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In solid, energy bands means the 1s orbital with spin up and down become 3 states when there are 3 atoms.

But then. If you look at hydrogen atoms, the allowed bound states have energies equal to En=Ry/(n2) and n is the main quantum number ranging from 1 to infinity. In other words, there is an infinite number of available electronic states in a single hydrogen atom.

When we talk about energy bands. Which of the above is usually referred to? The few states orbitals or the almost infinite allowed bound states?

For the latter, how do you characterize the energy bands in liquid like water? What kind of bands are formed in such cases?

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    $\begingroup$ Liquids, and especially water, are a bad example for discussing band structure of semiconductors. $\endgroup$ Dec 15 '20 at 5:10
  • $\begingroup$ No i chose semiconductor bec i cant find the energy band tag. But my question is about energy bands of liquid. What tag must i use? I just changed the tag to liquid $\endgroup$
    – Jtl
    Dec 15 '20 at 5:28
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When we talk about energy bands, we are talking about mixing of orbitals on different atoms (AOs) with similar energies. First, assume simple models. In the solid, the structure is very regular and all atoms have AOs with identical matching energies. This makes the number of orbitals available for mixing in solids very large, and mixing results in bands. In a liquid, on the other hand, structure is highly irregular. Orbitals on different atoms often have energy or overlap mismatches, but occassionally match and mix, and this results in much local mixing (say due to hydrogen bonding) but not extended delocalized orbitals. Because of energy mismatch you can also end up with bands in liquids but their origin (in this model) is different than in solids.

The difference between the origin of bands in liquids and solids is similar to the different cause of homogeneous and inhomogeneous broadening. In both cases you get bands (a distribution of energies), but in one case broadening involves delocalization over resonantly matched orbitals on different atoms/molecules and in the latter it involves variations in the local environment due to randomly fluctuating perturbations caused by neighboring molecules ("the solvent"). Extensive orbital mixing and Pauli's exclusion principle results in homogeneous broadening in the first case. Motional jostling in the case of a liquid results in inhomogeneous broadening.

Energy levels of individual molecules in liquid water are "perturbed" by neighboring molecules. These perturbations are small compared to the ionization energy of the valence electrons. The stability conferred by an H-bond in water is in the order of 20 kJ/mol. Compare that to the ionization energy, ~1200 kJ/mol. Therefore electrons remain bound to local waters even as their energies are slightly perturbed by interactions with other molecules. H-bonds do contain covalent character, that is, involve sharing of electrons between two molecules (specifically between H-bond donor and acceptor atoms). However this does not imply electrons in liquid water are extensively delocalized beyond local bonds, certainly not on the spatial scale possible in solids.

As a note of interest ice under pressure exhibits metallic phases.

Note perturbations due to the solvent are evident in the difference between vibrational spectra obtained in different phases.

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    $\begingroup$ What if there is no solvent. Let's just take the case of pure H2O. We mostly have hydrogen bonding, how do the electrons behave in water? That's what I wanted to know in the original question. $\endgroup$
    – Jtl
    Dec 15 '20 at 10:49
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    $\begingroup$ I included some additional comments. In pure H2O the solvent can be regarded as waters surrounding one molecule being inspected. $\endgroup$
    – Buck Thorn
    Dec 15 '20 at 20:41
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    $\begingroup$ I think this makes it seem that there are very different reasons for the electronic structure of solids and liquids - there aren't, within an orbital view of the world they both exhibit bands ultimately due to same thing, and those bands can be found by exactly the same method - solving an appropriate approximation to the Schrodinger equation. Yes the topological disorder does result in differences in the solutions in certain cases, for instance Anderson localisation en.wikipedia.org/wiki/Anderson_localization, but the effective Hamiltonian is the same. $\endgroup$
    – Ian Bush
    Dec 16 '20 at 16:18
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    $\begingroup$ @IanBush I am attempting to convey the difference between a regular extended bonding network in a solid involving relatively stable bonding as opposed to continuously rearranging bonds in a liquid. Bands in a solid are basically regarded as due to highly delocalized orbitals rather than localized covalent bonds. In a liquid the energy is altered not only by bonding but also due to perturbations by other (not necessarily bonding) molecules which do not share orbitals. $\endgroup$
    – Buck Thorn
    Dec 16 '20 at 17:44
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    $\begingroup$ OK - will do, prob at the weekend now. But note in the same AO basis the Hamiltonian matrix for a solid and a liquid at the same density will have approximately the same sparsity, Translational symmetry means that there exists a basis within which the Hamiltonian becomes block diagonal, and that is related to the original basis by a unitary transformation. And as only the solid has translational symmetry this new basis only exists in that case. In any other basis, almost certainly including an AO based one, the matrices will look very similar. $\endgroup$
    – Ian Bush
    Dec 17 '20 at 9:02

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