Bonding

In this section, you will learn about hybridisation, how it shapes molecules, and the types of bonds in them. You will explore how to predict molecule shapes (e.g., straight, bent, or triangular) and measure bond angles. You will also understand the two types of bonds, sigma and pi, and explain how atoms combine their “building blocks” (orbitals) to create different shapes and structures.

KEY IDEAS

• Valence Shell Electron Pair Repulsion (VSEPR) Theory is a model used to predict the geometry of molecules based on repulsion between electron pairs.

• Molecular geometry is the 3D arrangement of atoms in a molecule.

• Bond angle: is the angle formed between two covalent bonds at the atom where they meet.

• Sigma bond is a bond formed by the head-on overlap of orbitals, allowing free rotation.

• Pi (π) bond is a bond formed by the side-by-side overlap of p orbitals, restricting rotation.

• Hybridisation is the mixing of atomic orbitals to form new hybrid orbitals of equivalent energy.

Shape and bond angles for molecular compounds Electronegativity is a way to describe how strongly an atom can pull electrons toward itself when it bonds with another atom. It helps us understand why some atoms share electrons equally while others do not.

Linus Pauling came up with this idea and created a scale called the Pauling scale to measure how “electron-hungry” atoms are. The lowest value is 0.7 (like caesium, which does not pull electrons much). The highest value is 4.0 (fluorine, which pulls electrons very strongly).

These numbers do not have units—they just compare how atoms behave.

Electronegativity increases across a row or period because atoms get smaller (due to increased number of protons, which exert a greater electrostatic force of attraction on the surrounding electrons) and pull electrons more strongly.

Electronegativity decreases down a column (group) because atoms get bigger (due to the addition of more electron shells), and their outer electrons are farther from the centre, making it harder to pull more electrons.

Electronegativity helps us predict how atoms bond and how molecules behave in chemical reactions.

Bond Polarity

The difference in electronegativity between two bonded atoms determines the bond type and its polarity:

1. Non-Polar Covalent Bonds

a. Electronegativity Difference: Less than 0.5

b. Electron Sharing: Equal sharing of electrons.

c. Charge Distribution: No partial charges, bond is neutral.

Example: Hydrogen molecule (H₂): Both hydrogen atoms have the same electronegativity, resulting in a non-polar bond.

2. Polar Covalent Bonds

a. Electronegativity Difference: Between 0.5 and 1.7

b. Electron Sharing: Unequal sharing of electrons.

c. Charge Distribution:

The more electronegative atom becomes partially negative (δ−). The electronegative atom becomes partially positive (δ+).

Example: Water (H₂O): Oxygen is more electronegative than hydrogen, causing a dipole with oxygen as δ− and hydrogen as δ+.

3. Ionic Bonds

a. Electronegativity Difference: Greater than 1.7

b. Electron Sharing: Electrons are transferred, not shared.

c. Charge Distribution:

The more electronegative atom gains electrons, becoming negatively charged (anion).

The less electronegative atom loses electrons, becoming positively charged (cation).

Example: Sodium chloride (NaCl): Sodium (Na) loses an electron to become Na+, and chlorine (Cl) gains an electron to become Cl−.

Dipole Moments

A dipole moment happens in a polar bond when there is a separation of positive and negative charges. It shows how much the electrons are pulled toward one atom.

An arrow points from the positive side to the negative side of the bond to show the direction of the electron pull. See Figure 7.1.

Dipole moments are measured in Debye units (D). It tells us how uneven the charge distribution is in a bond.

Figure 7.1: Dipole Moment

Examples

1. Hydrogen Chloride (HCl)

In hydrogen chloride (HCl):

Electronegativity Difference: Chlorine (3.0) is more electronegative than hydrogen (2.1), creating a polar covalent bond.

Chlorine pulls the shared electrons closer, becoming partially negative (δ–).

Hydrogen becomes partially positive (δ+).

A dipole moment is created, showing the separation of charges.

An arrow points from hydrogen (positive) to chlorine (negative), showing the electron pull direction. See Figure 7.2. This dipole moment makes HCl a polar molecule.

Figure 7.2: Polarity of HCl

2. Methane (CH₄) In methane (CH₄):

The difference between carbon (2.5) and hydrogen (2.1) is 0.4, forming non-polar covalent bonds.

The molecule has a tetrahedral shape, which is symmetrical.

The dipole moments of the C-H bonds cancel out because they point in opposite directions. See Figure 7.3.

As a result, methane has no net dipole moment, making it a non-polar molecule.

Figure 7.3: Polarity of CH₄

3. Carbon Dioxide (CO₂)

In carbon dioxide (CO₂):

The difference between carbon (2.5) and oxygen (3.5) is 1.0, forming polar C= O bonds.

CO₂has a linear shape, which is symmetrical.

The dipole moments of the two C= O bonds point in opposite directions and cancel each other out. As a result, CO₂has no net dipole moment, making it a non-polar molecule. See Figure 7.4.

