My confusion arises from the fact that bonding and antibonding orbitals are created simultaneously. I have read this answer that clarifies about the existence of the molecular orbitals itself.

But now I want to know what does it mean for an electron to be in a bonding or antibonding orbital. Are phases a property of the electrons or of the orbitals? If it is a property of the orbital itself, then how are they constructive and destructive at the same time? If it is a property of the electron, how can a single electron occupy an orbital and still be antibonding/bonding, even though it is not comparing phase difference with anything?

What if the electrons of the two atoms forming the bond have the same spin? Will one electron occupy bonding orbital and the other the antibonding orbital? Or will the covalent bond not be formed at all?

I believe I've understood the mathematical reason for this involving orbital functions (only the square of the function is physically significant), but I want it to understand the physical implications of that on the electrons

  • $\begingroup$ I admit is not a great piece but does this comment which I took from the thread you have mentioned says something to you? “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
    Nov 27, 2021 at 13:02
  • $\begingroup$ @Alchimista I'm not sure if I understand it. Are you talking about how only the bonding orbital is usually occupied in the molecule? $\endgroup$
    – Balu
    Nov 27, 2021 at 13:30
  • $\begingroup$ yes, there is nothing simultaneously happening. $\endgroup$
    – Alchimista
    Nov 27, 2021 at 15:54
  • $\begingroup$ @Alchimista Well, that's the usual scenario. But I've read that it's even possible for scenarios like $\ce{H2-}$ ion to exist with 2 electrons in the bonding orbital and 1 in the antibonding orbital $\endgroup$
    – Balu
    Nov 27, 2021 at 15:59
  • $\begingroup$ how this relates to the question? $\endgroup$
    – Alchimista
    Nov 27, 2021 at 18:44

1 Answer 1


I do not understand your question, so the best that I can do is show you how I think about orbitals. Broadly speaking, we want to minimize the energy of our total electronic ground state wavefunction $\Psi$. This condition leads ultimately to a set of molecular orbitals with fixed phases and the atomic orbitals serve as input for this procedure. I'll try to give you a rough idea of the procedure in the following.

The wavefunction itself can be expressed as a linear combination of products of molecular orbitals. $$ \Psi(r_1,r_2) = \sum_{ij} c_{ij}\psi_i(r_1)\psi_j(r_2) $$ The $\psi_i(r),\psi_j(r)$ are chosen from the set of molecular orbitals. This set includes the bonding and antibonding molecular orbitals. These molecular orbitals are themselves constructed by linear combination of atomic orbitals that we know from solving the hydrogen atom. We usually start with a set of atomic orbitals based on the atoms in our molecule and this atomic orbital basis must provide at least one orbital per electron. Lets limit ourselves to a minimal basis set of atomic orbitals. We have 2 Hydrogen atoms so only 2 electrons in total. We use the $s$-orbitals of both atoms as our atomic basis. Lets call the $s$-orbital which is centered on hydrogen atom A, $s_A(r)$ and the other $s$-orbital, centered on atom B, $s_B(r)$. Our atomic basis is thus the set with two elements, $\text{Atomic orbital basis} = \{s_A(r), s_B(r)\}$.

The problem that we want to solve is finding the electronic groundstate of our hydrogen molecule H2. This problem gives us equations and conditions for the optimal orbitals, the so called molecular orbitals. Covering the details would go to far. But we can use for starters the so called Hartree-Fock method to determine which orbitals would give us the lowest energy. When you solve the Hartree-Fock equations you obtain the set of molecular orbitals. Using our minimal atomic orbital basis as input to the HF-calculation, we would obtain the following spatial molecular orbitals, $$ \psi_1(r) = \frac{1}{\sqrt 2} (s_A(r)+s_B(r))\\ \psi_2(r) = \frac{1}{\sqrt 2} (s_A(r)-s_B(r))\\ $$ These functions form the $\text{molecular orbital basis} = \{\psi_1,\psi_2\}$. Up to this point I have neglected spin. The orbitals that I have constructed so far do not include spin. Without going into detail we can again use these spatial molecular orbitals to construct spin-orbitals, that include spin. To do so, we simply multiply the spatial orbital with a spin up or spin down function,

$$ \phi_1(r,\omega) = \alpha(\omega)\psi_1(r)\\ \bar \phi_1(r,\omega) = \beta(\omega)\psi_1(r)\\ \phi_2(r,\omega) = \alpha(\omega)\psi_2(r)\\ \bar \phi_2(r,\omega) = \beta(\omega)\psi_2(r)\\ $$ and lets update my Ansatz from above to include these spin-orbitals

$$ \Psi(r_1,r_2) = \sum_{ij} c_{ij}\phi_i(r_1,\omega_1)\phi_j(r_2,\omega_2) $$

Our set of molecular spin-orbital basis functions is thus $\{\phi_1(r,\omega), \bar\phi_1(r,\omega), \phi_2(r,\omega), \bar\phi_2(r,\omega) \}$ and the indices $i,j$ move over this set.

