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What is the difference between $\mathrm{dsp^3}$ and $\mathrm{sp^3d}$ hybridization? Are they one in the same?
Also, I have a book that says that in compounds where the central atom is $\mathrm{dsp^3}$ hybridized, it's shape is square pyramidal, like that of $\ce{BrF5},$ but the hybridization of $\ce{Br}$ in $\ce{BrF5}$ is $\mathrm{sp^3d^2}.$ Why is it so? Is my book wrong or am I missing something?
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.
@Jan While your answer is technically correct and is the method that would most likely be used in an upper division/graduate inorganic or physical chemistry course, I think the OP's question was from the perspective of general chemistry where things are kept a bit more simplified. For example, in the general chemistry textbook I use by Petrucci, et.al., they discuss the hybridization model for both main group and transition metal chemistry. They also point out that experimental data indicates difficulties with $\ce{sp^3d^2}$ hybridization for $\ce{SF6}$ and suggest an alternative which involves 4 covalent bonds to the sulfur and 2 ionic bonds, i.e. $\ce{SF4^2+(F^-)2}$. No general chemistry textbook that I have seen (and I have seen quite a few) discusses 4-electron-3-center bonds because it becomes too complicated at the general chemistry level and not necessary for those students who are not chemistry majors. MO theory is discussed at the general chemistry level but only for diatomic molecules and ions from the second period, again because more atoms or higher periods makes it more complicated and it's just meant to give them a taste of what it is.
