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Chemical Bonding and Molecular Structure - Valence Bond Theory

Grade 11ICSEChemistry

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

🔑Concepts

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Valence Bond Theory (VBT) was proposed by Heitler and London and further developed by Pauling and Slater. It states that a covalent bond is formed by the partial overlap of two half-filled atomic orbitals containing electrons with opposite spins.

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The strength of a bond depends on the extent of orbital overlap; greater overlap leads to a stronger and more stable bond.

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σ\sigma (Sigma) Bond: Formed by the end-to-end (axial) overlap of bonding orbitals along the internuclear axis. Overlap can be s−ss-s, s−pzs-p_z, or pz−pzp_z-p_z.

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π\pi (Pi) Bond: Formed by the lateral (sideways) overlap of atomic orbitals. The electron density is concentrated above and below the plane of the nuclei.

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Hybridization is the process of intermixing atomic orbitals of slightly different energies to produce a new set of equivalent orbitals known as hybrid orbitals, such as spsp, sp2sp^2, sp3sp^3, sp3dsp^3d, and sp3d2sp^3d^2.

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Types of Hybridization and Geometry: spsp (Linear, 180∘180^\circ), sp2sp^2 (Trigonal Planar, 120∘120^\circ), sp3sp^3 (Tetrahedral, 109.5∘109.5^\circ), sp3dsp^3d (Trigonal Bipyramidal), sp3d2sp^3d^2 (Octahedral).

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Limitations of VBT: It fails to explain the paramagnetic nature of oxygen (O2O_2) and does not account for the coordinate covalent bond where both electrons come from one atom.

📐Formulae

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

Where: H=Number of hybrid orbitals\text{Where: } H = \text{Number of hybrid orbitals}

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

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

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

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

Extent of Overlap∝Bond Strength∝1Bond Length\text{Extent of Overlap} \propto \text{Bond Strength} \propto \frac{1}{\text{Bond Length}}

💡Examples

Problem 1:

Explain the hybridization and geometry of Methane (CH4CH_4) using Valence Bond Theory.

Solution:

In CH4CH_4, the central carbon atom has the ground state configuration 1s22s22p21s^2 2s^2 2p^2. In the excited state, one electron from 2s2s jumps to the empty 2pz2p_z orbital: 1s22s12px12py12pz11s^2 2s^1 2p_x^1 2p_y^1 2p_z^1. One 2s2s and three 2p2p orbitals undergo sp3sp^3 hybridization.

Explanation:

The four sp3sp^3 hybrid orbitals are directed towards the corners of a regular tetrahedron with bond angles of 109.5∘109.5^\circ. Each hybrid orbital overlaps with the 1s1s orbital of a Hydrogen atom to form four C−HC-H σ\sigma bonds.

Problem 2:

Describe the bonding in Ethene (C2H4C_2H_4).

Solution:

Each Carbon atom in C2H4C_2H_4 undergoes sp2sp^2 hybridization. Two sp2sp^2 orbitals of each carbon form σ\sigma bonds with HH atoms (1s1s), and the third sp2sp^2 orbital forms a C−CC-C σ\sigma bond.

Explanation:

The unhybridized 2pz2p_z orbitals on each carbon atom overlap laterally to form a π\pi bond. Thus, the C=CC=C double bond consists of one σ\sigma bond and one π\pi bond.

Problem 3:

Determine the hybridization of Phosphorus in PCl5PCl_5.

Solution:

Using the formula H=12[V+M−C+A]H = \frac{1}{2}[V + M - C + A], for PP (Z=15Z=15), V=5V=5. M=5M=5 (for five ClCl atoms). H=12[5+5−0+0]=5H = \frac{1}{2}[5 + 5 - 0 + 0] = 5.

Explanation:

Since H=5H=5, the hybridization is sp3dsp^3d. The geometry is Trigonal Bipyramidal, with three equatorial P−ClP-Cl bonds and two axial P−ClP-Cl bonds.

Valence Bond Theory Class 11 Notes & Examples