Hey everyone,

 

I was just wondering in the VO(acac)2 which is a Vandium (IV) complex the V=O bond length is significantly shorter than V-O bond length so the discription that it has a square-pyramidal geometry is somewhat misleading. what do you guys thing?

 

I know that in books and at various places it is stated to be square-pyramidal. I am also suspecting some distortion because of the Jahn-teller effect?

 

your opinion on that is welcome.

 

PS. I just realised I said Jahn-teller distortion which is obviuosly not possible for this complex.

Edited November 2, 200916 yr by ChemSiddiqui

acac is shorthand for a 2,4-Pentanedione ligand--common name ACetylACetone. This ligand is bidentate and has two donor atoms--the two oxygens. This suggests two points of attachment and you have placed two acac groups on the V central atom--accounting for all four of the equatorial bonds. I'm fairly certain both acac ligands take on an aromatic character when bonding to the metal as well--I have the proton NMR's for a Co(acac)3 which has a singlet integrating to 1, shifted down to just shy of 6. The lone oxygen will occupy an axial position and it leaves room for a sixth ligand to come in from the bottom... square pyramidal is actually quite a common geometry for this type of molecule.

 

Wikipedia has a pretty decent picture of it where you can more clearly see the square-pyramidal structure:

http://en.wikipedia.org/wiki/Vanadyl_acetylacetonate

Edited November 2, 200916 yr by aeontide

Thanks aeontde for that but that is what I already know. I merely wanted opinion on the geometry, because in this complex vanadium can adopt the 5 or 6 coordination. but anyway, you are right acac- do have the aromatic character. I got the bond length data and similar bond lengths suggest delocalisiation.

 

thanks again.