What is the Lewis structure for $\ce{MgB2}$? I just read that it is a superconducting material (up to a relatively high critical temperature, 39 K) and am wondering if this has to do with its electronic structure.

In $\ce{MgB2}$, $\ce{Mg}$ has 2 valence electrons and each $\ce{B}$ has 3 for a total of 8. I know that $\ce{B}$ is "satisfied" with 6 valence electrons, but I don't see how to get 8 around $\ce{Mg}$ with 6 on each $\ce{B}$.


It's not really a molecule, so you don't really have a Lewis structure consisting of one magnesium and two boron atoms. For a lattice structure see Wikipedia. The lattice structure is hexagonal with alternating magnesium and boron layers, the boron layers having twice as many atoms per unit area as the magnesium layers.

The referenced picture is given at https://en.m.wikipedia.org/wiki/File:Magnesium-diboride-3D-balls.png and consists of green balls (magnesium) plus pink balls (boron), the latter covalently linked with each other. If we model the magnesium as $2+$ ions (yes, the electronegativity difference is only 0.8, but de la Mora at al. indicates that the magnesium-boron bonding is predominantly ionic), then each boron atom has one negative charge and so becomes isoelectronic with carbon. Then each of the hexagonal, covalently bonded boron layers is isoelectronic and isostructural with a graphite layer.

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    $\begingroup$ Thank you, I'll review the Wikipedia page with your comments in mind. I appreciate the quick response. $\endgroup$ – Steve Schmidt Aug 10 '18 at 1:47
  • $\begingroup$ @SteveSchmidt also check out the Wikipedia article on borane clusters. $\endgroup$ – A.K. Aug 11 '18 at 3:56

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