I'm looking at the melting temperature of metallic elements, and notice that the metals with high melting temperature are all grouped in some lower-left corner of the $\mathrm{d}$-block. If I take for example the periodic table with physical state indicated at $\pu{2165 K}$:

enter image description here

I see that (apart from boron and carbon) the only elements still solid at that temperature form a rather well-defined block around tungsten (which melts at $\pu{3695 K}$). So what makes this group of metals melt at such high temperature?

  • $\begingroup$ Look at Carbon and Boron and think again. $\endgroup$
    – Georg
    Commented Apr 29, 2012 at 7:20
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    $\begingroup$ @Georg sorry but I don't find your comment very useful… could you be any more explicit? I know about carbon and boron, and I'm asking about metals… $\endgroup$
    – F'x
    Commented May 5, 2012 at 12:55
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    $\begingroup$ @F'x, I really like your question. It's hard! Georg, I'm not quite sure of your direction either, but the very tight covalent bonding of boron and carbon is so different from the metallic bonding typical of all metals that I'm pretty sure the key insight to this one lies elsewhere. I would also observe that atomic density (not quite the same as mass density) likely plays an important role, as the island of metals you have pointed out contains (I think) some of the most atoms-per-cc dense metals. Does that very cool tool you are using (?) by any chance show that also? $\endgroup$ Commented May 9, 2012 at 22:43
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    $\begingroup$ @TerryBollinger there is a flash tool rsc.org/periodic-table $\endgroup$
    – Hauser
    Commented May 10, 2012 at 23:22
  • $\begingroup$ A pure metallic bond makes rather soft and malleable things like sodium or copper. The metals in that central area (tungsten and co) do have some localized bonds mixed with metallic bond! Think of the W-Type structure! A real metal should have some closest packing, nothing else. $\endgroup$
    – Georg
    Commented May 15, 2012 at 21:26

2 Answers 2


Some factors were hinted, but let me put them in an order of importance and mention some more:

  • metals generally have a high melting point, because metallic interatomic bonding by delocalized electrons ($\ce{Li}$ having only a few electrons for this "electron sea") between core atoms is pretty effective in those pure element solids compared to alternative bonding types (ionic $\pu{6-20 eV/atom}$ bond energy, covalent 1-7, metallic 1-5, van-der-Waals much lower). Also, ionic lattices like $\ce{NaCl}$ have a higher lattice and bonding energy, they have weak interatomic long-range bonding, unlike most metals. They break apart or are easily solvable, metals are malleable but don't break, the electron sea is the reason for their welding ability.

  • the crystal structure and mass play an inferior role among your filtered elements (just look up the crystal structure of those elements), as metallic bonding is not directional unlike covalent bonding (orbital symmetry). Metals often have half filled $\mathrm{s}$ and $\mathrm{p}$ bands (stronger delocalized than $\mathrm{d}$ and $\mathrm{f}$) at the Fermi-edge (meaning high conductivity) and therefore many delocalised electrons which can move into unoccupied energy states yielding the biggest electron sea with half or less fill bands.

  • noble metals like $\ce{Au,Ag}$ have a full $\mathrm{d}$ orbital, therefore low reactivity/electronegativity and are often used as contact materials (high conductivity because of "very fluid" electron sea consisting only of $\mathrm{s}$-orbital electrons. Unlike tungsten with half or less occupied $\mathrm{d}$-orbitals they show no interatomic $\mathrm{d-d}$ bonding by delocalized $\mathrm{d}$-electrons, and more importantly, a half filled $\mathrm{d}$-orbital contributes 5 electrons to the energy band, while a $\mathrm{s}$ only 1, $\mathrm{p}$ only 3, the electron sea is bigger among the $\mathrm{d}$-group.

  • The "packaging" of core atoms in the lattice (interatomic distance) among the high $Z$ atoms (compared to e.g. $\ce{Li}$) is denser (more protons, stronger attraction of shell electrons, smaller interatomic radius), means stronger interatomic bonding transmitted by the electron sea:

enter image description here

You can see here that in each series ($\ce{Li,\ Na,\ K}$) the melting points rise to a maximum and then decrease with increasing atomic number (lacking unoccupied energy states for delocalized $\mathrm{d}$-electrons), bigger electron sea being here a stronger factor than a bit more dense packaging.

  • Boron as a semi-metal shows metallic and covalent bonding, Carbon strong directional covalent bonding and is able to build a network of bonds unlike other non-metal elements showing covalent intramolecular bonding, e.g., in diatomic molecules but not strong intermolecular bonding in macromolecules because of lacking unpaired electrons.

So there are some bigger trends for melting points explaining the high melting points of $\mathrm{d}$-metals, but also some minor exceptions to the rule like $\ce{Mn}$.

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    $\begingroup$ Actually " shows metallic and covalent bonding" is rather inaccurate description to highly non-trivial bonding in elementary boron. $\endgroup$
    – Mithoron
    Commented Sep 14, 2019 at 12:28

I think there are two properties at play here: atomic velocity and stability of the lattice structure.

Recall that temperature is a measure of the average kinetic energy of the molecules, so $v \propto \sqrt{\frac{T}{m}}$, or to achieve the same velocity, $T \propto m$. So at the same temperature, two heavier atoms will move more slowly past each other than two lighter molecules, giving them more time to interact.

Now recall that in a solid, molecules are held in a lattice by intermolecular forces, and in a liquid, atoms have enough energy that the forces between them are no longer strong enough to hold the atoms in a lattice. So the more stability a substance's atoms gain from being in their solid, crystalline arrangement, the higher its melting point will be. Now, why would the metals in that area gain more stability in their solid forms? I'm fairly sure the answer has to do with the completeness of orbitals and half-orbitals. I'm not sure of the particulars of the delocalized electron cloud in metals, but I think it's likely that it allows these metals to in some manner fill or empty their incomplete orbitals.

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    $\begingroup$ Yes, they're heavy, but that's only half of the story, right? Cause it's not all the heavy atoms (mercury is heavy, and it's a liquid at ambient conditions…) $\endgroup$
    – F'x
    Commented Apr 28, 2012 at 21:30
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    $\begingroup$ Weight of atoms plays very little if any role in transitions from solid to liquid. Only energy is important when it comes to bonds breaking and temperature is a measure of average kinetic energy not speed. $\endgroup$
    – Vytenis
    Commented Apr 29, 2012 at 14:36

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