Quote from: Mitch on October 17, 2008, 10:56:15 AM
Obviously the questions aren't easy, or we would answer them readily. I'll give the answer for why WF6 has a lower boiling point then MoF6.
The general argument is that the d-orbitals of Tungsten are more diffuse and thus bond better with fluorine. A more covalent molecule is usually a more volatile one, and thus it has a lower boiling point. We can examine it further by looking at their bond dissociation energies, the more covalent the bond the higher the energy needed to dissociate that bond[/b]
Mo - F 107 kcal/mol
W - F 120 kcal/mol
Hope this helps.
However the Si-Cl bond is stronger than the C-Cl bond but surely it is not more covalent!
All known hexafluorides are volatile because however polar the bonds it is impossible to produce a polymeric or ionic lattice without excessively high co-ordination numbers on the central atom. The same applies to tetrachlorides of the smaller elements, and to all known tetroxides (Ru, Xe, Os).
That compounds with highly polar bonds can be volatile if co-ordination numbers are restricted is also shown by such compounds as lanthanide bis(trimethylsilyl)amides Ln{[(CH3)3Si]2N}3, where the large size of the anions restricts co-ordination numbers.
It used to be thought that free quarks might exist and bind to nuclei. Consider imaginary 'quafline' Qf in which an up quark, charge +2/3, has bound to a fluorine nucleus. This would be most stable as Qf(-!/3) which would have an extremely high electron binding energy and would be most reluctant to form covalent bonds. MgQf6 would be bound almost purely ionically, but it would be gaseous - Mg cannot have a higher co-ordination number than 6 so this cluster of ions cannot polymerise.
Therefore the polarity of the bonds in substances such as MoF6 and WF6 should have little effect on their volatility.