# 2.7: Three-center Bonding

- Page ID
- 183906

Many (but not all) of the problems we will solve with MO theory derive from the MO diagram of the H_{2} molecule (Fig. **2.1.5**), which is a case of two-center bonding. The rest we will solve by analogy to the H_{3}^{+} ion, which introduces the concept of three-center bonding.

We can draw the H_{3}^{+} ion (and also H_{3} and H_{3}^{-}) in either a **linear** or **triangular** geometry.

**Walsh correlation diagram for H _{3}^{+}:**

*A few important points about this diagram:*

- For the linear form of the ion, the highest and lowest MO’s are symmetric with respect to the inversion center in the molecule. Note that the central 1s orbital has
**g symmetry**, so by symmetry it has**zero overlap**with the**u combination**of the two 1s orbitals on the ends. This makes the σ_{u}orbital a**nonbonding**orbital.

- In the triangular form of the molecule, the orbitals that derive from σ
_{u}and σ*_{g}become degenerate (i.e., they have identically the same energy by symmetry). The term symbol “e” means**doubly degenerate**. We will see later that “t” means triply degenerate. Note that we drop the “g” and “u” for the triangular orbitals because a triangle does not have an inversion center.

- The
**triangular form is most stable**because the two electrons in H_{3}^{+}have lower energy in the lowest orbital. Bending the molecule creates a third bonding interaction between the 1s orbitals on the ends.

**MO diagram for XH _{2} (X = Be, B, C…):**

*Some key points about this MO diagram:*

- In the linear form of the molecule, which has inversion symmetry, the 2s and 2p orbitals of the X atom factor into three symmetry classes:

- 2s = σ
_{g} - 2p
_{z}= σ_{u} - 2p
_{x}, 2p_{y}= π_{u}

- Similarly, we can see that the two H 1s orbitals make two linear combinations, one with σ
_{g}symmetry and one with σ_{u}symmetry. They look like the bonding and antibonding MO’s of the**H**(which is why we say we use that problem to solve this one)._{2}molecule

- The π
_{u}orbitals must be**non-bonding**because there is no combination of the H 1s orbitals that has π_{u}symmetry.

- In the MO diagram, we make bonding and antibonding combinations of the σ
_{g}’s and the σ_{u}’s. For BeH_{2}, we then populate the lowest two orbitals with the four valence electrons and discover (not surprisingly) that the molecule has**two bonds**and can be written**H-Be-H**. The correlation diagram shows that a bent form of the molecule should be**less stable**.

An interesting story about this MO diagram is that it is difficult to predict a priori whether CH_{2} should be linear or bent. In 1970, Charles Bender and Henry Schaefer, using quantum chemical calculations, predicted that the ground state should be a bent triplet with an H-C-H angle of 135°.^{[4]} The best experiments at the time suggested that methylene was a linear singlet, and the theorists argued that the experimental result was wrong. *Later experiments proved them right!*

“A theory is something nobody believes, except the person who made it. An experiment is something everybody believes, except the person who made it.” – Einstein