# Bond lengths in P₄O₁₀

According to JD Lee's Concise Inorganic Chemistry:

The double bond $$\ce{P=O}$$ in $$\ce{P4O10}$$ contains p(π)–d(π) backbonding, where a full p-orbital on oxygen overlaps with an empty d-orbital on $$\ce{P}$$.

The four oxygens around each $$\ce{P}$$ form a tetrahedral shape. Then one lone pair from $$\ce{P}$$ in it's $$\mathrm{sp^3}$$ orbital will donate to an empty p-orbital on $$\ce{O}$$ (created by pairing of electrons in it's remaining two p-orbitals), only then can the "doubly-bonded" oxygen attain it's octet. There is no "single electron from each" bond here. Is my understanding correct?

In earlier similar questions (Hybridisation of POCl3) and (Bonding in the phosphate ion) the d-orbital theory has been said to be disproven, but then how do you explain the different bond lengths in the double (1.43 Å) and single (1.65 Å) $$\ce{P-O}$$ bonds?