How do you determine which orbitals are used in hybridization?
Answer 1:
Yes. For general chemistry, we just count the number of electron groups around the central atom, and assume that the orbitals used are in order of angular momentum \(l\)
“NORMAL”, GENERAL CHEMISTRY WAY
We assume an ordering of
s,p,p,p,d,d,d,d,d,f,f,f,f,f,f,f
or
\(s p^{3} d^{5} f^{7}\).
For example, in \(\mathrm{NH}_{3} \text {, three } \mathrm{N}-\mathrm{H}\), bonds are made and one lone pair of electrons is leftover, so we assume that the hybridization is \(s+p+p+p \rightarrow s p^{3}\)
(You can think of it as the central atom “anticipating” future bonds it can possibly make.)
Or, in \(\mathrm{BH}_{3} \text {, three } \mathrm{B}-\mathrm{H}\) bonds are made, and there are no other lone pairs or bonds. So, we assume the hybridization is \(s+p+p \rightarrow s p^{2}\).
In short, count the orbitals from left to right in the list above until you have the same number of orbitals as the number of electron groups.
MORE RIGOROUS WAY
There is a more rigorous way to do it using Group Theory. You do not have to know this for general chemistry, but I am just providing it for the interested reader.