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I am an organic chemistry student learning how atomic orbitals interfere to give rise to molecular obitals. The image below suggests that each of the hydrogens' atomic orbitals interfere both constructively AND destructively. How is this possible? More fundamentally, how does the 1s atomic orbital of a hydrogen atom have both positive and negative phase in the first place? I realize that there is a similar question on this topic but I wasn't able to understand the answer to that post due to my limited background in quantum mechanics and was hence was looking for someone who could explain in simple terms.

enter image description here

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    $\begingroup$ As there are two atomic orbitals (a0) to begin with there are two MO's, we make the hypothesis that they are made up as linear combinations $ao_1\pm ao_2 $ as you have drawn. The total energy is the same as the two ao's, one MO up in energy the other lower, which makes a difference when only two electrons are added but not when four are. Most phys. chem. textbooks explain this in detail but that is the main idea. $\endgroup$ – porphyrin May 10 at 12:25
  • $\begingroup$ While this does not explain why the orbitals combine that way, it might still be helpful if you take in mind that the molecule is hold by only one. In this sense the molecule is not formed by both constructive and destructive interference, that concerns the orbitals as functions, which include l a phase like in the sketch. $\endgroup$ – Alchimista May 10 at 13:23
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Part of the issue is that you're treating orbitals as objects. They aren't objects, they are quantum mechanical wavefunctions. The electrons which occupy the orbitals are objects, but the orbitals themselves are decidedly not objects.

When we talk about quantum mechanical states, the overall phase of an individual state has no meaning. It is a fundamental postulate of quantum mechanics that the wavefunction $\psi(x)$ and the wavefunction $k\cdot\psi(x)$ represent exactly the same state, so the phase $k$ is completely meaningless. 1s orbitals are states as well, so it is meaningless to talk about the phase of a 1s orbital.

However, if you have a state that is made up of two different states (like a molecular orbital), then the relative phases of the two constituent orbitals matter. ​That's why the bonding and antibonding MOs are different: it's because the two constituent parts don't have the same relative phase in the two MOs (in the bonding it's the same, in the antibonding they're opposite).


There are very few, and possibly no, quick and dirty ways to explain MO theory, in my opinion. However, if you have studied linear algebra before,* then the transition from AOs and MOs can be viewed as a change of basis. A basis refers to a minimal set of vectors, from which any arbitrary vector may be constructed through linear combination. If you have a vector $(a, b)$, you can express this as: $$\begin{pmatrix}a\\b\end{pmatrix} = a\begin{pmatrix}1\\0\end{pmatrix} + b\begin{pmatrix}0\\1\end{pmatrix}.$$

It doesn't matter what $a$ and $b$ are, you can always express $(a, b)$ as a linear combination of the two basis vectors. You can think of the basis vectors $(1, 0)$ and $(0,1)$ as being the 1s orbitals on both individual atoms.

The MOs correspond to using the basis $(0.5, 0.5)$ and $(0.5, -0.5)$.† Notice that this is still a valid basis, because we can still express any arbitrary vector $(a, b)$ as a linear combination of these:

$$\begin{pmatrix}a\\b\end{pmatrix} = (a+b)\begin{pmatrix}0.5\\0.5\end{pmatrix} + (a-b)\begin{pmatrix}0.5\\-0.5\end{pmatrix}.$$

If we had rejected the "antibonding" combination $(0.5, -0.5)$, and only taken the "bonding" combination $(0.5, 0.5)$, then we would not be able to form a valid basis. For example, we could not express the vector $(1, 2)$ as just a linear combination of $(0.5, 0.5)$.

From this point of view, AOs and MOs are simply different bases.‡ The existence of both the bonding and antibonding MO is not only logical, but also mandatory in order to form a complete basis.

The familiar AOs that you have studied (1s, 2s, 2p, ...) are simply a series of basis states for the hydrogen atom. In the $\ce{H2}$ molecule, there are two sets of basis states, one for each atom. The transition from "2 sets of AOs" to "1 set of MOs" is just a change of basis: we go from two sets of atomic bases to one single molecular basis.


The proper answer is, of course, to pick up a QM textbook and study it. By QM I do indeed mean quantum mechanics, not just MO theory. It is a long journey, and you may not necessarily be fully prepared for it right now, but it will reward you with a better understanding of how this works. MO theory fundamentally relies on quantum mechanics, and to learn MO theory without understanding quantum mechanics is what leads to confusion and questions like yours.


* The analogy I use above is slightly simplified, and might look slightly contrived, but the link between QM and linear algebra is actually very deep. QM is somewhat analogous to linear algebra but on an infinite-dimensional vector space (technically, a Hilbert space).

† Technically this should be $1/\sqrt{2}$, not $0.5$. In QM we are nearly always concerned with orthonormal bases, i.e. ones where each basis vector has a length of $1$ and forms a scalar product of $0$ with every other basis vector. (Mathematically, we have a basis $\{|i\rangle\}$ for which $\langle i | j \rangle = \delta_{ij}$.)

‡ "Bases" being the plural of "basis".

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  • $\begingroup$ Good explanation. I had a rough time learning about SALC's and AO -> MO transitions, the identity tables etc - but I had never seen the maths before. After learning what linear combination actually IS, and what it basically DOES - it was much easier to understand what is going on. A learning that happened to me much later in life, but was very appreciated in the raw "A-HA!" value. $\endgroup$ – Stian Yttervik May 11 at 12:00
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Unfortunately, in quantum mechanics there is rarely an explanation "in simple terms." Quantum mechanics is a mathematical construct that so far seems to predict the results of all experiments that have been done to test it, but explaining those maths in terms of anything we have learned from the larger-than-quantum scale "classical" systems has so far eluded us. In fact, attempting to understand QM by applying intuition from classical physics generally causes more confusion, as seems to be the case here.

It seems that you are considering the two H-atom 1s orbitals to be static standing wave systems that can only combine constructively or destructively, which would be the case if these were classical standing waves. But they aren't. They're quantum mechanical systems that live by different rules. When the two atoms come together, the whole system is changed, and the wave function describing the new system is not just an overlay of the previously separated "waves". Conceptually, it may be helpful to think of the new system as formed by constructive and destructive interference, but only in the vaguest conceptual sense. Pushing that metaphor too far gets you into the confusion you are in now.

As unsatisfying as it will be, it might be best at this point to accept that the conceptual description is qualitatively consistent with the mathematics and so is a guide to intuitively constructing molecular orbitals from atomic orbitals, but it isn't an explanation of why?. If you continue farther in your study of quantum chemistry, you'll learn how to construct these molecular orbitals more quantitatively, and you'll get better answers of how the math works to give certain answers, but you'll never get a good answer of why.

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