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So molecular orbitals are represented as a weighted sum over atomic orbitals. Of course I reckon that the total sum of the weights per atomic orbital has to equal 1, when summed over all molecular orbitals. But why then should n atomic orbitals results in exactly n molecular orbitals? Is there a fundamental reason for this I am not seeing?

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marked as duplicate by Mithoron, A.K., a-cyclohexane-molecule, Todd Minehardt, Tyberius Jul 19 '18 at 21:13

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  • $\begingroup$ When you combine two orbitals you add and subtract them. $\endgroup$ – ringo Jul 19 '18 at 13:16
  • $\begingroup$ The notion is that two atomic orbitals combine to give you a molecular bonding orbital and a molecular antibonding orbital. $\endgroup$ – MaxW Jul 19 '18 at 13:22
  • $\begingroup$ Possible duplicate of The “rules” for LCAOs in Molecular Orbital Theory - fundamentally this arises because an $n \times n$ matrix has exactly $n$ eigenvectors (MOs) with associated eigenvalues (energies) $\endgroup$ – orthocresol Jul 19 '18 at 13:26
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The orbitals, both atomic and molecular, must be orthogonal, that is, they have no net overlap or correlation: equal constructive and destructive interference. You can think of them as forming the basis set of an n-dimensional vector space. If you think of the atomic orbitals as the standard basis...

Example for a 4-dimensional space:

$\langle 1,0,0,0 \rangle$, $\langle 0,1,0,0 \rangle$, $\langle 0,0,1,0 \rangle$, $\langle 0,0,0,1 \rangle$

...Then the molecular orbitals are just a rotated basis. In other words, they take linear combinations of the atomic orbitals to span the same space.
Ex:
$\langle \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \rangle$, $\langle \frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2} \rangle$, $\langle \frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2} \rangle$, $\langle \frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2} \rangle$

The same n-dimensional space requires at least the same number (n) of vectors to span it. This is exactly the same phenomenon as rehybridising individual s and p orbitals to form sp$^3$ orbitals.

You can find explicit formulae for the orbital vectors with the actual numbers (one possible representation), along with fantastic illustrations here Darmstadt, LCAO Theory.

There are two normalisation conditions when combining atomic orbitals to make either hybridised orbitals or molecular orbitals. The first is that every atomic orbital must be used completely. That is, the sum of its contributions to all the molecular orbitals must be 1. The other is that each of the new orbitals cannot have more than 100% orbital. Although I think that it is this second part that you were questioning the reasoning for.

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