The electronic singlet groundstate is then obtained by taking the determinant of the product of the lowest energy molecular spin orbitals, where we have to include one orbital for each electron and each spin-orbital may only appear once.

$$\begin{aligned} \Psi(r_1,\omega_1, r_2,\omega_2) &= \frac{1}{\sqrt 2}\det \left[\phi_1(r_1,\omega_1)\bar \phi_1(r_2,\omega_2) \right] = \frac{1}{\sqrt 2} \left[\phi_1(x_1)\bar\phi_1(x_2) - \bar\phi_1(x_1)\phi_1(x_2)\right]\\ &= \frac{1}{\sqrt 2}\psi_1(r_1)\psi_1(r_2)[\alpha(\omega_1)\beta(\omega_2)-\beta(\omega_1)\alpha(\omega_2)]\\ \Psi(r_1,r_2) &\propto \psi_1(r_1)\psi_1(r_2) \end{aligned}$$ Lets neglect spin again for a second and look only at the spatial parts. You can see that this groundstate function only includes the spatial orbital $\psi_1\propto s_A+s_B$.

We can compare the energy of this wavefunction with a simple ansatz where we keep the electrons in the atom centered orbitals $$ \Psi_{\text{non-bonding}}(r_1,r_2) \propto s_A(r_1)s_B(r_2) $$ This would describe two Hydrogen atoms that are not interacting and each electron is always located on a single hydrogen atom. Unlike the actual ground state where each electron is in a binding spatial molecular orbital, a spatial superposition state with equal probability for an electron to be found on either hydrogen atom and in-between both atoms, due to the positive sign/constructive interference of the underlying linear combination of atomic orbitals.

After this very lengthy exposition I'd like to address bonding and antibonding molecular orbitals. You saw that we obtained two spatial molecular orbitals, $\psi_1$ and $\psi_2$. The antibonding orbital $\psi_2$ is a mathematical necessity when we construct the set of spatial molecular orbitals from the set of atomic orbitals. The whole procedure is just a change of basis sets and such a change does not change the number of basis functions. We used the atomic orbital basis $\{s_A, s_B\}$ to construct our new molecular orbital basis $\{\phi_1, \phi_2\}$. You will always have the same number of functions in both sets. The coefficients that define the new molecular basis functions in terms of old atomic basis functions are fixed by the condition to minimize the energy of the final total groundstate wavefunction and the equations that follow from this condition.

If you truly want to understand orbitals I recommend that you try to get a good understanding of linear algebra, especially basis sets and how a change of basis works, then you need some basic quantum mechanics especially the hydrogen atom and what happens when you go from a single fermion system to systems with multiple fermions. Then you should take a closer look at the Hartree Fock method. Qualitative hand wavy explanations will always leave some confusion and not a quantitative understanding. My understanding of orbitals improved by going through the math. All the qualitative explanations that I had gotten before typically only caused more confusion.

  • $\begingroup$ Thanks, this was helpful to some extent. My question was basically about what is the physical significance of the phases and the interactions between them. And I'll look into the things you've mentioned, I've already started reading a book on the basics of quantum mechanics $\endgroup$
    – Balu
    Nov 27, 2021 at 16:15
  • $\begingroup$ @Balu phase has no direct physical meaning (similar to the wavefunction itself). But it allows for interference, as observed. By the way I do not see how this answer above can be useful to you but at least its author said that at the very beginning. $\endgroup$
    – Alchimista
    Nov 27, 2021 at 19:32
  • $\begingroup$ @Alchimista Yeah, I meant the physical implications. Like wave function can be used for probability density. The above answer didn't answer my question, but it helped me understand the mathematical part better $\endgroup$
    – Balu
    Nov 28, 2021 at 19:10
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    $\begingroup$ @Balu and at least the phase account for interference without affecting spatially the distribution of probability for the base orbitals. I know it is not high level, but it is the implication of phase. $\endgroup$
    – Alchimista
    Nov 29, 2021 at 8:21

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