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).


