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If four orbitals on one atom overlap with four orbitals on another atom, how many molecular orbitals will form?

Why are there eight molecular orbitals and not four? Is not, in molecular orbital theory, each atom's orbital overlaps with another atom's orbital to form a bond?

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    $\begingroup$ If you have eight orbitals to start, you will have eight orbitals to finish. $\endgroup$ Nov 28, 2016 at 4:24
  • $\begingroup$ Can you give me an intuitive explanation as to what's going on? $\endgroup$ Nov 28, 2016 at 7:09
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    $\begingroup$ Technically speaking, each individual atom has infinitely many orbitals for its electrons, most of which are unoccupied. Infinity + Infinity = Infinity. $\endgroup$ Nov 28, 2016 at 12:30

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The question itself deals with molecular orbital theory and by extension with the approximation of combining atomic orbitals to form these molecular orbitals; in quantum chemistry this approach is called LCAO (Linear combination of atomic orbitals).

Let us look at one of the simple cases, the dihydrogen molecule. You have one 1s orbital at the first hydrogen, i.e. $\chi_A$, and one 1s orbital at the second hydrogen, i.e. $\chi_B$. You can now combine these orbitals to form a bonding $\eqref{eq:bonding}$ and an antibonding $\eqref{eq:antibonding}$ combination to create the molecular orbitals $\phi_{A+B}$ and $\phi_{A-B}$. \begin{align} \phi_{A+B} \equiv 1\sigma_\mathrm{g} &= \chi_A + \chi_B \tag1\label{eq:bonding}\\ \phi_{A-B} \equiv 1\sigma_\mathrm{u} &= \chi_A - \chi_B \tag2\label{eq:antibonding} \end{align} A scheme of this can be found on Wikipedia created by the user CCoil. MO diagram by CCoil from Wikipedia

This does in principle mean that for every bonding configuration of a molecular orbital you will always have an antibonding configuration. In other words the number of atomic orbitals is the same as the number of molecular orbitals.

In your case, when there are four orbitals on one atom and four orbitals on the other atom, you will end up with eight molecular orbitals. A quite complicated case of bonding that shows this is the $\ce{C2}$ molecule, which has previously been discussed here: Bonding in diatomic C2, a carbon-carbon quadruple bond?

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As it usually said, the number of orbitals is conserved in LCAO-MO. Intuitively it should be clear that the number of electrons is conserved when forming MOs from AOs. And since each orbital (atomic or molecular) holds up to 1 or 2 electrons (depending on whether are we talking about spin or spatial orbitals), the number of orbitals should also be (at least) conserved, for otherwise you might have not enough MOs to place all the electrons that populated AOs.

Specifically, for OP's example 4 spatial AOs of one atom can hold up to 8 electrons, so 4 spatial AOs of two atoms can hold up to 16 electrons. Now, if only 4 spatial MOs are built out of these 8 spatial AOs, they could hold only up to 8 electrons and there is no room for any of the possibly remaining 8 electrons.

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