Question

Hello! I am working on some homework and came across this question, which completely confuses me for various reasons.

Experimental evidence suggests that the nitrogen atom in ammonia, NH3, has four identical orbitals in the shape of a pyramid or tetrahedron.
a). Draw an energy-level diagram to show the formation of these hybrid orbitals. (Hint: no electron promotion is required).

b). Name the type of hybrid orbitals found in NH3. Of the four hybrid orbitals on the N atom, how many will take part in bonding? Explain.

Answer 1

I thought I had a relatively good understanding on the concept of orbital hybridization, so my first question is why are we even talking about hybridization in the case of NH3? Nitrogen has 5 valence electrons, and so it can bond with the 3 from hydrogen to form a full octet.

Secondly, if I somehow get over this initial problem, and draw the energy-level diagram, then sp3 hybrid orbitals form. But why are there supposed to be 4? The 2s orbital electrons and the 2p electrons are supposed to hybridize and form 5 sp3 hybrid orbitals, not 4. Unless I missed a point somewhere. So you see, completely confused now. The Lewis structure tells me that there are 3 bonds with hydrogen and one lone pair on the nitrogen. So how are they getting 4 instead of 5 hybrid orbitals? 2 electrons are supposed to be paired up, so that would leave 3 unpaired and ready to bond with electrons from the 3 hydrogen atoms (aka. 3 hydrogen electrons). Please help! :)

Answer 2

1 Px orbital, 1 Py orbital, 1 Pz orbital and the 2s orbital = 4 orbitals for bonding

the lone pair doesn't bond, if it did bond, you would have an ammonium cation (NH4)

The hybridized bonds are called sigma bonds.

the lone pair on top makes it a tetrahedral

Answer 3

edit: 1 2Px orbital, 1 2Py orbital, 1 2Pz orbital and the 1 2s orbital = 4 orbitals for bonding