# Covalent bonding in cuprates

In high temperature cuprate superconductors like YBCO, there are intermediate copper-oxide planes, where $$Cu$$ and $$O$$ atoms are arranged alternatively in a square lattice. In this arrangement, the $$Cu$$ atoms generally make 4 in-plane and 2 axial bonds, and each $$O$$ makes 2 in-plane bonds (see attached image).

Oxygen's electron configuration $$[He]2s^2 2p^4$$, and can indeed form two bonds. However, copper has the electronic configuration $$[Ar] 3d^{10} 4s^1$$ which in the crystal field of the lattice becomes $$[Ar] 4s^2 3d^9$$, with the $$3d_{x^2-y^2}$$ orbital half-filled.

What hybridisation does the $$Cu$$ orbitals undergo to be able to form four in-plane bonds. To form 6 bonds, it seems that the five 3d and one 4s orbitals will hybridise, but the $$3d_{x^2-y^2}$$ orbital is well separated from the other 3d orbitals in energy.

• I'm voting to close this question as off-topic because it's a good question but one that belongs in Chemistry SE. – Gert Sep 2 '19 at 20:09
• @Gert Typical comment here on questions about solid-state physics. There is a very unwelcoming attitude here. – Pieter Sep 2 '19 at 20:16
• Agree that this is solid state physics. Not quite sure how crystal electonic structure gets lumped with chemistry... – Jon Custer Sep 2 '19 at 20:22
• @Gert - wait, crystal bonding is now ‘quantum chemistry’? If it were just fcc copper would that still be true? Why didn’t Ashcroft and Mermin ever get told this isn’t Solid State Physics? – Jon Custer Sep 3 '19 at 0:18
• @Gert - fire held! I just see too many good materials/semiconductor physics questions downvoted or closed around here (well, and lots of bad ones too - they can go away). Molecules? I'm fine with them on Chemistry. Crystalline solids? They should be here on Physics. – Jon Custer Sep 3 '19 at 13:46

The square-planar coordination is typical for a Cu(II) ion. Also when there are six ligands which could form an octahedral crystal field, the Jahn-Teller effect on a $$3d^9$$ configuration generally leads to the hole occupying the $$3d_{x^2-y^2}$$ orbital, and the bonds in the $$z$$-direction becoming longer.