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There's a graph of the melting points of period three oxides. The melting point of magnesium oxide is several hundred Kelvin higher than aluminiumoxide. I can't find any explanations for this on the Internet or in any of my A-level textbooks — perhaps this is something more advanced I haven't come across.

The reason I am questioning the value of its melting point is I'd expect aluminium oxide to be higher. Aluminium is a 3+ ion whereas magnesium is only 2+. Aluminium is also a smaller ion so the effective electrostatic attraction on the valence electrons should be greater. I'm also pretty sure aluminum oxide has greater covalent charterer than magnesium oxide. All of these would lead me to believe it has a higher melting point, so what is the explanation for it being less?

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    $\begingroup$ You could apply Kapustinskii's Equation to obtain the lattice energy and see if that provides an explanation. $\endgroup$
    – Tyberius ♦
    Apr 15, 2017 at 19:38
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    $\begingroup$ For what it's worth, based on your plot of melting points for these 3rd row oxides, MgO appears to fall right in line and $\ce{Na2O}$ seems to be the odd-ball. $\endgroup$
    – airhuff
    Apr 15, 2017 at 19:46

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Two ideas to consider:

  1. Alumina is not fully ionic. Neither is magnesia, but magnesia has more ionic character and we may see the greatest amount of ionic attraction in magnesia versus alumina which is less ionic, or soda which has only singly charged cations.

  2. When metal oxides are melted they do not necessarily produce free metal and oxide ions. As described by Shi et al. [1], molten alumina retains mostly four-and five-coordination of oxygen to aluminum, rather than forming "free" aluminum and oxide ions and also contrasting with the six-coordination of aluminum in the solid conundrum phase. Presumably magnesia would behave similarly to alumina when melted, but the greater ionic bonding character in magnesia makes reduced coordination less favorable and thus enhances retaining the fully octahedrally-coordinated solid phase.

Reference

1. Caijuan Shi, Oliver L. G. Alderman, Diana Berman, Jincheng Du, Joerg Neuefeind, Anthony Tamalonis, J. K. Richard Weber, Jinglin You and Chris J. Benmore, "The Structure of Amorphous and Deeply Supercooled Liquid Alumina", Front. Mater. 6:38 (19 March 2019), https://doi.org/10.3389/fmats.2019.00038.

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  • $\begingroup$ If magnesium oxide has more ionic character shouldn't that make it's melting point lower. I would expect a greater covalent charecter to increase the melting point. Also your idea about polyatomic species - would enough of these be present in molten aluminium oxide before it's reached it's quoted melting point? $\endgroup$
    – Michael Harding
    Apr 16, 2017 at 13:12
  • $\begingroup$ You have to break some ionic bonds. Solid magnesia has six of each atom/ion surrounding an opposite counterpart. Melted magnesia will have less coordination. $\endgroup$
    – Oscar Lanzi
    Apr 16, 2017 at 13:29
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Here is a simple minded answer. High melting point compounds tend to follow the octet rule - the number of valence electrons sums to 8. The Group IVB carbides (TiC, ZrC, HfC) have an 8 electron sum (4+4) and have very high melting points - for HfC it is approximately 7100F. The Group IIIB nitrides and phosphides have an 8 electron sum (3+5) and are the highest melting point compounds among the rare earth metals. The Group IIA metals also follow this pattern with MgO (2+6) as the prime example. While these compounds all have the rock salt structure, the same valence count pattern extends to other crystal structures, such as with the monoborides of the Group VB metals - TaB (5+3) does not have the NaCl structure (atom sizes fall outside of the Haag rule) yet has the highest melting point in the Ta-B system. Following this trend, Al nitride and phosphide (3+5 electrons) also follow this pattern - they are the highest melting Al compounds. Why does this pattern exist? I don't know except to say that at this magic 8 valence electron sum, the electronic structure is optimized. In density of states calculations, that valence electron sum fully fills the "bonding states" and leaves the "anti-bonding states" unpopulated, and the Fermi level occurs at a deep well. I am not a solid state physics expert, so this is about as far as my explanation goes.

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