Predicting Fermi Surface from Chemical Formula

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SUMMARY

The discussion centers on predicting the Fermi surface of the ionic compound LaFeAsO based on its chemical formula. Participants clarify the oxidation states of the elements: La (3+), Fe (3+), As (3-), and O (2-), which influence the band structure. The primary contributor to the band crossings is identified as Fe due to its partially filled 3d orbitals, specifically 3d5 for Fe 3+. The conversation emphasizes the importance of understanding valence orbitals and their occupancy in predicting electronic properties.

PREREQUISITES
  • Understanding of oxidation states in inorganic chemistry
  • Familiarity with band theory and electronic structure
  • Knowledge of ligand field theory
  • Basic concepts of solid-state physics
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  • Study the principles of band theory in solid-state physics
  • Learn about ligand field theory and its applications in predicting electronic properties
  • Explore the role of oxidation states in determining chemical bonding and structure
  • Investigate the electronic configurations of transition metals in ionic compounds
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Solid-state physicists, inorganic chemists, and materials scientists interested in predicting electronic properties of ionic compounds and understanding band structure.

tylerscott
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Hi, I was hoping I could get some things cleared up. Recently my Solid State professor mentioned that we could simply, from the chemical formula, predict where the band crossings are going to be. For example, take LaFeAsO. Since La has a valency of +3, Fe of +3, As of -3, and O of -2, he deducted from the lattice (which is more or less cubic), that the band crossings are going to be primarily from the Fe. However, I'm completely lost as to why. Any help is appreciated!
 
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This is a mainly ionic compound. Did you pay attention in your chemistry class? Then you should be able to name the valence orbitals of these elements and their occupation in the ions. For which element in your list aren't they neither full nor empty?​
 
So, La is 5d1, Fe is 3d6, As is 4p3, and O is 2p4. So Fe is close to half filling. But, so is As and O. So can we just say the bandstructure will compose of mostly Fe because of the half filling? Therefore, it is highly metallic and also the binding energy of the 3d6 should be much smaller?
 
You didn't count well. Take in mind that these are ions.
 
So Fe is 3d8. La is 5d3. On the right track?
 
No. Neutral La has 5d1 6s2, so La 3+ has none. Neutral Fe has 3d6 4s2, so what do you get for Fe 3+?
 
I'm afraid I'm quite rusty in my chemistry here. So, Fe 3+ would be 3d5. However, I'm not sure where the 3+ for La is coming from (unless it's just because its a common oxidation state). How can one assume that the most common oxidation state is the right one to choose?
 
Solid state physics is in some respect inorganic chemistry specialized to molecules of infinite size. So you better polish your inorganic chemistry.
Lanthanum has almost exclusively oxidation state 3+ just like oxygen mostly 2-. Iron mostly 2+ or 3+ and arsenic 3-, 3+ or 5+.
The only reasonable combination to make this compound neutral is thus O 2-, As 3-, La 3+ and Fe 3+. Hence the valence shells on O and As are completely filled while the one of La is completely empty. If you form bands from these atomic states, they will be filled or empty, too. This leaves as the main suscpect only Fe whose d-orbitals are only partially filled and partially degenerate in a cubic environment (cf Ligand field theory).
 

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