Abstract
A new oxide containing Bi, Mg, and V has been prepared, and its structure was determined from single-crystal X-ray diffraction data. The space group is R3 with hexagonal cell dimensions of a=10.1985(9) and c=20.4875(14) Å. The formula may be written as BiMg2(MgV)V18O38 to indicate that one crystallographic site is occupied by a 1:1 mixture of Mg and V. The vanadium on that site appears to be in an oxidation state of 2. Thus, the average oxidation state of vanadium on the other three vanadium sites would be 3.72+. The site occupied exclusively by magnesium is in tetrahedral coordination to oxygen. All sites occupied by vanadium are octahedrally coordinated by oxygen. The low electrical resistivity of this compound suggests itinerant electron behavior.
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