Question

The hybridization and shape of $N{H}_2^{-}$ ion are:
(a) $s{p}^2$ and angular
(b) $s{p}^3$ and angular
(c) $s{p}^3$ and linear
(d) $sp$ and linear

Answer 1

Hint: Hybridization is the notion that newly hybridized orbitals are formed from the fusion of atomic orbitals. This in turn, influences the molecular bonding and geometry properties of the compound.

Complete answer:
* The energy of the hybridized orbitals are lower than the energy compared to their separated, unhybridized counterparts. Due to this more stable compounds are formed.
* The hybridization also has a greater role in determining the shape or the molecular geometry of the compound. the shapes can be: linear, angular or V or bent, tetrahedral, seesaw, square pyramidal and many more.
* Hybridization is an expansion of valence bond theory.
* The Hybridization of any compound can be calculated as follows:

$H=\frac{1}{2}[V+M-C+A]$

where, H = Number of orbitals involved in hybridization. V = Valence electrons of the central atom. M = Number of monovalent atoms linked to the central atom. C = Charge of the cation. A = Charge of the anion.

* Now, in the case of $N{H}_2^{-}$, V=5, M=2, A=1. Substituting these values in the above equation, we get,

$H=\frac{1}{2}[5+2+1]$

$⟹H=4$

Therefore, the hybridization of $N{H}_2^{-}$ is $s{p}^3$ and has the shape bent or angular .
Thus, the correct option is (b).

Note: In the above mentioned case the compound does not have any cationic charge and therefore, C=0 but has an anionic charge thus A=1. And note that the shape of $N{H}_2^{-}$ is an angular shape even though the hybridization is $s{p}^3$, this is due to the lone pair-bonding pair repulsion which as a result pushes the two bonding pairs closer together.