Is it possible that a transititon metal M=O FTIR peak split to doublet when M=O is hyrolyzed to form HO-M-OH?
or
Any other condition under which M=O ftir peak split to form doublet? The peak at 924.11 is M=O split to form two peaks at 928.63 and 906.73 (ATTACH FILE). The Metal-complex sample is water soluble and also changes its color showing solvent (H2O) to metal charge transfer behavior.
what can be the possible reasons?
When M=O became HO-M-OH, there is no longer M=O group and no M=O mode should be observed. If you are looking for M-OH mode, for example OH stretch of the M-OH, there should be in theory two bands, symmetric and antisymmetric OH stretching modes. However, these groups are typically hydrogen bonded and the structural breadth is so broad, you will not be able to resolve those; however, theoretical calculation should show two modes.
Aside from HO-M-OH, M=O mode can sometime split if the conditions are met. For example, Fermi resonance. If the main mode, say M=O stretch is almost the same as one of the overtones or combination modes, this M=O stretch could split. The best known example is the NH stretch of a primary amine for which only two NH stretch is expected, but actually show three. One of the two band interact with the overtone mode and the energy level split into two. Adding another non split mode, the total becomes three. If this is the case, look around 462 cm-1 and see if you find rather strong band. You may find M=O bending mode there.
Another possibility of split, which is possible for only solid samples, is Davidov split (otherwise known as the crystal field split). When more than one group is in a crystal lattice, the mode related to this group can split into the same number of the group in the lattice. The best known example is the 1414 cm-1 band of CH2 in-plane bending mode of polyethylene, since polyethylene crystal lattice contains two chains of polyethylene in one lattice.
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