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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).

Copper-oxide planes in high Tc superconductors

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.

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  • $\begingroup$ I'm voting to close this question as off-topic because it's a good question but one that belongs in Chemistry SE. $\endgroup$ – Gert Sep 2 '19 at 20:09
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    $\begingroup$ @Gert Typical comment here on questions about solid-state physics. There is a very unwelcoming attitude here. $\endgroup$ – Pieter Sep 2 '19 at 20:16
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    $\begingroup$ Agree that this is solid state physics. Not quite sure how crystal electonic structure gets lumped with chemistry... $\endgroup$ – Jon Custer Sep 2 '19 at 20:22
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    $\begingroup$ @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? $\endgroup$ – Jon Custer Sep 3 '19 at 0:18
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    $\begingroup$ @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. $\endgroup$ – Jon Custer Sep 3 '19 at 13:46
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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.

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