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Even though certain atomic orbitals would support the observed number of bonds in a molecule, they don’t always support the observed bond angles. For instance, in nh3, there are three unpaired electrons in nitrogen’s atomic orbitals that correctly support the three h bonds; however, these three bonds would be 90 degrees to one another, something vesper disproves because of the distortion the lone pair creates. Therefore, hybridization is necessary.
I had similar confusion on this topic. Can someone please explain to me why the three h bonds without hybridization would be at 90 degree angles and how hybridization changes these angles to the correct bond angles according to VESPR?
The three unpaired electrons in nitrogen are in the p orbitals, which are in the x, y, and z planes, which are 90 degrees to each other (orthogonal). Ammonia (NH3) has a tetrahedral electron density geometry, so you need 4 sp3 hybrid orbitals.
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