# What are the angles between sp2 orbitals?

Dec 29, 2015

With the awareness that there may be some exceptions, we can say that the angle between ideal $s {p}^{2}$ orbitals is ${120}^{\circ}$, due to three coplanar electron groups evenly spanning ${360}^{\circ}$.

One example of when $s {p}^{2}$ hybridization can occur is when a certain number of electrons need to be donated to other atoms to form a bond, but not all electrons needed are available in the highest-energy orbitals.

Let's say that we wanted to describe the hybridization of one of the carbons in ethene:

Carbon atom normally has the electron configuration $1 {s}^{2} 2 {s}^{2} 2 {p}^{2}$, so it has four valence electrons, but the $2 p$ orbitals are a lot higher in energy than hydrogen's $1 s$ orbital.

To lower the energy of the $2 p$ orbitals so that they can interact with the $1 s$ orbital of hydrogen, they can hybridize with the $2 s$ orbitals to achieve an energy level that is in between those of the pure $2 s$ and $2 p$ energy levels.

This allows access to the $2 s$ electrons.

Now, we have three degenerate $\setminus m a t h b f \left(s {p}^{2}\right)$ hybridized orbitals, formed from one $2 s$ and two $2 p$ atomic orbitals (hence $s {p}^{2}$).

Although the resultant energy is technically closer to that of the pure $2 p$ than the pure $2 s$, this particular energy level is nevertheless low enough that hybridized bonding becomes favorable, and carbon can now bond with hydrogen with the electrons that originally belonged to the pure $2 s$ orbital. (And carbon can double bond with another carbon to finish its octet.)

We should also notice that there are 3 electron groups surrounding the $s {p}^{2}$ carbon. Additionally, the $s {p}^{2}$ carbon is planar due to the rigid sidelong $\setminus m a t h b f \left(p\right)$ orbital overlap of the two carbons.

Therefore, to distribute the atoms evenly in space, while also keeping the planar structure, the angle should be close to or precisely ${120}^{\circ}$.