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 M+lattice. Going towards group 2 and group 3 elements, one can expect to find a M2+and M3+ lattice, and so on. However, this does not mean that that, for example, metallic osmium is expected to be made out of Os8+ 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 4th period metals are in general a little bit underwhelming compared their 5th and 6th period counterparts when looking at the boiling points. This can be attributed to the fact that the 3dorbitals 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 4th period metal atom tries to delocalize its 3delectrons, 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 Cr, Mn and Tc.
As has been suggested, one is tempted to take into consideration the fact that they all have d5configurations, 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 Cr doesn't follow Mn and Tc, more on that soon). If this is the case, then the delocalization of the third electron (which would create Mn3+ and Tc3+ 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, Crdoesn't show the same electronic behaviour as Mn
and Tc, even though it has a d5
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 Cr
is due to a combination of increased effective nuclear charge and the anomalously small size of 3d orbitals. When going from V to 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 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.