Question

The hybridization and shape of \[NH_2^ - \]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: There are a total of four valence electrons attached to the central nitrogen atom as two lone pairs in a given compound. It has bent molecular geometry.

Complete Step by Step Solution:
1. The chemical formula \[NH_2^ - \]represents an azanide ion and it is also referred to as an amide.
2. The total number of valence electrons of nitrogen is five and each hydrogen has one electron so the total number of this compound is eight including the one negative charge.
3. The two valence electrons are used to form each covalent bond between the nitrogen and hydrogen atoms.
4. There is a total of four remaining valence electrons which form as two lone pairs on the nitrogen atom.
6. It has a cation and an anion with a non-neutral ion and it also does not exist because of the incomplete valence of a nitrogen atom.
7. The formula to calculate the hybridization of a compound is as follows:
\[H = \dfrac{1}{2}[V + M - C + A]\]
Where,
H = Number of hybridised orbitals
V = Central atom valence electrons
M = Number of monovalent atoms which are linked to the central atom
C = Cation charge
A = Anion charge
Apply the values of the given ion to the above equations.
\[H = \dfrac{1}{2}[5 + 2 - 1 + 2] = 4\]
8. The hybridization of the given ion is \[s{p^3}\]that contains one 2s and three 2p-orbitals.
The option (B) is correct.

Additional information:
1. The shape of the azanide ion is a tetrahedral electronic shape with bond angles of \[{104.5^ \circ }\].
2. According to Valence bond theory, orbitals of similar energy fuse together to form a new hybrid orbital which is also known as hybridization.

Note: The central nitrogen atom is attached to the two hydrogen atoms through two sigma bonds and the repulsion between the two hydrogen atoms forms a V-shape of the ion.