# How to derive the total number of molecular orbitals based on the given number of atomic orbitals?

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?

• If you have eight orbitals to start, you will have eight orbitals to finish. Nov 28 '16 at 4:24
• Can you give me an intuitive explanation as to what's going on? Nov 28 '16 at 7:09
• Technically speaking, each individual atom has infinitely many orbitals for its electrons, most of which are unoccupied. Infinity + Infinity = Infinity. Nov 28 '16 at 12:30

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.
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?