"The shoulder at ~3250 cm -1 [here in the Fig. , the blue component on the left] corresponds to the asymmetric -OH stretch, and the most intense feature at ~3410 cm -1 [magenta] corresponds to the symmetric -OH stretch. Even more intriguing, there is a small shoulder at ~3630 cm "The OH stretching vibration is resolved into a total of four bands, two bands resulting from a fully hydrogen bonded system and two from a partially bonded system. These four vibrational modes are postulated to be the result of the splitting of the v1 symmetrical stretching vibration into four distinct vibrations. The lowest, at approximately 3240 cm -l , corresponds to a fully symmetric 4-hydrogen bonded tetrahedral water molecule where all neighboring water molecules are vibrating together in phase. This is analogous with that observed in ice. The next v1 vibration at ~3385 cm -l represents the loss of vibrational harmony between neighboring molecules. Although still fully tetrahedrally bound, this vibration is the result of neighboring water molecules stretching; but no longer in phase with each other. The two next v1 vibrations at ~3490 and ~3620 cm -l represents the loss of a hydrogen bond, leaving one hydrogen atom "dangling" while the other is fully hydrogen bonded. The water molecule then, due to loss of equivalency of its two hydrogen atoms, is no longer C2v symmetry; but of the lower Cs symmetry group. It is now best to view the two OH bonds as nonequivalent local mode oscillators. The ~3490 cm -1 peak represents the vibration of the OH bond where the hydrogen atom is hydrogen bonded, while the ~3620 cm -l peak represents the vibration of the dangling OH bond. As temperature increases a new v1 vibration will appear at ~3670 cm -l . This vibration represents free OH stretching, unhindered by neighboring hydrogen bonds, as observed in the vapor state" .