1
$\begingroup$

I just want to make sure my basic understanding of this is correct.

When the internuclear distance is finite, the symmetric wavefunction of the electron is lower in energy because it experiences a potential energy minimum. This potential energy minimum comes from the fact that because the electron can be at both proton A and proton B (of a hydrogen+ ion), their is a repulsion between the two protons. Thus when you solve the Hamiltonian there is a negative potential energy that considers this repulsion.

The antisymmetric wavefunction is larger in energy because the electron is only found on either proton A or proton B. This is where I get confused! Its one electron, but it acts like there are two electrons so much so that one electron is at proton A and the same electron is at proton B at the exact same time. So essentially you can consider both of these scenarios as a neutral atom. If they are neutral atoms, there are no repulsion/attraction between the two 'atoms' and so there is no minimum in potential energy. Thus, the overall energy of the antisymmetric wavefunction is higher. So as they act as 'neutral atoms'?? they will fall apart easily because there is nothing holding them together - the molecule will dissociate.

Is this the correct theory?

$\endgroup$

2 Answers 2

4
$\begingroup$

Its one electron, but it acts like there are two electrons so much so that one electron is at proton A and the same electron is at proton B at the exact same time.

It sounds like you are still under the impression that an electron has a well-defined position, which is not the case here. Remember that in quantum mechanics, states of a single particle are associated to wavefunctions which are extended throughout space. The absolute square $|\psi|^2$ of the (so-called position space) wavefunction $\psi$ tells us the probability of finding the electron in a particular region, but we should not think of it as having a single, well-defined position.

I've shown $|\psi|^2$ for the symmetric and antisymmetric 1s states of the H$_2$ molecule below. Notice that the symmetric $|\psi|^2$ is larger in between the atoms, whereas the antisymmetric $|\psi|^2$ has larger peaks at the locations of the atoms. enter image description here

Intuitively speaking, the fact that the antisymmetric state is sharply peaked at the locations of the individual atoms and is small in the region in between could be loosely interpreted through a semi-classical lens as saying that the electrons spend more time near the individual nuclei and less time in the space in-between. As a result, the electrons in the antisymmetric state feel the attractive potential due to one nucleus or the other, but tend not to be in a position (i.e. the center) where they can strongly feel the potential due to both.

Conversely, an electron in the symmetric state "spends more time" in between the two, and so tends to feel the attractive potential due to both nuclei at the same time. Since the attractive potential is negative, this means that the average potential experienced by electrons in the symmetric state is more negative than the average potential experienced by electrons in the antisymmetric state.

At the same time, the average kinetic energy in a state $\psi$ is related to how rapidly $\psi$ changes - specifically, $$\langle KE \rangle_\psi = \frac{\hbar^2}{2m}\int |\nabla \psi(\mathbf x)|^2 \mathrm d^3x$$

Its easy to see that the antisymmetric wavefunction varies more sharply than the symmetric wavefunction, which means the average kinetic energy of an electron in the antisymmetric state is larger than the average kinetic energy of an electron in the symmetric state.

Putting these facts together, an electron in the antisymmetric state has more (positive) kinetic energy and less (negative) kinetic energy, which means that the total energy in the antisymmetric state is larger. As a result, the ground state of this system consists of two electrons (with opposing spins) each occupying the symmetric state. This is what we mean when we say that the two constituent hydrogen atoms "share" electrons.

In turn, the energy of this state depends on the separation distance between the two nuclei. It exhibits a well-defined minimum at a specific distance $d_0$. At distances larger than $d_0$, the (negative) potential energy contribution becomes weaker, and at distances smaller than $d_0$ the spatial variation in the wavefunction (i.e. the kinetic energy) becomes stronger. This equilibrium distance is the bond length of the H$_2$ molecule, which turns out to be approximately $d_0 = 7.4 \times 10^{-11}$ m.

$\endgroup$
0
$\begingroup$

It can be also seen as a question of energetic stability. For a molecule to be in a state where they share the electrons then it must be more energetic stable to be in a state where the electron is shared (attraction; symmetric wave-function; lower energy) with more probability to him to be in between the atoms, than to be in a state where the electron is more connected to the one of the atoms since in this case there is no stability whatsoever because the electron is no longer shared as much as in the other case. Thus, in the anti-symmetric case (higher energy; repulsion of the atoms) the molecule is no longer tightly bound as the symmetric state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.