Why can't Sp3 hybrid orbitals can form pi bonds?

A single bond consists of a `sigma` bond, which is defined as the head-on overlap between two compatible orbitals.

Because hybrid orbitals are made to match the symmetry of the incoming (i.e. not-yet-bonded) atom's atomic orbital for the sake of being able to bond at all (which is usually if not always "totally symmetric about the internuclear axis"), they always have to overlap head-on.

Thus, a hybrid orbital has to make a `sigma` bond.

Not just `sp^3`, but any hybrid orbital. Even in a triple bond, like in acetylene (`"H"-"C"-="C"-"H"`), the `pi` bonds are made by the `p_x` and `p_y` orbitals (or any qualified equivalent sidelong orbital overlap), while the `sigma` bonds are made with the hybrid orbitals, which consist of only the `p_z` and `s` orbitals.

![http://2012books.lardbucket.org/]()

Here, the `z` axis is defined as the internuclear axis, whereas the `x` and `y` axes are vertical and towards us, respectively.

As an alternative, consider ethene (`"H"_2"C"="CH"_2`), where the double bonds lie on the `xy`-plane in the plane of the screen/paper, and the `z` axis protrudes outward towards us.

If we consider the `sp^2` hybridization of carbon in ethene, carbon actually uses three `sp^2` hybrid orbitals: one each to `sigma` bond with the hydrogens, and one to `sigma` bond with the other carbon. (You would get that if you drew the molecule with all single bonds and no double bonds.)

It made each hybrid orbital by mixing the `2s`, `2p_x`, and `2p_y` orbitals together and distributing their angular positions evenly on the `xy`-plane since the original `2p` orbitals used to construct them were actually aligned in the `x` and `y` directions, respectively. (The notation `sp^2` comes from mixing one `s` and two `p` orbitals together to get `33%` `s` character and `66%` `p` character.)

Hence, the `\mathbf(sigma)` bonds of ethene (shown in yellow) lie on a single plane.

In addition, we have the `\mathbf(2p_z)` atomic orbital (perpendicular to the plane of the molecule, in purple) from each carbon that is used to `\mathbf(pi)` bond (sidelong overlap) with the other carbon, thus constructing the second component of the double bond (one `sigma` + one `pi` bond = one double bond).

It was not used in the orbital hybridization, and it remains as a different, incompatible orbital (with respect to the `\mathbf(2p_x)` and `\mathbf(2p_y)`) for `\mathbf(sigma)` bonding within the molecule. The only thing it can do at this point is `pi` bond because it is oriented precisely to do so.