# NMR of trigonal bipyramidal molecules

I seem to remember at school that if a molecule is symmetrical in $$\ce{^1H}$$ NMR, that only one peak would be shown e.g. propane's terminal $$\ce{-CH3}$$.

Consider the two molecules shown below.

My professor remarked that both these isomers would show a doublet and triplet in their proton-decoupled $$\ce{^31P}$$ NMR and I was struggling to work out why.

Consider the molecule on the left as molecule A and the right as molecule B.

My train of logic was that the axial $$\ce{Ph3P}$$ group on molecule A would be split by each $$\ce{P}$$ atom that is equatorial, so using the expression $$2nI + 1$$ where $$I = \frac{1}{2}$$ so our singlet would be split into a triplet.

Likewise, the equatorial $$\ce{Ph3P}$$ groups would both only see one axial $$\ce{Ph3P}$$ group, hence only a singlet would be observed.

I used the same logic on molecule B.

Is this the correct reasoning for the doublet and triplets observed in both molecules?