# Hybridisation of Azide ion

In azide ion, $$\ce{N3-}$$, the hybridisation is $$\mathrm{sp}$$ because number of hybrid orbitals = steric number, the central atom $$(\ce{N})$$ has 6 bonds with other $$\ce{N}$$ atom, 2 of which are sigma bonds, and 4 pi bonds. But it doesn't make any sense to me, like if there are 6 bonds then the hybridisation should be $$\mathrm{sp^3d^2}$$ thus the hybrid orbitals along the principal axis make sigma bonds with the $$\mathrm{p}$$ orbital (which lie on principal axis) of $$(\ce{N})$$ atom (terminal $$(\ce{N})$$ atoms) and the other four $$\mathrm{sp^3d^2}$$ hybrid orbitals (which are not on principal axis) make pi bonds with $$\mathrm{p}$$ orbitals (which doesn't lie on principal axis) of $$(\ce{N})$$ atoms (terminal $$(\ce{N})$$ atoms).

• Consider question reformulation and formatting, as users have to read it. // There are no 6 bonds of nitrogen, it can form 4 bonds max. Aug 28, 2021 at 8:23
• One of readability recommendations for scientific texts is to avoid sentences longer than 35 words, keeping the average 20-25 words. Aug 28, 2021 at 8:59
• Pi bonds are not hybridized. No not even theoretically! Aug 28, 2021 at 9:04