 Solution

Three PO${}_4$ tetrahedra share corners in a ring. Each phosphorus has two ring-bridging oxygens and two terminal oxygens, sois one conventional Lewis representation of trimetaphosphate. Its three P and nine O atoms consist of three bridging and six terminal O atoms. Within each phosphate unit the two terminal bonds have delocalized character; drawing one P=O and one P–O${}^{-}$ is not a claim that every free anion has permanently different terminal oxygens.

A puckered six-membered P–O ring is expected. The tetrahedral phosphorus geometry and P–O–P torsional preferences allow folding without losing a required continuous perpendicular-p orbital system. This differs from the planar trigonal-boron arrangement. Tetrahedral P alone is not a geometrical proof of nonplanarity, since its two ring bonds could lie in a plane; the argument is that puckering is energetically accessible and relieves conformational strain without sacrificing a strong planar pi-conjugation benefit. The precise chair-like or boat-like distortion depends on environment and counterions.