will be around I in the compound IF7 ?​

Answer 1

Answer:

know that the VSEPR theory explains IF7 (iodine heptafluoride)'s structure as a pentagonal bi-pyramidal one. The valence bond theory can be used to say that it has sp3d3 hybridisation (I think). How can the bonding in it be explained using the LCAO (Linear Combination of atomic orbitals) method in molecular orbital theory?

I got this doubt because there are only two "axial" d-orbitals (dz2 and dx2−y2) among the 5 present, which can have head-on overlap with p-orbitals (non-axial orbitals can only have sideways overlap). In the excited state, the iodine atom has 7 half filled atomic orbitals (an s-orbital, three p-orbitals, and 3 d-orbitals) and according to LCAO each of them has to combine with one p-orbital of each fluorine atom to form molecular orbitals by σ and π bonds. But, the formation of only 6 σ bonds can happen between an iodine and 6 fluorine atoms, as σ bonds involve head-on overlap of orbitals, and there are only 6 orbitals capable of doing that (the s and p orbitals, and the two 'axial' d-orbitals). Therefore, the seventh fluorine atom must be in pπ-dπ bond with fluorine.

This doesn't explain "five equivalent bonds and two bonds of equal length (all σ bonds)" as predicted by VSEPR theory. How many σ bonds does it really have?

Answer 2

Answer:

Answer: Iodine heptafluoride, IF7, is a good example of a pentagonal bipyramidal geometry. The molecule XeF6 is an interesting case. As with IF7, application of VSEPR rules suggests seven electron pairs

Explanation:

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