No. Electron spin is either + or - and is one of the quantum numbers which characterize the orbital. The full set of quantum numbers (analogous to the atomic Quantum Numbers) 'completely' characterize the orbital (not literally 'completely', since nuclear charge (among other things) isn't part of them). No two electrons may have the same quantum numbers (in the same location) and since spin is the easiest (lowest energy) QN to change, it works quite well to pretend that there is this 'thing' called an orbital which can be occupied by two electrons (as long as they have opposite spin.) Bonding and Antibonding Molecular Orbitals are made up of a LCAO (linear combination of atomic orbitals) and have a 'phase'. The phase can be thought of as the sign of the wave-function ψ. You may recall that ψ² is the electron probability density (the probability of a electron being at a particular point in space). But ψ×ψ = (-ψ)×(-ψ), so the phase of the wave isn't relevant to where the electron is. You don't want to go into the quantum mechanics to try to understand the reasons why, I think. When you compile a set of LCAO, the 'housekeeping' requires you to have equal numbers of both +ψ and -ψ, so that your MOs are made up of the overlap of +ψₐ with +ψᵦ, +ψₐ with -ψᵦ, -ψₐ with +ψᵦ, and -ψₐ with -ψᵦ. You can consider each 'lobe' of an atomic orbital to have either + or - phase. The simplest example of how this effects bonding is by using two atomic orbitals...say atom A has an s orbital, which has no nodes and is everywhere positive (this is only true for a 1s orbital, but is a great approximation for 2s, etc.) You probably know that the 'shape' of the orbital is a sphere, right? Ok, now let's say atom B has a p orbital. You (hopefully) know that there are 3 p orbitals, px, py, and pz and each is 'shaped' like a dumbell (or figure 8). Well, that's correct, but it is also true that one lobe has + phase and the other -. You'll have to believe me on this (although I simplify the situation). So picture O ⟷ 8 moving together. If you accept that +ψₐ with +ψᵦ results in bonding and +ψₐ with -ψᵦ results in antibonding, then you may be able to see that as the two orbitals overlap, the amount of ++space is exactly the same as the amount of +- space? This results in non-bonding (a non-bonding MO). Same process but now let's change the p orbital to this ∞, where the left lobe is +. As you move them together O ⟷ ∞ the result is a lot of ++ overlap and no +- overlap (unless you shove the p orbital into the O so far that the nucleus is inside the other atom's radius). This is bonding. Antibonding is just as easy to picture, just flip the p orbital end-over-end and approach the O orbital with a negative lobe.
