# dsp3 and sp3d hybridization — what is the difference and which applies to square pyramidal?

What is the difference between dsp3 and sp3d Hybridization? Are they one in the same?

Also, I have a book that says that in compounds where the central atom is dsp3 hybridized, it's shape is square pyramidal, like that of $\ce{BrF5}$, but the hybridization of Br in $\ce{BrF5}$ is sp3d2. Why is it so? Is my book wrong or am I missing something?

• $dsp^3$ hybridization cannot result from square pyramidal geometry because it only produces five hybrid orbitals whereas square pyramidal requires six. $sp^3d^2$ is a possible hybridization for square pyramidal geometry. – bon Nov 14 '15 at 11:20
• So , my book was wrong and I was right , right ? Even if so , then what is the difference between dsp3 and sp3d , if there is one ? – user22729 Nov 14 '15 at 12:19
• Well, your book is wrong. Both hybridisations don’t really exist. Main group elements only hybridise s and p-orbitals. – Jan Nov 14 '15 at 13:29
• So there are no sp3d , sp3d2 or sp3d3 hybridizations ? – user22729 Nov 14 '15 at 14:06
• @quintopia No it doesn’t. Xenon is essentially unhybridised as is fluorine. Xenon’s $\mathrm{p}_z$ orbital forms a four-electron-three-centre bond with the two fluorine atoms. Xenon’s 4d orbitals are fully populated and belong to the core orbitals; they won’t participate in bonding. Xenon’s 5d orbitals are empty and energetically remove; the 6s orbitals are closer. Thus, xenon has no available d electrons for ‘hybridisation’. – Jan Nov 24 '16 at 1:08

We can split this into two cases: main group elements and transition elements.

Main group elements do not use their d-orbitals for bonds and thus cannot be hybridised in any d-containing way. Bromine in $\ce{BrF5}$, to cite your example, is unhybridised. The $\mathrm{p}_z$ orbital forms a traditional two-electron-two-centre bond with one of the fluorines. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals each form a four-electron-three-centre bond with two additional fluorines for a total count of five fluorines. The bond order of these four-electron-three-centre bonds is $\frac12$.

Bromine’s 3d orbitals are completely populated and effectively belong to the core orbitals. They do not participate in bonding. Bromine’s 4d orbitals are energetically even further removed than the 5s orbital is — remember the aufbau principle. Thus, neither the lower nor the higher d-orbitals can participate in bonding to any notable extent.

This extends well to all main group elements as hinted above. Practically no main group element compounds will include d-orbitals.

For transition metals the case is slightly different since we are in the midst of the d-block where d-orbitals can play a role. However, invoking hybridisation for transition metals puts a big blanket over the bonding situation declaring things equal which really aren’t. A much better description for transition metals would be to use a molecular orbital scheme as I have done in countless answers of mine. Most importantly, the bonding situation can, in my opinion, not be understood without this depiction.

Additionally, many transition metal geometries provide unintuitive ‘hybridisation’ solutions — octahedral being the notable exception which actually hybridises six orbitals (three d-orbitals are nonbonding). For example, a tetrahedral metal complex should be understood as using all orbitals for bonding. Likewise, the square-planar ligand arrangement only has one nonbonding d-orbital.

My suggestion is never to use hybridisation approaches for transition metal complexes.

Formally, one may say that putting the d before s and p implies a lower-shell d-orbital. That would be the transition metal case above. And placing the d behind s and p would use d-orbitals from the same shell. That would be the main group case. Thus, the difference between the two (tl;dr) is:

The first is very strongly discouraged, the second is wrong.