22
$\begingroup$

The melting and boiling points of transition elements increases from scandium ($1530~\mathrm{^\circ C}$) to vanadium ($1917~\mathrm{^\circ C}$). They increase because as we go across the group, we have more unpaired (free) electrons.

But at chromium ($1890~\mathrm{^\circ C}$) however, the melting point decreases even though it has more unpaired electrons than the previous atoms. Why does this happen?

$\endgroup$
  • $\begingroup$ Could it be related to the half-filled d-orbital of Chromium, which is special because it 'borrows' from the s-orbital? (see en.wikipedia.org/w/…). $\endgroup$ – Michiel Apr 21 '13 at 10:47
  • $\begingroup$ the melting point of chromium is $1907^oC$ not $1890^oC$ - see en.wikipedia.org/wiki/Chromium (Check the general properties section on the right side for melting point). This makes it almost the same as Vanadium (which has melting point $1910^oC$). $\endgroup$ – kaliaden Apr 21 '13 at 11:35
  • 2
    $\begingroup$ What I find more peculiar is the massive drop in melting point from Cr to Mn. Mn seems to be the most anomalous, though Cr and Tc also are outliers. Here are graphs for the melting point and boiling point. Can anyone explain the dips satisfactorily? $\endgroup$ – Nicolau Saker Neto Apr 21 '13 at 14:51
  • $\begingroup$ To answer the question one has to put crystal structure and band structure on same picture. $\endgroup$ – permeakra Apr 21 '13 at 16:14
  • 1
    $\begingroup$ That would be a part of the explanation. If you can elaborate, please do! Though it is not enough on its own, since some boiling points are also anomalous, and they cannot be due to structural characteristics in the solid. $\endgroup$ – Nicolau Saker Neto Apr 21 '13 at 16:48
8
$\begingroup$

To make an argument based on electron configuration, one could see that the orbital occupation does affect the melting point:

Element     Z   Electron Config.    Melting Point / K
-----------------------------------------------------
Vanadium    23  [Ar] 4s2 3d3        2183
Chromium    24  [Ar] 4s1 3d5        2180
Manganese   25  [Ar] 4s2 3d5        1519
Iron        26  [Ar] 4s2 3d6        1811
---
Copper      29  [Ar] 4s1 3d10       1358
Zinc        30  [Ar] 4s2 3d10        693

This is apparent in both Mn and Zn: the filled s-orbitals and (half-)filled d-orbitals impact the bonding character of the elements. This is also apparent in the 2nd and 3rd row transition metals, though to a lesser extent due to more diffuse orbitals, greater shielding of the valence electrons and relativistic effects from heavier nuclei. If you look at @NicolauSakerNeto's graph of the melting points, notice that there is still a dip in the graph when the d-orbitals are (half-)filled (i.e., for elements Tc, Cd, Os and Hg).

Essentially, melting point boils (haha!) down to bonding: more unpaired electrons lead to the creation of stronger bonds and thus the higher melting points.

$\endgroup$
  • 3
    $\begingroup$ But there are 5 unpaired electrons in the d shell of Magnese so, why does it have low melting point? $\endgroup$ – Rafique Apr 25 '13 at 23:48
  • 1
    $\begingroup$ The half filled d-orbital allows for extra stability compared to the other partially filled states. Read about Hund's Rule and the Aufbau Principal. $\endgroup$ – Chauncey Garrett May 1 '13 at 17:41
7
$\begingroup$

I don't really like parts of this explanation (was hoping to find a better one here actually), but it's the best I know. I'll build on suggestions by @michielim and @vrtcl1dvoshun.

(Note: I think most of this argument can be transposed to the orbital overlap picture of bonding, but it might be slightly trickier to describe.)

