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Chemical Bonding and Molecular Structure - Hybridization

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Hybridization is the process of intermixing atomic orbitals of slightly different energies belonging to the same atom to redistribute their energy, resulting in the formation of a new set of orbitals of equivalent energies and identical shapes.

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The number of hybrid orbitals produced is always equal to the number of atomic orbitals that undergo hybridization.

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Hybrid orbitals are always equivalent in energy and shape. They are more effective in forming stable bonds than pure atomic orbitals due to better overlapping.

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spsp Hybridization: Involves one ss and one pp orbital. It results in a linear geometry with a bond angle of 180∘180^\circ (e.g., BeCl2BeCl_2, C2H2C_2H_2).

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sp2sp^2 Hybridization: Involves one ss and two pp orbitals. It results in a trigonal planar geometry with bond angles of 120∘120^\circ (e.g., BF3BF_3, C2H4C_2H_4).

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sp3sp^3 Hybridization: Involves one ss and three pp orbitals. It results in a tetrahedral geometry with bond angles of 109∘28′109^\circ 28' (e.g., CH4CH_4, NH4+NH_4^+).

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sp3dsp^3d Hybridization: Involves one ss, three pp, and one dd orbital. It results in trigonal bipyramidal geometry (e.g., PCl5PCl_5).

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sp3d2sp^3d^2 Hybridization: Involves one ss, three pp, and two dd orbitals. It results in octahedral geometry (e.g., SF6SF_6).

📐Formulae

H=12[V+M−C+A]H = \frac{1}{2} [V + M - C + A]

Where:\text{Where:}

V=Number of valence electrons of the central atomV = \text{Number of valence electrons of the central atom}

M=Number of monovalent atoms surrounding the central atomM = \text{Number of monovalent atoms surrounding the central atom}

C=Charge on the cationC = \text{Charge on the cation}

A=Charge on the anionA = \text{Charge on the anion}

Steric Number=(Number of σ bonds)+(Number of lone pairs on central atom)\text{Steric Number} = (\text{Number of } \sigma \text{ bonds}) + (\text{Number of lone pairs on central atom})

💡Examples

Problem 1:

Determine the hybridization and geometry of the Phosphorus atom in PCl5PCl_5.

Solution:

Using the formula H=12[V+M−C+A]H = \frac{1}{2} [V + M - C + A]: For PP (Phosphorus), V=5V = 5. For ClCl (monovalent atoms), M=5M = 5. Since it is neutral, C=0C = 0 and A=0A = 0. Therefore, H=12[5+5−0+0]=5H = \frac{1}{2} [5 + 5 - 0 + 0] = 5.

Explanation:

A value of H=5H = 5 corresponds to sp3dsp^3d hybridization. The geometry of the PCl5PCl_5 molecule is trigonal bipyramidal.

Problem 2:

Explain the hybridization in Ethene (C2H4C_2H_4).

Solution:

Each Carbon atom in C2H4C_2H_4 is bonded to two Hydrogen atoms and one Carbon atom via a double bond. In terms of steric number, each Carbon has 3 σ\sigma bonds and 0 lone pairs. Steric Number = 3+0=33 + 0 = 3.

Explanation:

A steric number of 3 indicates sp2sp^2 hybridization. This results in a trigonal planar arrangement around each carbon atom with bond angles of approximately 120∘120^\circ. The unhybridized 2pz2p_z orbital forms the π\pi bond.

Problem 3:

Find the hybridization of the central atom in SF6SF_6.

Solution:

For Sulfur (SS), V=6V = 6. For Fluorine (FF, monovalent), M=6M = 6. H=12[6+6]=6H = \frac{1}{2} [6 + 6] = 6.

Explanation:

An HH value of 6 corresponds to sp3d2sp^3d^2 hybridization. The spatial arrangement of the six sp3d2sp^3d^2 hybrid orbitals is octahedral, and the F−S−FF-S-F bond angles are 90∘90^\circ.

Hybridization - Revision Notes & Key Formulas | ICSE Class 11 Chemistry