Hybridization Of NH₃ (Ammonia)

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What Is The Hybridization Of NH3?

Before going to understand the hybridization of Ammonia (NH3), we have to sift through the Nitrogen areas. If we notice the atomic number of nitrogen, it is 7, and if we consider its ground state, it is 1s2, 2s2, 2p3.

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During the ammonia formation, one 2s orbital and three 2p orbitals of nitrogen combine to produce four hybrid orbitals with equivalent energy, which is therefore considered an sp3 type of hybridization.

Furthermore, if we look at the NH3 molecule, we will notice that the three half-filled sp3 orbitals of nitrogen form bonds to hydrogen’s three atoms. The fourth sp3 orbital that is present, however, is a nonbonding pair of hybridized orbital and is used for holding the lone pair regularly.

The central atom in the ammonia molecule is sp3 hybridized. Let us look at the determination of NH3 hybridization.

Name Of The Molecule


Molecular Formula


Hybridization Type


Bond Angle



Pyramidal or Distorted Tetrahedral

NH3 Molecular Geometry Of sp3 Hybridization

The sp3 hybridized orbitals repel each other and they are directed towards four corners of a regular tetrahedron. The geometry of the molecule is tetrahedral (non-polar) and the angle between them 109.5°. The sp3 hybridization can be shown as follows.

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If we notice Ammonia's molecular geometry, the molecular shape of NH3 is a distorted tetrahedral or trigonal pyramidal structure. This is primarily due to the presence of a lone nonbonding pair, which exerts a greater repulsion, usually on the bonding orbitals. Here, in all, nitrogen lies at the centre and the three identical hydrogen atoms create the base while one pair of electrons produces the apex of the pyramid.

The NH3 bond angle or the ammonia bond angle is 107°.

Lewis Structure Of NH3

Ammonia, otherwise called Nitrogen Trihydride, is a form of a colourless gas. On the periodic table, nitrogen is considered in group 15 and it has five valence electrons. The hydrogen falls under category one. So, we can say that it has only one valence electron. Since we have three hydrogens in NH3, this valence electron is to be multiplied by three.

Thus, we should add these electrons of hydrogen and nitrogen to get the total number of atoms. Since there are five nitrogen electrons, one multiplied by three i.e. three hydrogen electrons, the outcome becomes eight.

Here, we should make a note that the hydrogen goes on the outside always. So, keep it away and place the nitrogen in the centre. As we have three hydrogens in NH3, all of them will be set around the central atom of nitrogen and all the total eight valence electrons will form chemical bonds with them.

Hydrogen is used to set with simply two valence electrons to create an outer shell. Since they have two for each of them, six will be the final result. However, as we have calculated, there are eight valence electrons because there are 5 Nitrogen + 3(1) Hydrogen. So, place all of them here, and we will find out that the nitrogen has eight valence electrons, and the hydrogen has two valence electrons, and the octet became full now.

Electron Geometry Of NH3

As it has four electrons groups, NH3 electron geometry is ‘Tetrahedral.’ Where one group has an unshared pair of electrons. ‘N’ has a tetrahedral electronic geometry. Therefore, Ammonia is an example of the molecule, where the central atom has both shared and an unshared pair of electrons as well.

Calculating The Number Of Hybrid Orbitals

The NH3 molecule is acetylene (ethyne) and both carbons are sp hybridized. It means an s orbital has been hybridized with a single p orbital to form two sp hybrids on each carbon atom and each of these hybrids contains one electron each. The remaining two electrons are in unhybridized p orbitals that can eventually be used to produce two pi bonds.

Hence each of the carbon atoms has two unhybridized p orbitals and two sp hybrids. So, the total number of hybrid orbitals present on each carbon atom is two.

In order to calculate the hybridization, it is a bit more complicated but can be simplified simultaneously. The sum of the attached atoms plus the lone electron pairs will provide the hybridization. If the sum is,

2 = sp

3 = sp2

4 = sp3

Each carbon is directly attached to 1 hydrogen and 1 carbon for acetylene. So, 1+1=2, and therefore, each carbon is an sp hybridized.

For a formaldehyde molecule, CH2O, the carbon is attached to 3 atoms and has no lone pair electrons, so it should be sp2 hybridized.

Besides, in formaldehyde, the oxygen is attached to carbon and has two pairs of electrons, which also count as attachments. So, it has 1 carbon and 2 lone pairs, equals three attachments. Thus, oxygen should be sp2 hybridized.

FAQ (Frequently Asked Questions)

1. How to Calculate the Hybridization?

Ans: First, we have to write the Lewis structure to get an idea about the structure of a molecule and bonding pattern. We can use the valence concept to do so. The further step is to calculate the sigma (σ) bonds count in that molecular structure. After that, find the lone pair numbers on a given atom by using the formula,

Number of lone pairs =v-b-c ∕2

Where v = number of valence electrons in the concerned atom in a free state

c = charge on the atom

b = number of bonds formed by a concerned atom

We can calculate the steric number as,

Steric number = number of σ-bonds + number of lone pairs and then based on the steric number Assign hybridization and shape of a molecule

2. Why do Hybrid Orbitals Form Stronger Bonds?

Ans: The hybrid orbitals have identical energy. They also contain more electron density in a particular lobe compared to the other lobe. The pure atomic orbitals contain uniform charge density throughout the orbital.

The hybrid orbital overlaps the central atom with the atomic orbital of the bonded atom resulting in a stronger bond compared to the overlap of two pure atomic orbitals.

For example, an sp3 hybrid orbital has 75% of electron density in a lobe and 25% in the other lobe. The 75% electron density lobe overlap with an atomic orbital of the bonded atom results in the formation of a strong bond because of the large overlapping region.

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