Re: why is vanadium (v) oxide coloured?

Patrick, you have asked one that is well beyond my field of
expertise. I'm basically an organicker :-) so I
threw this out on the MadSci listserv.  The following
response was from Dr. John Christie. (Thanks John)
--------
"Transition metal spectroscopy is a fairly complicated and
esoteric realm, and you would need a good reference book.
"Inorganic Chemistry" by James E Huheey is a possible good
start, but may not go far enough.

A few pointers:

-- Although the formula of V2O5 indicates a +5 oxidation state
for vanadium, it does not indicate that V5+ ions are involved.
The bonding is largely covalent. There is probably some d-pi
to p-pi bonding involved. So there are d electrons involved
in the bonding at least.

-- Whether or not a compound is coloured depends very simply
on whether there is an accessible excited electronic state
separated from the ground state by an energy gap corresponding
to a visible wavelength. Accessible means that an optical transition
must be allowed by symmetry rules, or at least not strongly forbidden.
d to d transitions are weakly forbidden. But in the later elements
of the transition series (after Sc & Ti) there is the possibility
of promotion from a non-bonding or weakly bonding 3d or 4s
orbital into 4p, which is allowed.

-- V2O5 is by no means alone. Cr(VI) compounds are strongly
coloured (CrO3, CrO2Cl2 are red, K2Cr2O7 orange, and K2CrO4 yellow),
Mn(VII) as in KMnO4 is intense violet, and Os(VIII) in OsO4
is deep red. The change seems to occur between Ti and V. TiO2 is,
of course, a major white pigment used in paints, etc.

-- Even some compounds of the heavier main group elements that
seem to have neatly filled shells form coloured compounds, e.g.
PbCl4 is yellow.

That information does not provide an answer -- some research
effort may be needed to go much further. I hope it is a useful
pointer in the right direction."

-------------------------
Thanks again to Dr. Christie for helping out an organicker who
was well out of his realm on this question :-)

Richard