Orbital Hybridization: Definition, Types, Geometry, and More

During the formation of covalent bonds, atomic orbitals overlap according to the valence bond theory. The process of combining atomic orbitals due to overlapping to form newly hybridised orbitals is called hybridization. The same energy level atomic orbitals usually participate in hybridization. These hybrid orbitals have distinct shapes and orientations based on overlapping orientation. For examples, in methane $\left({\mathrm{C}\mathrm{H}}_4\right)$ , the carbon atom undergoes hybridisation to form four equivalent bonds with hydrogen, despite having different types of atomic orbitals ( 2 s and 2 p ).

There are three key things involved in hybridization process:

• Overlapping of atomic orbitals of similar energy level to form hybrid orbitals.
• The energy of hybrid orbitals gets redestributed as an average of the original orbitals, making them degenerate.
• Geometric Arrangement of hybrid orbitals determines the molecule's shape, such as linear, tetrahedral, or trigonal planar.

Here are the key features of the hybridization, which you must remember during the exams. Read below:

• Atomic orbitals with equal energies take part in the hybridization process.
• Hybrid orbitals have equivalent energy, which ensures stable bond formation.
• Hybrid orbitals overlap with orbitals of other atoms head-to-head to form sigma bonds, which are strong and directional.
• Hybrid orbitals overlap with orbitals of other atoms sideways to form pi bonds, which are weaker.
• In many compounds like ${\mathrm{N}\mathrm{H}}_3$ or ${\mathrm{H}}_2\mathrm{O},{\mathrm{s}\mathrm{p}}^3$ hybrid orbitals may contain lone pairs, which affect the molecular geometry.

You can follow these simple steps to determine the type of hybridisation in a chemical compounds along with its molecular structure. Read below.

• Step 1: Find out the electronic configuration of all the elements taking part in the formation of compound. (For example, in case of NH₃, do the elctronic configuration of Nitrogen and Hydrogen.)
• Step 2: Calculate the number of electrons in the valence shell and the valence shell orbital (s,p,d or f).
• Step 3: Calculate the steric number and orbitals overlapping to form the compound.
• Step 4: Match the steric number with the associated molecular geometry, use the table given below for ease.

Table for Identification of Hybridization and Molecular Geometry

Steric Number	Hybrid Type	Orbitals Mixed	Geometry
2	sp	1 s + 1 p → 2 sp	Linear
3	sp²	1 s + 2 p → 3 sp²	Trigonal planar
4	sp³	1 s + 3 p → 4 sp³	Tetrahedral
4	sp²d (square planar)	1 d + 1 s + 2 p → 4 dsp²	Square planar
5	sp³d (dsp³)	1 s + 3 p + 1 d → 5 sp³d	Trigonal bipyramidal
6	sp³d²	1 s + 3 p + 2 d → 6 sp³d²	Octahedral
7	sp³d³	1 s + 3 p + 3 d → 7 sp³d³	Pentagonal bipyramidal

There are various type of hybridised orbitals and hybridization based on orbital orientation. Read complete information realted to the types of the hybridization along with self-explanatory diagrams below.

• sp Hybridisation

The sp hybridisation involves overlapping of one s and one p orbital to form two sp hybrid orbitals. This hybrid orbital will be linearlly oriented ${180}^{\circ }$ apart. The hybrid orbital contains one electron available for bonding.

Comibation: 1s + 1p orbital at 180° angle, linear overlapping

 

Example: We can take the example of ${\mathrm{B}\mathrm{e}\mathrm{C}\mathrm{l}}_2$ (beryllium chloride). Beryllium has two valence electrons . To form two bonds, one 2 s electron is promoted to a 2 p orbital, and the 2 s and 2 p orbitals hybridise to form two sp orbitals. These overlap with chlorine's p orbitals, forming a linear molecule (Cl-Be-Cl). 

• ${\mathbf{s}\mathbf{p}}^2$Hybridisation

In $s{p}^2$ hybridization, one s and two p orbitals overlap to combine and form three $s{p}^2$ hybrid orbitals. These $s{p}^2$ hybrid orbitals are arranged in a trigonal planar fashion ( ${120}^{\circ }$ bond angles). This is common in molecules with three electron domains around the central atom.

Comibation: overlapping of 1s and 2p orbital at 120° angle

Example: Taking the example of  ${\mathrm{B}\mathrm{F}}_3$ (boron trifluoride). Boron's 2 s and 2 p orbitals hybridise to form three $s{p}^2$ orbitals, each forming a sigma bond with fluorine's $2p$ orbital. The molecule adopts a trigonal planar shape. 

