Hybrid lobes of sp3d2 and sp3d3 [closed]

Question

It's known that $\ce{sp^3d}$ hybridization has 5 lobes; 3 equatorial with $\ce{sp^2}$ lobes and 2 axial with $\ce{dp}$ lobes. Do we have such splitting in $\ce{sp^3d^2}$ and $\ce{sp^3d^3}$ hybridization states too?
What I think is $\ce{sp^3d^2}$ splits into $4 \ce{dsp^2}$ lobes and $2 \ce{dp}$ lobes.

Is this correct?

Answer 1

To answer this, consider the purpose of the mathematical exercise we call "hybridization". The goal is to carve up the total electron density of the molecule into discrete segments (bonding orbitals) that are localized mostly to the area between two nuclei and which have some type of symmetry (usually either sigma or pi) with respect to the axis connecting the two nuclei.

A final important criterion is that these hybrid orbitals should be consistent with our intuitive understanding of equivalent bonds. For example, the four C-H bonds in methane are indistinguishable. Thus, the four hybrid bonding orbitals should also be indistinguishable.

With that in mind, consider that the six metal-ligand bonds in an octahedral complex are also indistinguishable (assuming all six ligands are identical). It does not make sense, then, to use four dsp² hybrid orbitals and two dp, since that is not match six indistinguishable bonds. Instead, we extend the methane example and make six identical sp³d² hybrids.

With odd numbers of ligands like 5 or 7, the metal ligand bonds are not all equivalent, so it makes sense to have a separate hybridization construct for each type of metal-ligand bond. For example, in a trigonal bipyramid, the equatorial bonds form one group and the axial bonds another.