\[s{p^3}\] hybridization is found in
A) \[{\rm{C}}{{\rm{O}}_{\rm{3}}}^{2 - }\]
B) \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\]
C) \[{\rm{N}}{{\rm{O}}_{\rm{3}}}^ - \]
D) \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]

Hint: Hybridization is the process of intermixing of orbitals possessing similar energy to give new orbitals having different shapes and energies. Here, to check the hybridization of the given compounds, we will use a formula.

Formula Used:The formula of hybridization is,
\[H = \dfrac{{V + X - C + A}}{2}\]
Here, V stands for the count of valence electrons, X stands for count of monovalent atoms of the compound, C stands for charge of cation and A is for the charge of anion.

Complete step by step solution:Let’s find out the hybridization one by one.
Option A is \[{\rm{C}}{{\rm{O}}_{\rm{3}}}^{2 - }\]. The count of valence electrons of a Carbon atom is four and the count of monovalent atoms is zero and the anionic charge is 2.
\[H = \dfrac{{4 + 2}}{2} = 3\]
So, the hybridization of \[{\rm{C}}{{\rm{O}}_{\rm{3}}}^{2 - }\]is \[s{p^2}\] .
Option B is \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\]. The count of valence electrons of Boron is three and the count of monovalent atoms is three \[H = \dfrac{{3 + 3}}{2} = 3\]
So, the hybridization of\[{\rm{B}}{{\rm{F}}_{\rm{3}}}\] is \[s{p^2}\] .
Option C is \[{\rm{N}}{{\rm{O}}_{\rm{3}}}^ - \]. The count of valence electrons of Nitrogen is five and the count of monovalent atoms is zero and the anionic charge is 1.
\[H = \dfrac{{5 + 1}}{2} = 3\]
So, the hybridization of\[{\rm{N}}{{\rm{O}}_{\rm{3}}}^ - \] is \[s{p^2}\] .
Option D is \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]. The count of valence electrons of nitrogen is five and the count of monovalent atoms is three \[H = \dfrac{{5 + 3}}{2} = 4\]
So, the hybridization of\[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] is \[s{p^3}\] .

Therefore, option D is right.

Note: Always remember that, if the H value is 4, the molecule is \[s{p^3}\] hybridized. The value of H is 3 means that the molecule is \[s{p^2}\] hybridized. And the value of H is 2 indicates that the molecule is \[sp\] hybridized.