Yes, I'll add a couple of exampes to illustrate, paraphrase and elaborate on Astrokel's method (I always encourage students to understand the *meaningfulness* behind any mathematical formula). Here, students should appreciate the *purpose* of hybridization.

1) Ethyne.

To minimize electron repulsion, the VSEPR molecular geometry is linear, and the VSEPR electron geometry is also linear (since there are no lone pairs). To achieve linear VSEPR electron geometry, hybridization of C orbitals must be sp (without hybridization, the bond angles would result in significant electron repulsion and an unstable geometry). This leaves 2 p orbitals (s s p p --> s p p p --> sp sp p p) for overlap (with the other C's 2p orbitals) to form the 2 pi bonds (over the sigma bond; hence triple bond between the 2 Cs).

2) Ammonia.

The molecular VSEPR geometry is trigonal pyramidal, but the electron geometry is tetrahedral. Hence hybridization is sp3 (to achieve tetrahedral electron VSEPR geometry).

A related question to this :

**Identify and explain whether PH3 is a stronger or weaker nucleophile and base, as compared to NH3.**

The solution, is to recognize that P is one electron shell larger than N; consequently the electron repulsion (between electron pairs, regardless lone or bond) is less in PH3 as compared to NH3; consequently there is less need and therefore less extent of hybridization of electron orbitals; consequently the bond angles in PH3 are almost orthogonal as compared to (slightly less than, due to greater repulsion between lone pair and bond pairs) tetrahedral bond angles in NH3; which implies that the lone pair in PH3 is largely the s orbital as compared to sp3 orbital in NH3; consequently the s orbital lone pair of PH3 is less available for nucleophilic attack and/or accepting a proton, as compared to NH3.