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Chemical Bonding and Molecular Structure - Valence Bond Theory

Grade 11CBSEChemistry

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

🔑Concepts

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Valence Bond Theory (VBT) was introduced by Heitler and London and developed further by Pauling. It describes bond formation based on the overlapping of atomic orbitals.

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A covalent bond is formed by the overlap of half-filled atomic orbitals of the valence shell of two atoms, containing electrons with opposite spins. The strength of the bond depends on the extent of overlap.

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Types of Overlapping: σ\sigma (sigma) bond is formed by the end-to-end (head-on) overlap of atomic orbitals along the internuclear axis (s−ss-s, s−ps-p, or p−pp-p overlap). π\pi (pi) bond is formed by the lateral (sideways) overlap of pp-orbitals.

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Strength Comparison: A σ\sigma bond is stronger than a π\pi bond because the extent of orbital overlap is greater in the case of axial overlapping compared to lateral overlapping.

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Hybridization: The process of intermixing of atomic orbitals of slightly different energies of the same atom to form a new set of equivalent orbitals of identical shapes and energies. Types include spsp, sp2sp^2, sp3sp^3, sp3dsp^3d, and sp3d2sp^3d^2.

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Directional Properties: Atomic orbitals (except ss-orbitals) have specific directions in space, which explains the definite geometry and bond angles of polyatomic molecules like H2OH_2O and NH3NH_3.

📐Formulae

X=12[V+M−C+A]X = \frac{1}{2} [V + M - C + A] (where V=V = valence electrons of central atom, M=M = number of monovalent atoms, C=C = cationic charge, A=A = anionic charge)

Extent of overlap∝1Bond length∝Bond Strength\text{Extent of overlap} \propto \frac{1}{\text{Bond length}} \propto \text{Bond Strength}

Percentage s-character in spn hybrid orbital=11+n×100\text{Percentage } s \text{-character in } sp^n \text{ hybrid orbital} = \frac{1}{1+n} \times 100

💡Examples

Problem 1:

Explain the formation of the H2H_2 molecule based on VBT and potential energy.

Solution:

When two HH atoms approach, new attractive forces (between nucleus of one and electron of another) and repulsive forces (between nuclei or electrons) act. Bond forms at a distance of 74 pm74\,pm where potential energy is minimum (−435.8 kJ mol−1-435.8\,kJ\,mol^{-1}).

Explanation:

The stability of the H2H_2 molecule is reached when the net force of attraction balances the force of repulsion, resulting in the overlapping of 1s1s orbitals.

Problem 2:

Determine the hybridization and geometry of CH4CH_4 using Valence Bond Theory.

Solution:

In CH4CH_4, the central Carbon atom (1s22s22p21s^2 2s^2 2p^2) undergoes sp3sp^3 hybridization. One 2s2s and three 2p2p orbitals mix to form four equivalent sp3sp^3 hybrid orbitals.

Explanation:

The four sp3sp^3 hybrid orbitals are directed towards the corners of a regular tetrahedron with a bond angle of 109.5∘109.5^\circ to minimize electronic repulsion.

Problem 3:

Why is a σ\sigma bond stronger than a π\pi bond?

Solution:

The σ\sigma bond is formed by head-on overlap along the internuclear axis, whereas the π\pi bond is formed by lateral overlap.

Explanation:

According to VBT, the strength of a covalent bond is directly proportional to the extent of orbital overlap. Head-on overlap allows for a much larger region of electron density between nuclei than sideways overlap.

Valence Bond Theory - Revision Notes & Key Formulas | CBSE Class 11 Chemistry