According to the following formula:
Type of hybridisation/steric no. = (no. of sigma bonds + no. of lone pairs)
it should be $\ce{sp^3d^3}$, however according to my textbook it is $\ce{sp^3d^2}$ in solid state. That is how it was written(in my textbook), emphasising on the fact that "in solid state it is $\ce{sp^3d^2}$". How does solid state of a compound changes its hybridisation?
Answer
Forget about applying hybridization outside the second row, especially in 'hypervalent' compounds. I know, that it is common to use and sometimes works, but it is incorrect.
The $\ce{XeF6}$ molecule is a hard spot. While, indeed, experimental data suggest that it adopts distorted octahedral geometry in the gas phase, there is evidence that the minimum is very shallow.
The octahedral structure of $\ce{XeF6}$ (which is probably a local minimum, or at least found to be one in more than one calculation) can be pretty easily described in terms of three-center four-electron bonds.
There is no good agreement on the nature of the stereo-active electron pair in $\ce{XeF6}$. It seems, that repulsion with core d-shells is important, but the recent articles I could get my hands on has little to no rationalization on the fact. Still, it must be noted that the minimum is very shallow, meaning that rough qualitative theories like VSEPR and MO LCAO are too rough for a good rationalization anyway.
The solid state typically involves a lot more interaction, and, as result, compounds may adopt very different structures. The simplest one example I can think of, I could point to $\ce{SbF5}$ for example. A monomer adopts trigonal-bipyramidal structure. However, in the solid state, a tetramer with octahedrally coordinated antimony and four bridging fluorines is formed. In terms of VSEPR, it is a move from $\mathrm{sp^3d}$ to $\mathrm{sp^3d^2}$. Similarly, $\ce{XeF6}$ has a crystal phase (one of six) that involve bridging atoms.
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