Why isn't cyclobutane planar?

Question

In short, cyclobutane adopts a certain, non-planar, configuration in order to reduce ring strain caused by the fact that carbons are $\\mathrm{sp^3}$ hybridised. That much is understandable. In an exercise it was suggested to propose another kind of hybrisation for the carbons in order to justify a (theoretical) planar configuration. I thought that carbons could be $\\mathrm{sp}$ hybridised. This way, $\\mathrm{p}$ orbitals could form the $\\ce{C-C}$ bonds and $\\mathrm{sp}$ orbitals could form the $\\ce{C-H}$ bonds. I checked the answer and I was right. However, I couldn't think of something that could justify why *this* particular structure is not possible. That exactly is my question.

Answer 1

In short, any carbon with four single bonds can be considered sp3 which always has a tetrahedral geometry keeping all bonded atoms from being in different planes from one another with a 109.5 degree separation in three dimensions. This also applies to cyclohexane. However, when we consider something like benzene and it's six carbons, it is flat because it is composed of six sp2 carbon where all 3 substituents (C on each side and 1 H per benzene ring carbon) are 120 degrees apart and in the same plane.

Let me know if this helps clarify.