Question

Why is the hybridisation of N in borazine sp2 ?

N in borazine forms 3 sigma bonds and has one lone pair so its steric number is 4. Hence, the hybridisation should be `sp^3`.

Answer 1

Because there are only three electron groups that are coplanar with each other. The lone pair is part of the `pi` system. You may want to read up on that here.

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The point of hybridization is to generate identical orbitals for sigma (and ONLY sigma!) interactions.

Regardless of how many actual bonds there are, what matters is the number of bonding directions that preserve symmetry.

Borazine is a benzene analog; not a perfect one, but enough that the lone pair is perpendicular to the plane of the ring, and is in the `pi` system.

The resonance structures where the electrons are delocalized are favored.

Never forget... it's not the number of electron groups that determine the geometry, it's their placement in space.

Furthermore, the ring itself is not a perfect hexagon, since the bond angles are smaller at boron and larger at nitrogen, although the `"B"-"N"` bond lengths are all the same.

As a result, the point group is probably `D_(3h)`, rather than `D_(6h)`.