Hybridisation

Master Chemical Bonding! Pauling's Valence Bond Theory could not explain the exact bond angles and shapes of many polyatomic molecules (like Methane's 109.5° angle). To solve this, the revolutionary concept of Hybridization was introduced.

Hybridization: Salient Features, Conditions, and Formula

Hybridization is defined as the process of intermixing atomic orbitals of slightly different energies to form an entirely new set of equivalent orbitals (called hybrid orbitals) with identical shapes and energies.

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Figure 1: Salient features defining the concept of hybridization.

Salient Features of Hybridization

1. Conservation of Orbitals: The number of hybrid orbitals generated is always equal to the number of atomic orbitals that get hybridized.
2. Equivalence: The newly formed hybridized orbitals are always equivalent in both energy and shape.
3. Bond Strength: Hybrid orbitals are far more effective in forming stable bonds than pure atomic orbitals.
4. Molecular Geometry: Hybrid orbitals direct themselves in space in specific directions to minimize electron-pair repulsions. Therefore, the type of hybridization directly dictates the geometry of the molecule.

Figure 2: Essential conditions and mathematical formula for finding hybridization.

Important Conditions for Hybridization

Not just any orbitals can intermix. Specific criteria must be met:

• Valence Shell Only: Only the orbitals belonging to the outermost (valence) shell of the central atom undergo hybridization.
• Similar Energies: The orbitals undergoing hybridization should have almost equal energies (e.g., 2s and 2p).
• Excitation Not Mandatory: The promotion or excitation of an electron to a higher energy level is not an essential condition prior to hybridization.
• Involvement of Lone Pairs: It is a common misconception that only half-filled orbitals participate. Fully filled orbitals (containing lone pairs) can also undergo hybridization!

The Super Trick: Formula to Find Hybridization

In competitive exams like JEE and NEET, drawing orbital diagrams is too time-consuming. You can instantly find the hybridization state of a central atom using this simple mathematical formula:

Hybridization Number (H) =

H = ½ [ V + M - C + A ]

V = Number of Valence electrons of the central atom.
M = Number of Monovalent atoms attached to it (like H, F, Cl, Br, I).
C = Charge of Cation (subtract it).
A = Charge of Anion (add it).

Interpreting the Result (H):

• If H = 2 → sp hybridization (Linear)
• If H = 3 → sp² hybridization (Trigonal Planar)
• If H = 4 → sp³ hybridization (Tetrahedral)
• If H = 5 → sp³d hybridization (Trigonal Bipyramidal)
• If H = 6 → sp³d² hybridization (Octahedral)
• If H = 7 → sp³d³ hybridization (Pentagonal Bipyramidal)

Example Calculation: Ammonia (NH₃)

Let's calculate the hybridization of Nitrogen in NH₃.

• Central Atom = Nitrogen (N). Its valence electrons (V) = 5.
• Number of Monovalent atoms attached (M) = 3 (three Hydrogens).
• No charge on the molecule, so C = 0, A = 0.

Calculation: H = ½ [ 5 + 3 - 0 + 0 ] = ½ [ 8 ] = 4.

Since H = 4, the hybridization of Nitrogen in Ammonia is sp³.

Frequently Asked Questions (FAQs)

Does hybridization occur in isolated atoms?

No, hybridization does not occur in isolated gaseous atoms. It is a theoretical concept that takes place only during the process of bond formation to explain the experimentally observed shapes and bond angles of molecules.

Do pi (π) bonds participate in hybridization?

No, pi (π) bonds never participate in hybridization. Only sigma (σ) bonds and lone pairs of electrons in the valence shell of the central atom are considered when determining the hybridization state.

Is the promotion of an electron necessary for hybridization?

No, the promotion or excitation of an electron to a higher energy level is not an essential condition. Even fully filled orbitals (like the 2s orbital containing a lone pair in Oxygen or Nitrogen) undergo hybridization.

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