From the physicists' "electron sea" point of view of metal bonding, the higher the ionic charge the metal atom can support, the higher the element's melting and boiling points. This explains why group 1 metals such as sodium have quite low melting/boiling points, since the metal would be composed of electrons delocalized in a $\ce{M}^+$ lattice. Going towards group 2 and group 3 elements, one can expect to find a $\ce{M}^{2+}$ and $\ce{M}^{3+}$ lattice, and so on. However, this does not mean that that, for example, metallic osmium is expected to be made out of $\ce{Os}^{8+}$ cations surrounded by a dense electron sea; as each successive "ionization" of the metal atom is performed, it becomes more unlikely that the next electron can become delocalized, since it has to fight against a higher effective nuclear charge relative to the electron before it. Therefore, we can expect some "optimal average charge" for the metal ions in the pure metal. Since melting and boiling points tend to increase towards groups 7 and 8, we can safely assume that this optimal average charge is at least greater than $+3$, and it is likely higher for the heavier elements since they have significantly higher melting/boiling points.

If we can somehow figure out how easily electrons can be removed from their parent atoms to create the electron sea, we may have an argument to explain the trends and apparent discontinuities. First, notice that the $4^{\ce{th}}$ period metals are in general a little bit underwhelming compared their $5^{\ce{th}}$ and 6$^{\ce{th}}$ period counterparts when looking at the boiling points. This can be attributed to the fact that the $3d$ orbitals are anomalously compact relative to higher $nd$ orbitals. This happens because all first occurrences of a given sublevel (i.e. $1s$, $2p$, $3d$, $4f$, $5g$, etc.) are composed of orbitals whose wavefunctions have no radial node. Since the wavefunction's value is monotonically increasing up to its maximum value, it tapers off quickly. Thus, when a 4$^{\ce{th}}$ period metal atom tries to delocalize its $3d$ electrons, it meets a slight increase in resistance due to its stronger interaction with the nucleus, meaning that the optimal average charge of the ions is slightly lower and leading to lesser interatomic interactions holding the metal together.

Now, for the sudden dips, most visible for $\ce{Cr,\, Mn}$ and $\ce{Tc}.$ As has been suggested, one is tempted to take into consideration the fact that they all have $d^5$ configurations, which represent a half-filled $d$ subshell. It can be argued that they are particularly stable due to exchange energy, though Cann suggests it is not the most adequate explanation and instead forwards the "parallel spin avoidance factor".

In half-filled subshell configurations, there is a maximum in the effective nuclear charge felt by the electrons (compared to the previous elements with no doubly-occupied orbitals) combined with relatively low interorbital repulsions due to the Pauli exclusion principle. The element that follows a half-filled subshell must now put an electron in a previously occupied orbital, creating stronger, intraorbital electron repulsion. These effects combine to suggest that removing an electron from or forcing an electron into half-filled configurations is particularly difficult (notice that $\ce{Cr}$ doesn't follow $\ce{Mn}$ and $\ce{Tc},$ more on that soon). If this is the case, then the delocalization of the third electron (which would create $\ce{Mn}^{3+}$ and $\ce{Tc}^{3+}$ ions) and all those beyond it would be somewhat suppressed. This would effectively reduce the optimal average charge of the ions in the metal, and decrease the strength of interactions.

Finally, $\ce{Cr}$ doesn't show the same electronic behaviour as $\ce{Mn}$ and $\ce{Tc},$ even though it has a $d^5$ configuration, and yet it is still anomalous. What gives? Well, according to Greenwood and Earnshaw (Chemistry of the Elements, 1997), the low boiling point of $\ce{Cr}$ is due to a combination of increased effective nuclear charge and the anomalously small size of $3d$ orbitals. When going from $\ce{V}$ to $\ce{Cr},$ the increase in nuclear charge is enough to pull the already small $3d$ orbitals quite close to the nucleus, disfavouring electron delocalization to the extent that a sudden drop appears. Mo has the same electron configuration as $\ce{Cr},$ yet it does not show nearly as large a dip because its $4d$ orbitals are already significantly more diffuse, and aren't so easily suppressed from delocalizing.

$\endgroup$
  • 2
    $\begingroup$ Since writing this post, I have learned that, according to the book Solid State Physics (Ashcroft, Mermin), the Drude model as originally proposed gives no theoretical basis for finding how many electrons per atom in a metal are delocalized, so it has no way to suggest how positively charged the metallic ions are in the lattice. $\endgroup$ – Nicolau Saker Neto Jul 10 '15 at 22:17

protected by orthocresol Feb 3 '18 at 16:14

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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