Because there exist exactly three p orbitals per quantum level n, and because the (n−1)d orbital is closer in energy to the (n−1)s orbital than an np orbital.
It sounds like a misunderstanding of the hybridization notation.
HYBRIDIZATION NOTATION
Having sp3 hybridization tells you that one ns orbital mixes with three np orbitals to generate four identical (in symmetry), degenerate (in energy) sp3 orbitals. That way, the given compound can form up to four identical bonds if it wishes.
From the explanation above, using n=4, it follows that sp4 implies the existence of a fourth (n−1)p orbital, or implies an np orbital somehow participates as the fourth p orbital.
However, that is not reasonable, given that an np orbital is too far away in energy to interact in place of using the (n−1)d orbital for n=4. Furthermore, there exist only three (n−1)p orbitals, not four---remember that for l=1 (for all p orbitals), ml={−1,0,+1}.
A MORE EXPLICIT EXAMPLE
The 3d orbital, for instance, is closer in energy to the 3s orbital than the 4p is to the 3s for a given atom.
That means it is more favorable to mix one 3s, three 3p, and one 3d if the compound wishes to form five bonds using sp3d hybridization.
An example of sp3d would be PF5.
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