Valence Bond Theory�Orbital Overlap as a Chemical Bond

Valence Bond Theory

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• Covalent bonding occurs when atoms share electrons, concentrating electron density between nuclei.
• If we combine this idea with quantum mechanics and atomic orbitals, a covalent bond can be visualized as the overlap of atomic orbitals of adjacent atoms.
• This model of chemical bonding is referred to valence bond theory.

H

H

H₂

Overlap

(Bond)

Valence Bond Theory – Basics

• According to valence bond theory, a bond forms between two atoms when the following conditions are met:
1. Two orbitals of adjacent atoms overlap.
2. The total number of electrons in both orbitals is no more than two.

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Overlap

Bond

No Overlap

No Bond

A bond cannot form here as it violates Pauli’s Exclusion Principle, which states each orbital can hold a maximum of two electrons of opposite spin.

Hydrogen

Chlorine

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Valence Bond Theory – Basic Theory

Chlorine

3s

3p

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Chlorine

Valence Bond Theory – Basic Theory

Chlorine

3s

3p

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Chlorine

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Valence Bond Theory – Basic Theory

Chlorine

3s

3p

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Chlorine

Overlap

(Bond)

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Valence Bond Theory – Basic Theory

Chlorine

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Overlap

(Bond)

3s

3p

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Chlorine

Valence Bond Theory

• Beryllium difluoride (BeF₂) forms a linear molecule.
• Fluorine atoms have an unpaired electron in their p sublevel available for bonding (1s²2s²2p⁵).
• Beryllium, on the other hand, has all its electrons paired (1s²2s²).
• How can beryllium form two bonds without unpaired electrons?

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Be

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Paired?

Unpaired

How can BeF₂ form?

Electron Promotion

• Even though beryllium in the ground state can’t form bonds, if an electron can be excited from the 2s to the 2p then it could form two bonds.
• This electron excitation is referred to as promotion and is used to explain how atoms can form more bonds than their ground state would predict.

Be

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Paired

Ground State

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Unpaired

Excited State

Electron Promotion

1s²2s¹2p¹

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VSEPR and Electron Promotion

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Be^*

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VSEPR and Electron Promotion

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Be^*

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VSEPR and Electron Promotion

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Be^*

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How can beryllium make a linear molecule with two equal bonds while using two different orbitals?

Orbital Hybridization

• While electron promotion can explain how beryllium makes two bonds with fluorine, it still can’t explain how it makes a linear molecule with two bonds of equal length, strength, and spacing.
• To explain this dilemma, we can propose that beryllium’s occupied s and p orbitals hybridize to create orbitals of identical shape and energy.
• These hybrid orbitals were made with one s orbital and one p orbital, so they are called sp orbitals.

2sp

Hybridization

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sp Hybridization

2s

2p

Energy

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sp Hybridization

s orbital

p_x orbital

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2s

2p

Energy

sp Hybridization

2s

Energy

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2p

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2p

sp Hybridization

180°

sp

Energy

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2p

sp

sp Hybridization

sp orbital

sp orbital

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sp Hybridization

Be^*

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sp Hybridization

Be^*

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p Orbitals 3D

p orbitals are much more bulbous than normally depicted, but they are drawn skinnier for easier visualization.

sp Hybrids 3D

Likewise, sp hybrids are also much more bulbous than commonly depicted.

Quick Recap

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• To explain the number of bonds beryllium forms and the shape of BeF₂, we used two concepts:
• Promotion – an electron is excited to higher energy level allowing for more bonding.
• Hybridization – orbitals hybridize to produce orbitals of identical energy and size.
• The number of hybrid orbitals formed is equal to the number orbitals that were used.
• For example, one ‘s’ orbital and one ‘p’ orbital are hybridized to produce two ‘sp’ orbitals.
• This is called sp hybridization and can explain the linear shape of molecules.

hybridization

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sp hybridization

s + p

Two sp orbitals

Linear

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sp

sp

180°

sp 3D

sp 3D

Trigonal Planar Molecules

How can boron make three identical bonds with the following electron configuration?

sp² Hybrid Orbitals

Electron Promotion

Hybridization

• We can extend our analysis to situations where more than two atomic orbitals mix to explain the bonding of elements such as boron (B).
• The ground state configuration for boron would predict it can make one bond.
• It can actually make three bonds of equal strength, length, and spacing due to promotion and hybridization.
• This is called sp² hybridization.

