Question

Trimethyl phosphine oxide can be represented by two canonical forms as shown but trimethylamine oxide can't be. It is due to:

• A. phosphorus accepts charge density form oxygen and accommodates it in its vacant d-orbitals
• B. nitrogen does not possess d-orbitals and thus can't form the $\pi$-bonded structure
• C. both $A$ and $B$
• D. none of the above

Answer 1

The correct option is C both $A$ and $B$

$P$ can expand its valency due to a presence of d-orbitals and here, phosphorus accepts charge density from oxygen and accommodates it in its vacant d-orbitals. Due to an absence of vacant d-orbitals, $N$ is unable to expand its valency beyond $3$.

$(C{H}_3{)}_3\stackrel{+}{P}-\stackrel{-}{O}\leftrightharpoons (C{H}_3{)}_{3}P=O$