Hafnium carbide, Tungsten and Molybdenum are the most refractorymetals known

Hi Sumit,

the trend to evaporate is based on entropy, formation of gaseous species increases the entropy, according to Gibbs dG = dH -TdS formation of volatile species is favored at high temperature. Thats a general trend.

Concernig your question: The transition metals of group V-VIII form high oxidation states. In these high oxidation states the bonds in compounds such as OsO4 or WO3 have a more covalent nature while the lower oxidation states have more ionic character. So the lattice energy of high oxidation state oxides is lower which makes them easier to evaporate.

best regards

Frank

To Frank Kern ’s answer I would add that the elements in question also have relatively small ionic radii in their highest oxidation state – specifically, ionic radii around or below 0.6 A, which puts their oxides in the region of preferred stability of coordination number 4 or lower, rather than the very common 6. Once you move to larger ions such as Nb(V) and Ta(V) (radii~0.7 A), the preferred coordination number is 6 or larger; the 3D oxide lattice is relatively more stable and doesn’t allow easy evaporation or even easy melting.

To evaporate easily a compound needs to generate small molecules easily. For oxidation state ≥5 to be compatible with a coordination number of ≤4, the small volatile molecules must contain one or more double (or “double”) M=O bonds. Not coincidentally, the elements we are discussing are known for their capability to form relatively stable multiple bonds. Contrast that with Si(IV) or B(III), which rarely if ever form bonds with multiple character and have refractory (in the sense of non-volatile) oxides despite their normal coordination number of 4 (or in boron’s case – mostly 3).

Oxides which have a low melting point, tend to be volatile, for example PbO, similarly sulphur dioxide. Hence I would try to look into the melting point data.

Alternately SiO2 will not be volatile.