B^*

2sp²

Excited State

sp² hybridization

sp² hybridization

s + p_x + p_y

Three sp² orbitals

Trigonal Planar

120°

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sp²

sp²

sp²

sp² Hybridization

2s

2p

Energy

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sp² Hybridization

2s

2p

Energy

s orbital

p_y orbital

p_x orbital

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sp² Hybridization

2s

Energy

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2p

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2p

sp² Hybridization

sp²

Energy

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2p

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120°

sp²

sp² Hybridization

sp² orbital

sp² orbital

sp² orbital

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sp² 3D

sp² 3D

Tetrahedral Molecules

How can carbon make four identical bonds with the following electron configuration?

sp³ Hybrid Orbitals

Electron Promotion

Hybridization

• To explain the bonding of carbon in molecules such as methane (CH₄), we can once again use the ideas of promotion and hybridization.
• The ground state configuration for carbon would predict it can make two bonds.
• It can actually make four bonds of equal strength, length, and spacing due to promotion and hybridization.
• This is called sp³ hybridization.

C^*

2sp³

Excited State

sp³ hybridization

sp³ hybridization

s + p_x + p_y + p_z

Four sp³ orbitals

Tetrahedral

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sp³

sp³

sp³

sp³

109.5°

sp³ Hybridization

2s

2p

Energy

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sp³ Hybridization

2s

2p

Energy

s orbital

p_y orbital

p_x orbital

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p_z orbital

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sp³ Hybridization

2p

2s

Energy

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sp³ Hybridization

sp³

Energy

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sp³

109.5°

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sp³ Hybridization

sp³ orbital

sp³ orbital

sp³ orbital

sp³ orbital

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sp³ 3D

sp³ 3D

Hybridization Without Promotion

90°

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• Hybridization can also be used to describe the bonding and shape of molecules without invoking the idea of promotion.
• The ground state for nitrogen accurately predicts it can make three bonds (e.g., NH₃).
• If these were formed from the three p orbitals, then the bond angles would be 90° apart.
• In fact, they are much closer to 109.5° (107°), which suggests sp³ hybridized orbitals.

107°

Hybridized Orbitals

N

2sp³

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VSEPR and Hybridization

• Hybridization is required to explain the molecular geometry of molecules predicted by VSEPR.
• More accurately, the electron domain geometry can be used to infer the type of hybridization, and vice versa.

Lewis

Structure

Octahedral

Number of electron domains

Number of bonding domains

VSEPR and Hybridization

Number of electron domains

Hybridization

Number of bonding domains

• If you determine the electron domain geometry of a molecule, you can infer the type of hybridization.

4

,

2

sp³

VSEPR and Hybridization Practice

• Determine the hybridization of the molecules below.

Types of Orbital Overlap

Types of Orbital Overlap

Types of Orbital Overlap

This sideways overlap makes more sense when you consider the true size of a p orbital.

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Types of Orbital Overlap

Types of Orbital Overlap

σ and π Bonds

• Sigma (σ) bonds concentrate electron density on the internuclear axis between the nuclei and result from head-on orbital overlap.
• They tend to be stronger since there is more orbital overlap.
• Pi (π) bonds concentrate electron density above and below the internuclear axis and result from sideways orbital overlap.
• They tend to be weaker since there is less orbital overlap.

Sigma (σ) Bond

Electron density between nuclei

Electron density above and below nuclei

Pi (π) Bond

σ and π Bonds Practice

• Determine if the bonds below are sigma (σ) bonds or pi (π) bonds.

Multiple Bonds

• Atoms can only make one σ bond between them – all other bonds (multiple bonds) are the result of π bonds between unhybridized orbitals.

σ

π

σ Overlap

σ

sp²

sp²

Multiple Bonds

σC

π C

• The carbons in ethene have an electron domain geometry of three (trigonal planar) and are therefore sp² hybridized.
• The double bond is the result of a π bond between the unhybridized p orbitals.

π

This is one π bond!

Multiple Bonds

π Bond of Ethene

Multiple Bonds

• The carbons in ethyne have an electron domain geometry of two (linear) and are therefore sp hybridized.
• The triple bond is the result of two π bonds between the unhybridized p orbitals aligned orthogonally two each other.

σC

π C

π C

Ethyne Orbital Overlap

π Bonds of Ethyne

Practice Problem

• Formaldehyde has the following Lewis structure:
• Describe how the bonds in formaldehyde are formed in terms of σ and π overlaps of hybridized and unhybridized orbitals.

Practice Problem

σ Bond Rotation

Side View

• Sigma bonds can rotate because rotation would not break the end-to-end overlap of the orbitals and the bond will be maintained.

Front View

Overlap

Rotation will not break overlap

π Bond Rotation

Side View

Front View

Overlap

• Pi bonds cannot rotate because it would require the sideways overlapping orbitals to break their overlap, which would break the bond.

σ and π Bond Rotation

Bond Rotation 3D