Question

The hybridization of atomic orbitals of nitrogen in ${\text{NO}}_{\text{2}}^{\text{ + }}$,${\text{NO}}_3^ -$ and ${\text{NH}}_4^ +$ respectively are:
A. ${\text{sp}}$,${\text{s}}{{\text{p}}^2}$and ${\text{s}}{{\text{p}}^{\text{3}}}$
B. ${\text{s}}{{\text{p}}^2}$,${\text{sp}}$and ${\text{s}}{{\text{p}}^{\text{3}}}$
C. ${\text{sp}}$,${\text{s}}{{\text{p}}^{\text{3}}}$and${\text{s}}{{\text{p}}^2}$
D. ${\text{s}}{{\text{p}}^2}$,${\text{s}}{{\text{p}}^{\text{3}}}$and ${\text{sp}}$

Answer 1

We will determine the geometry of the molecules by drawing the Lewis structures. Lewis structure is drawn by arranging the total valence electrons around central and surrounding atoms. The sum number of lone pairs at the central atom and the number of sigma bonds gives the type of hybridization.

Complete answer:
According to the valence bond theory, the orbitals of metals combine to form orbitals of the same energy. These orbitals are known as hybrid orbital. Each ligand donates an electron pair to a hybrid orbital.
Write the Lewis structure as follows:
1. Write the basic structure.
2. Write the central atom around which writes all atoms of the molecule. The least electronegative atom is the central atom.
3. Count total valence electrons.
4. Two electrons are used in the formation of a bond.
5. Count the total electron used in bond formation.
6. Subtracts the electrons used in bond formation from the total valence electrons.
7. Arrange the remaining electrons around each atom to complete the octet.
8. Based on the number of electron pairs the hybridization and shape is determined.

No. of electron pair around central atom	Hybridisation
2	$sp$
3	$sp^2$
4	$sp^3$

The Lewis structure of ${\text{NO}}_{\text{2}}^{\text{ + }}$is as follows:
Total valence electrons are as follows:
$= \,\left( {5 \times 1} \right) + \left( {6 \times 2} \right) - 1$
$= \,16$

The number of sigma bonds of central atom nitrogen is two so, the hybridization of ${\text{NO}}_{\text{2}}^{\text{ + }}$ is ${\text{sp}}$.
The Lewis structure of ${\text{NO}}_3^ -$is as follows:
Total valence electrons are as follows:
$= \,\left( {5 \times 1} \right) + \left( {6 \times 3} \right) + 1$
$= \,24$

The number of sigma bonds of central atom nitrogen is three so, the hybridization of ${\text{NO}}_3^ -$ is ${\text{s}}{{\text{p}}^2}$.
The Lewis structure of ${\text{NH}}_4^ +$ is as follows:
Total valence electrons are as follows:
$= \,\left( {5 \times 1} \right) + \left( {1 \times 4} \right) - 1$
$= \,8$

The number of sigma bonds of central atom nitrogen is three so, the hybridization of ${\text{NH}}_4^ +$ is ${\text{s}}{{\text{p}}^3}$.

**Therefore, option (A) ${\text{sp}}$,${\text{s}}{{\text{p}}^2}$ and ${\text{s}}{{\text{p}}^{\text{3}}}$, is correct.

Note:**
Valence electrons of nitrogen atoms are five, oxygen is six and hydrogen is one. Pi bonds do not include in the counting of electron pairs around the central atom. The number of hybrid orbitals depends upon the number of sigma bonds. During counting the valence electrons, one is subtracted for each positive charge because a positive charge means an electron is lost. One is added for each negative charge because a negative charge means an electron is gained.