Q.Why does R3P = O exist but R3N = O does not (R = alkyl group)?

Step 1: Bonding requirement for a true E=O double bond.

A genuine double bond between the central atom E and oxygen requires, in addition to the sigma bond, a pi bond formed by overlap of a filled p-orbital on O with an empty, appropriately-oriented orbital on E.

Step 2: Phosphorus's case.

Phosphorus (period 3) has empty, energetically accessible 3d orbitals in its valence shell. A filled p-orbital on oxygen can donate electron density into an empty 3d orbital on phosphorus, forming a $p\pi$–$d\pi$ back bond. Combined with the sigma bond, this gives $R_3P=O$ genuine double-bond character, so $R_3P=O$ (e.g. phosphine oxides) exist as stable, well-characterised compounds.

Step 3: Nitrogen's case.

Nitrogen (period 2) has no d-orbitals in its valence shell (only 2s, 2p are available, all used up in bonding/lone pair). It cannot accept a $\pi$-donation from oxygen's p-orbital, so no true $p\pi$–$d\pi$ N=O double bond can form.

Step 4: What nitrogen forms instead. …

Identify the structural requirement for a true double bond to oxygen (an empty orbital on the central atom for π back-donation from O), then show P has accessible 3d orbitals for this …

• Do not say nitrogen "cannot bond to oxygen" at all — it does bond, but only through a coordinate single bond ($R_3N^+$–$O^-$), not a true double bond; the distinction is bond order/type, not existence of any N–O interaction. …