# 2.9: Building up the MOs of More Complex Molecules- NH₃, P₄

- Page ID
- 184895

## MO diagram for NH_{3}

We can now attempt the MO diagram for NH_{3}, building on the result we obtained with triangular H_{3}^{+}.

*Notes on the MO diagram for ammonia:*

- Viewed end-on, a p-orbital or an sp
_{x}hybrid orbital looks just like an s-orbital. Hence we can use the solutions we developed with s-orbitals (for H_{3}^{+}) to set up the σ bonding and antibonding combinations of nitrogen sp^{3}orbitals with the H 1s orbitals.

- We now construct the sp
^{3}hybrid orbitals of the nitrogen atom and orient them so that one is “up” and the other three form the triangular base of the tetrahedron. The latter three, by analogy to the H_{3}^{+}ion, transform as one totally symmetric orbital (“a_{1}”) and an e-symmetry pair. The hybrid orbital at the top of the tetrahedron also has a_{1}symmetry.

- The three hydrogen 1s orbitals also make one a
_{1}and one (doubly degenerate) e combination. We make bonding and antibonding combinations with the nitrogen orbitals of the same symmetry. The remaining a_{1}orbital on N is non-bonding. The dotted lines show the correlation between the basis orbitals of a_{1}and e symmetry and the molecular orbitals

- The result in the 8-electron NH
_{3}molecule is**three N-H bonds**and**one lone pair**localized on N, the**same as the valence bond picture**(but much more work!).

## P_{4} molecule and P_{4}^{2+} ion:

By analogy to NH_{3} we can construct the MO picture for one vertex of the P4 tetrahedron, and then multiply the result by 4 to get the bonding picture for the molecule. An important difference is that there is relatively little s-p hybridization in P_{4}, so the lone pair orbitals have more s-character and are lower in energy than the bonding orbitals, which are primarily pσ.

Take away 2 electrons to make P_{4}^{2+}

Highest occupied MO is a bonding orbital → *break one bond, 5 bonds left*

Square form relieves ring strain, (60° → 90°)