3
$\begingroup$

Why is the pure tungsten ($\ce{W}$) melting point way higher than that of its oxides $\ce{WO3}$ or $\ce{WO2}$? And why does pure $\ce{Al}$ react differently? I.e. the oxides have a higher melting point than pure $\ce{Al}$.

$\endgroup$
4
  • $\begingroup$ Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the help center. Do you want to know more about this site, please take the tour. $\endgroup$ Feb 5, 2015 at 8:50
  • $\begingroup$ I think along with these, $\ce{W87O13}$ exists as well. $\endgroup$
    – M.A.R.
    Feb 5, 2015 at 13:38
  • $\begingroup$ The phase diagrams available at the ASM Alloy Phase Diagram Database (if your institution has access) show varying numbers of 'stable' phases. Most are likely various orderings on the oxygen sublattice (my guess). I can find no recent (last 20 years) work on the thermodynamics of the oxides. $\endgroup$
    – Jon Custer
    Feb 5, 2015 at 14:31
  • $\begingroup$ There is no reason to think that the oxide shold have a higher melting point. The bonding mechanism in the metal is totally different, you cannot reasonably compare their physical properties. $\endgroup$
    – Karl
    Jan 10, 2017 at 0:44

1 Answer 1

2
$\begingroup$

As far as I know,

$\ce{WO2}$ and $\ce{WO3}$ have a covalent characteristic, and thus lower melting point than the metal.

while, $\ce{Al2O3}$ have more ionic characteristic, and thus higher melting point than the metal.

$\ce{W}$ is a d-block metal from the $\ce{3^{rd}}$ row which make it possible to have covalent materials rather than ionic materials.

this article might prove this point http://pubs.acs.org/doi/abs/10.1021/jp963724z?journalCode=jpcbfk

$\endgroup$
1
  • 4
    $\begingroup$ Well, except that it doesn't explain why the titanium oxides, cobalt oxides, and chrome oxides, amongst others, have higher melting temperatures than their metals. Bottom line is: thermodynamics of solid phases may not be easy to predict based on simple heuristics. $\endgroup$
    – Jon Custer
    Feb 5, 2015 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.