9C question 5
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9C question 5
Can someone explain to me how oxalate could be bidentate? I think I might be drawing the lewis structure wrong or something. it's section 9C, question 5, part D.
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Re: 9C question 5
Oxalate is shaped like ethylene, and when both double bonds are on one side, both extra lone pairs are on the opposite side (on the single bonded oxygens)
When that's the case, the oxalate ion can donate both of those lone pairs to form two coordinate bonds to a single ion.
When that's the case, the oxalate ion can donate both of those lone pairs to form two coordinate bonds to a single ion.
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Re: 9C question 5
Ya just adding on to the previous answer, the 2 oxygens forming the bond with the transition metal must also contain single sigma bonds in order to be able to rotate into a position that allows for such bonding to occur
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Re: 9C question 5
Oxalate follows the general formula lone pair - spacer - spacer - lone pair, so it can be a polydentate ligand. In this case, it is a bidentate ligand because the two -1 charged oxygens can make coordinate covalent bonds with two transition metals, or form a chelate bonding to the same transition metal.
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