• $s{p}^3$ Hybridisation

In $s{p}^3$ hybridisation one s and three p orbitals overlap and combine to form four $s{p}^3$ hybrid orbitals. The orientation of $s{p}^3$ hybrid orbitals is tetrahedral (109.5 ° bond angles). This is commonly  found in 4 single bond compound attached to a central atom like methane.

Comibation: overlapping of 1s and 3p orbital at 109.5° angle

Example: The most prominent example is  ${\mathrm{C}\mathrm{H}}_4$ (methane). Carbon's 2 s and 2 p orbitals hybridise to form four ${\mathrm{s}\mathrm{p}}^3$ orbitals, each overlapping with hydrogen's 1s orbital, resulting in a tetrahedral structure. 

• ${\mathbf{s}\mathbf{p}}^3\mathbf{d}$ Hybridisation

The ${\mathbf{s}\mathbf{p}}^3\mathbf{d}$is complex type hybridization structure. This ${\mathbf{s}\mathbf{p}}^3\mathbf{d}$involves not just s and p but also te d orbital. In ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ hybridisation, one s, three p, and one d atomic orbitals overlap and combine to form five ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ hybrid orbitals. You can check the orbital overlapping and ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ diagram below.

${\mathbf{s}\mathbf{p}}^3{\mathbf{d}}^2$

• ${\mathbf{s}\mathbf{p}}^3{\mathbf{d}}^2$ Hybridization

The ${\mathbf{s}\mathbf{p}}^3{\mathbf{d}}^2$ has also a complex type hybridization stats, as both of atoms involve d orbital..six hybrid orbitals form instead of 5 like ${\mathbf{s}\mathbf{p}}^3\mathbf{d}$ hybridization. This type of hybrid orbital has an octahedral geometry. The shape of these orbitals will be a trigonal bipyramidal shape. check the ${\mathbf{s}\mathbf{p}}^3{\mathbf{d}}^2$ diagram below.

Examples: Phosphorus pentachloride  (PCl₅) uses one 3s, three 3p, and one 3d orbital to form five ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ orbitals.

Sulphur hexafluoride(SF₆) hybrid orbitals form six ${\mathrm{s}\mathrm{p}}^3{\mathrm{d}}^2$ orbitals, with an octahedral geometry.

Here are the clear differences in hybridisation and VSEPR theory.

Features	Hybridisation	VSEPR Theory
Fundamental Concept	Combination of Atomic Orbitals	Electron Pair Repulsion
Key Point	Orbital Overlapping	Electron Pair Repulsion
Explanation	Bond Formation	Molecular Geometry
Participating Element	s, p, d orbitals	Electron Domains
Example	CH₄ (sp³), BF₃ (sp²), BeCl₂ (sp)	CH₄ (tetrahedral), NH₃ (trigonal pyramidal), H₂O (bent)

Many students fall for the mistake due to urgency or the need to quickly solve problems. Here are a few important areas where you can make mistakes. You can check the point of mistake and tips to avoid them as well. Read below.

1. Calculating Steric Number: This is important to understand that the steric number is the sum of all electron domains or groups, rather sum of all electron pairs and the individual number of bonds. You need to carefully examine the electron domains, especially in molecules with multiple lone pairs or resonance involvement or multiple bonds. 
2. Confusing σ vs π Bonds Counting: When students count individual bonds instead of electron domains, they miscalculate the steric number. For example, if you are treating a double bond as two electron pairs instead of one electron domain, It will lead to wrong classification of hybridization.
3. Resonance and Delocalization Effect: Some atoms, such as nitrogen and oxygen, show resonance in certain conditions, which changes the hybridization state. You must carefully understand the concept of resonance and delocalization of electrons as per molecular orbital theory to avoid these mistakes. 

Tips to Avoid These Mistakes

• You should draw Lewis structures to identify the bond pairs, lone pairs, σ domains, and other electron domains.
• Count the Steric Number very carefully, especially count double/triple bonds as a single entity.
• Always check for resonance in the atom while overlapping; Forexample if there is delocalization of an electron, the central atom often uses sp² hybridization to leave a p orbital free.

JEE Main questions frequently include Hybridisation concepts, molecular geometry, bond angles, and chemical properties. You can check the type of questions asked based on this topic below: 

• Question related to the prediction of geometry
• Question based on bond angles of some specific compounds like water, methane.
• Bond strength and effects of hybridization in bond strength-based questions