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Be aware of Fermi level versus Fermi energy, as it seems to me the book speaks about the latter. The quotes below are from both Wikipedia links. In band structure theory, used in solid state physics to analyze the energy levels in a solid, the Fermi level can be considered to be a hypothetical energy level of an electron, such that at thermodynamic ...


5

This is actually a lot simpler than I initially thought. I'll be using the same example as previously explained in How to obtain the radial probability distribution function from a quantum chemical calculation? My test setup has slightly changed since then, i.e. I've tried it with Gaussian 16 B.01 and Multiwfn 3.6.0 and it worked the same. We start the same ...


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The reasons for the change in internuclear separation and the imporance of the Franck-Condon factors, as has been clearly pointed out in answers and comments. The FC factors determine the strengths of transitions from $M$ to $M^{+.}$ and to clarify this figure below shows a simple calculation based on harmonic oscillator wavefunctions of the effect of ...


4

Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state? It's not so much about whether the bond length in $\ce{M^.+}$ is larger or smaller than that in $\ce{M}$; it's more about whether ...


3

Can anyone explain why the exchange contribution to the total energy is negative? I find it misleading that exchange interaction is treated as something that changes total energy of the system. This lowering of energy is actually due to the Hartree-Fock scheme being in principle inexact, and is not really specific to indistinguishable particles. Let me ...


3

I'll try to give my interpretation of the "physical" explanation of why exchange would lower the energy. For the true wavefunction, the motion of all the electrons should be correlated, with the classical view being that the electrons are avoiding each other to minimize repulsion. With Hartree-Fock, we find an approximate wavefunction by solving for 1-...


3

I have been watching this question with keen interest, but wanted to let someone else go for the bounty -- However it expires tomorrow and no one has answered, so I will give my perspective. Why does the exchange interaction in Hartree-Fock theory lower the total energy? The answer is given in the question: This integral is always positive, and so ...


3

The only case in which electrons can be considered distinguishable is when they are so well separated that their wave functions do not overlap. Any electrons on the same atom or within the same molecule are considered indistinguishable. In chemistry we like to talk about electrons being in specific orbitals, but the indisinguishable nature of them means ...


2

The author of the presentation is talking about large, but not yet very large/infinite distance here. Note the annotations: $$ \frac{e^2}{4 \pi \epsilon R} = k/R $$ when $R \rightarrow \infty$: slowly to zero $$ S_{AB} $$ when $R \rightarrow \infty$: to zero $$ \left< 1s_A\left| k/r_A \right| 1s_B\right> $$ when $R \rightarrow \infty$: rapidly to ...


2

If you irradiate a metal with the exact amount of energy to produce a free electron with KE = 0, it will just stay free, in the "sea of electrons" of the metal, while the nucleus, with its positive charge, also sits in the same sea - near "other" electrons, each of which has KE due to temperature. Then, after a while, an electron drops in around that nucleus,...


2

It would have to be $\ce{Li^2+}$ or $\ce{He+}$ to have 1 electron only. And even for 1 electron only ions, energy levels depend on the kernel charge. For multielectron atoms/ions, energies of different orbitals with the same quantum number $n$ differ, due electron repulsion and kernel shielding. That is the reason why the orbital 4s is occupied before ...


2

Using ORCA and reverse engineering the calculation provided in the question (HF/STO-3G at a $\ce{H}$-$\ce{F}$ distance of 0.955464 Angstroem), I obtained the following output, to which I added the matrix product of the density and overlap matrices, dubbed the Mulliken Population Matrix. By comparing to the output of other programs, the reader can follow the ...


2

As pointed out in the comments the rigid rotating molecule only gains kinetic energy when it rotates not potential energy. The degeneracy describes the fact that some levels have exactly the same energy and this depends the value of the angular momentum rotational quantum number $J$. The number of degenerate levels is given by the multiplicity $2J+1$. The $...


2

I have two questions... Why does a molecule "gain potential energy" when it rotates? Does it want to stop rotating for some reason? What do degenerate energy levels correspond to physically in terms of the molecule's rotation? Do you remember the Newton's law of motion? A body in motion will always remain in motion until and unless there is a ...


1

Mulliken charges are not representative of charge transfer, I would argue, in any case. Mostly because of their enormous basis set dependency, as Geoff pointed out. In any case, several points are missing here; for instance whether the geometries were optimized with each method/basis set combination. This is not trivial. Assuming a single static geometry (...


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