Like the previous compound, the one above contains a triple bond so we form a lone pair hybrid. We will first attempt to determine the aromaticity of the entire compound by focusing on the heterocycle contained within it. There are two atoms of interest here (the oxygen and nitrogen). According to the inclusive approach both the nitrogen and oxygen are classified as sp 3 -hybridized (Oxygen; 2 sigma bonds + 2 lone pairs = 4, sp . Nitrogen; 3 sigma bonds + 1 lone pair = 4, sp 3 ), while under the non-inclusive approach they are classified as sp 2 -hybridized (Oxygen; 2 sigma bonds + 1 localized lone pair = 3, sp 2 . Nitrogen; 3 sigma bonds + 0 localized lone pairs = 3, sp 2 ) since the lone pairs on each atom (only one of the two lone pairs specifically in regards to the oxygen atom) are able to be considered delocalized. Under the non-inclusive approach, one may initially think the heterocycle is aromatic because it is cyclic, contains a conjugated pi system and Huckel number of 10 pi electrons (4 from the pi bonds, 4 from the delocalized lone pairs on the oxygen and nitrogen atoms, and 2 from the recognizable pi bond delocalization pattern), but actually drawing the resonance structure that supports an aromatic configuration proves problematic: