Abstract

Hybridization in aromatic molecules plays an important role in determining their electron transport properties. When an sp(2) bonded aromatic ring hybridizes with a carbon pair in either sp(3), sp(2), or sp form, only sp(2) and sp result in a delocalized state across the whole molecule, opening a channel for transport. At the same time negative differential resistance (NDR) is found only in the sp(2) and sp cases. An energy alignment scheme is adopted to elucidate the NDR process. This is constructed by comparing the positions of frontier molecular levels and the states of leads. It is revealed that the alignment between lead states located around the Fermi energy with the lowest unoccupied molecular orbital of the central molecule is the main cause of the NDR.

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