I was making the $\ce{H3BO3}$ structure and tried to think of hybridization of oxygen atoms and got confused between $\mathrm{sp^2}$ and $\mathrm{sp^3}$. Boron has empty orbitals, so the lone pairs of oxygen can do backbonding with boron, resulting in $\mathrm{sp^2}$ hybridization. But at a moment only one out of three oxygen atoms would do so. What about other two oxygens?

  • $\begingroup$ Whenever the lone pair electrons goes in resonance, hybridisation decreases $\endgroup$
    – Ritil
    Jul 14, 2022 at 17:49

1 Answer 1


All three oxygen atoms backdonate a π-electron pair and thus they are all $\mathrm{sp^2}$. When you construct the molecular orbitals, you find that the π orbitals are as follows:

  • One orbital bonds the boron atoms to all three oxygen atoms in a double Y shape — one Y on each side of the molecular plane.

  • Two orbitals are nonbonding because they have electron density on the oxygen atoms only; the boron is a node. (This corresponds to two oxygen atoms having a second lone pair whenever you draw any one of the contributing valence-bond structures.)

  • Finally, an antibonding orbital, which has the same threefold rotational symmetry as the bonding one but the oxygen atoms contribute out of phase from the boron atoms.

Since there are six π electrons (two from each oxygen atom but none from the electron-deficient boron), only the bonding and nonbonding orbitals are occupied and thus the molecule is bonded with the planar, $\mathrm{sp^2}$-hybridized structure.

  • 1
    $\begingroup$ Thank you very much , this cleared my doubt . $\endgroup$ Jul 15, 2022 at 8:07

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