Past exam of the chemistry B course of the University of Cambridge / 2023 / Paper 2 / B9 / a / i / Solution

The unequal P–N distances reflect unequal local bonding interactions in the alternating-chain conformation, rather than an extended conducting P=N pi system. nitrogen is much more electronegative than phosphorus, so the backbone sigma bonds are strongly polar. nitrogen lone-pair density can also donate into appropriately oriented antibonding P–substituent orbitals: negative hyperconjugation, schematicallyThe strengths of these interactions depend on the orientation of the donor and acceptor orbitals. In a planar cis–trans backbone, the P–N links are not all geometrically equivalent. Different overlap and charge redistribution give alternating greater and smaller effective P–N bond orders, hence shorter and longer bonds. A formal alternating P=N/P–N drawing is a useful shorthand for this inequality, but does not imply that isolated textbook single and double bonds are present with no electronic communication.

The historical island picture captures the limitation of pi interaction to local P–N–P regions rather than a continuous band along the whole chain. A modern description need not invoke strong phosphorus $3d$–nitrogen $2p$ bonding; polar bonding and donation into sigma-star orbitals provide the useful explanation. Calculations on oligomers found bond alternation for planar cis–trans conformations but not for optimized helices. Bond alternation is therefore conformation dependent, not a universal consequence of the repeat formula alone.^[ref]