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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).

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

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