isopolymolybdates heteropolytungstates formation and structure
TRANSCRIPT
ExamplesV10O28
2- Mo7O246-
Mo8O264- Ta6O198-
Al13O4(OH)24(H2O)127+
or AlO4Al12(OH)24(H2O)127+
PW12O403- P2W18O62
6-
Co4P4W30O112(H2O)216-
or Co4(H2O)2(P2W15O56)216-
As12Ln16(H2O)36W148O52476-
Chromate - DichromateCrO4
2- + 2 H+ → Cr2O72- + H2O
Think of CrO42- as being composed of
4 O2- with a central Cr6+
One O2- is neutralized by the acid.O2- + 2 H+ → H2O
Chromate - Dichromate
The Cr2O72- ion has each Cr at the center of a
tetrahedron with the two tetrahedrons sharing a corner.
Molybdates
Molybdates form polyions by edge sharing rather than corner sharing.
The Mo6+ ions are at the center of an octahedron.
Polymolybdates
Unlike chromates, molybdates are more complicated.
As oxides are neutralized, larger structures are formed.
MoO42- → Mo7O24
6- → Mo8O264- → Mo36O112
8-
Molybdates and Tungstates
What three structures can be built fromthree octahedrons by edge sharing?No corner sharing.No face sharing.
In the following structures there is an oxide at each corner and a metal ion in the center of each octahedron.
Electrostatic Repulsions
In the linear structure, the center metal ion has little space to move to be farther away from the other two metal ions.
In the 90° structure, there is more possibility of movement away from each other but still not as much as the 60° structure.
Electrostatic Repulsions
So, 60° structures are preferred over 90° structures, which are preferred over 180° structures.
The best structures will maximize 60° and minimize 180°. The building block is the 60° structure.
What’s in the Literature?Lots of errors in the older literature.1. Systems not at equilibrium when measured,
especially tungsten systems.2. Bad theory. Assumed that diffusion in solution
was only related to molecular weight (like gases).
SynthesisProduct depends on:1. stoichiometry of reactants2. pH3. temperature
A variety of ions can be put into the cavity.P(V), Si(IV), Co(II), Co(III), C(IV), . . .
SynthesisCan also have a heteroion replace one of the tungstens.A P(V) in the center and a Co(II) replacing a W.A Co(II) in the center and a Co(II) replacing a W.
Synthesis
Co2+ + WO42- → CoW11Co(H2O)O39
8-
precipitate potassium saltCoW11Co(H2O)O39
8- + H+ → CoW12O406- (aq)
this contains Co2+ in the centerOxidize to Co3+ → CoW12O40
5-
precipitate as potassium saltResult: tetrahedral Co(III) in the center
Lacunary Structures
Can have structures with a piece of the regular structure missing.
Remove a W3 unit from the Keggin structure and have a W9 structure remaining.
PW11O397-
Remove one W unit from the Keggin structure and have a W11 structure remaining.
Uses
PMo12O403-
Developing Agent for TLCacts as oxidizing agentsome molybdenums reduced from +6 to +5forming a blue color (“molybdenum blue”) as a visible spot.