Explain hybridization of the central atom in $NO_{3}^{-}$.

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Hint: Find the values to put it in the formula of hybridization(or no. of hybrid orbitals) like Number of valence electron, Number of Monovalent atoms, etc and put it in formula to find out the hybridization of $NO_{3}^{-}$.

Complete answer:
There are various types of hybridization involving s, p and d orbitals these atomic orbital combines to form a new set of equivalent orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation.
There are important conditions for hybridization which are as follows:
$\left( 1 \right)$ The orbital present in the valence shell of the atom is hybridised.
$\left( 2 \right)$ The orbitals which are taking part in hybridization should have similar energy.
$\left( 3 \right)$ Promotion of electrons is not a necessary condition as compared to hybridization.
$\left( 4 \right)$ It is not important that only half filled orbitals participate in hybridization. In some cases, even filled orbitals of valence shells take part in hybridization.

 Now let us take these important points and the formula for finding the hybridization to find the hybridization of Nitrogen (Centre atom) in $NO_{3}^{-}$.
In $NO_{3}^{-}$,
Number of valence electron is $5$
Number of Monovalent atoms is $0$
Number of cationic charges is also $0$
Number of anionic charges is $1$

Let us put it in formula
\[Hybridization=\dfrac{1}{2}\left( valence\text{ }electrons+No.\text{ }of\text{ }Monovalent\text{ }atoms-cationic\text{ }charge+anionic\text{ }charge \right)\]
\[Hybridization=\dfrac{1}{2}\left( 5+0-0+1 \right)\]
\[Hybridization=3\], Here we can say that the hybridization of $NO_{3}^{-}$is \[s{{p}^{2}}\].
Where there is involvement of one s and two p orbitals in order to form three equivalent \[s{{p}^{2}}\]hybridized orbitals.
 After hybridization by this table we can find the shape of the compound.
The table is given below: -

H= l.p.+ b.p.Hybridization stateShapes
3$s{{p}^{2}}$Trigonal planar
4$s{{p}^{3}}$Tetrahedral or pyramidal or v- shaped
5$s{{p}^{3}}d$Trigonal bipyramidal or T- shaped or linear
6$s{{p}^{3}}{{d}^{2}}$Octahedral or square pyramid or planner
7$s{{p}^{3}}{{d}^{3}}$Pentagonal bipyramidal or distorted or distorted pentagonal bipyramid

Here l.p refers to Lone pair electrons and b.p refers to bond pair electrons. Here in $NO_{3}^{-}$all\[A-X\] bond lengths are identical ( \[A-\]central atom, \[X-\]terminal atom).

Note: -The phenomenon is known as hybridization which can be defined as the process of intermixing of the orbital of slightly different energies so as to redistribute their energies to form new bonds.

-Check the valence electron properly while calculation of hybridisation
-Check the charge of cation and anions on the molecule. It should also be taken into account
-Check the hybridization properly for concluding the final geometry of the molecule.