How does Sp3 hybridization work for Nitrogen in NH_2^- ?

Mar 3, 2016

There are formally 4 electron pairs distributed around the nitrogen in the amide molecule; $s {p}^{3}$ hybridization would be the description.

Explanation:

Of course, there are likewise 4 electron pairs distributed around the nitrogen centre in ammonia, $: N {H}_{3}$. $s {p}^{3}$ hybridization would be the description, and the gross structure (of electron pairs) is that of a tetrahedron. But we describe molecular geometry on the basis of actual atoms, so the structure of ammonia is pyramidal.

Likewise for ${H}_{2} {N}^{-}$, the amide ion, the 4 electron pairs are distributed around nitrogen as a tetrahedron. Again, we describe structure on the basis of $N - H$ bonds only. The structure of the amide ion is thus akin to that of water, that is bent because there are also 2 lone pairs on the central oxygen/nitrogen atom.

Mar 3, 2016

Sp3 hybridization for Nitrogen in $N {H}_{2}^{-}$ has tetrahedrical geometry

Explanation:

Nitrogen has 5 outer electrons, in $N {H}_{2}^{-}$ it gains an extra electron. Of the 6 electrons, 2 are shared with 2 hydrogen atoms to form 2 sigma bonds, the remaining 4 eletrons fill 2 orbitals with 2 lonely electron pairs, similarly to ${H}_{2} O$.