Figure 7.4: Polarity of CO₂

Predicting Molecular Polarity with VSEPR Theory

The Valence-Shell Electron-Pair Repulsion (VSEPR) theory helps predict the shape of a molecule by considering how electron pairs repel each other. The goal is to arrange electron pairs around the central atom as far apart as possible to minimize repulsion.

Steps to Use VSEPR Theory

1. Draw the Lewis structure Show all bonding and lone pairs of electrons.

2. Count electron pairs around the central atom Include both bonding pairs (BP) and lone pairs (LP).

3. Account for repulsions Electron repulsions depend on the type of pairs:

Lone Pair–Lone Pair (LP-LP): Strongest repulsion.

Lone Pair–Bonding Pair (LP-BP): Medium repulsion.

Bonding Pair–Bonding Pair (BP-BP): Weakest repulsion.

Lone pairs take up more space than bonding pairs, so they push other pairs apart.

4. Adjust Molecular Shape

Modify the shape based on repulsions to minimize electron pair interactions.

For example: Lone pairs cause angles to shrink compared to a shape with only bonding pairs.

Double bonds occupy more space than single bonds.

1. Species with Four Negative Charge Centres

Example: Methane (CH₄)

Structure: Methane (CH₄) consists of one carbon atom bonded to four hydrogen atoms.

Carbon has four valence electrons, and each hydrogen atom has one electron.

Together, they form four single covalent bonds.

Electron Groups: The carbon atom has four electron groups (four bonding pairs) around it.

Shape: According to VSEPR theory, these four electron groups repel each equally and arrange themselves in a tetrahedral shape to minimise repulsion.

The bond angles in methane are approximately 109.5°.

In a 3D structure, the hydrogen atoms form a tetrahedron around the central carbon atom.

See Figure 7.5

Polarity: Methane is non-polar because The bond polarities cancel out due to its symmetrical tetrahedral shape.

This results in no net dipole moment.

Figure 7.5: Tetrahedral shape of CH₄.

2. Species with 3 Negative Charge Centres

Boron Trifluoride (BF3)

Structure: Boron trifluoride (BF₃) has one boron atom bonded to three fluorine atoms.

Boron has three valence electrons, and each fluorine atom has seven valence electrons.

They form three single covalent bonds.

Electron Groups: The boron atom has three electron groups (three bonding pairs) around it.

Shape: According to VSEPR theory, the three electron groups repel each other equally and arrange themselves in a trigonal planar shape to minimise repulsion.

The bond angles are approximately 120°.

In 3D, the fluorine atoms form an equilateral triangle around the central boron atom.

Polarity: Boron trifluoride is non-polar because:

The bond polarities cancel out due to its symmetrical trigonal planar shape. This results in no net dipole moment. See Figure 7.6.

Figure 7.6: Trigonal planar shape of BF₃

3. Species with 2 Negative Charge Centres

Carbon Dioxide (CO2)

Structure: Carbon dioxide (CO₂) has one carbon atom double-bonded to two atoms.

Carbon has four valence electrons, and each oxygen has six valence electrons.

This forms two double bonds.

Electron Groups: The carbon atom has two electron groups (two double bonds) around it.

Shape: According to VSEPR theory, the two electron groups repel each other and arrange in a linear shape to minimise repulsion. The bond angle is 180°.

In 3D, the oxygen atoms are aligned in a straight line on opposite sides of the carbon atom. See Figure 7.7.

Figure 7.7: Linear shape of CO₂.

Predicting Molecular Polarity

To know if a molecule is polar, consider:

• Bond Polarity: Check if the bonds are polar (unequal sharing of electrons).

• Shape of the Molecule: Look at the molecule’s symmetry.

• Symmetrical Molecules: If bond dipoles cancel out, the molecule is non- polar.

• Asymmetrical Molecules: If bond dipoles don’t cancel, the molecule is polar.

Example

1. Water (H₂O) Polar Bonds: The O-H bonds are polar because oxygen is more electronegative than hydrogen.

Oxygen has a partial negative charge (δ−), and hydrogen has a partial positive charge (δ+).

Shape: Water has a bent shape due to two lone pairs on the oxygen atom.

Bond angle: 104.5°.

Polarity: The bond dipoles do not cancel because of the bent shape, making water polar. See Figure 7.8. This polarity is responsible for many of water’s unique properties, such as its high boiling point and its ability to dissolve many substances.

Figure 7.8: Bent shape of H₂O

2. Carbon Tetrachloride (CCl₄)

Polar Bonds: The C-Cl bonds are polar because chlorine is more electronegative than carbon.

Shape: CCl₄has a tetrahedral shape with bond angles of 109.5°.

Polarity: Even though the bonds are polar, the molecule is symmetrical, so the dipoles cancel out. See Figure 7.9.

Result: CCl₄is a non-polar molecule.

Figure 7.9: Tetrahedral shape of CCl₄

3. Ammonia (NH₃) The N-H bonds are polar because nitrogen (electronegativity 3.04) is more electronegative than hydrogen (2.20). The molecule has a trigonal pyramidal shape with bond angles of about 107°, caused by the lone pair on nitrogen.

Due to its shape, the bond dipoles do not cancel, making ammonia polar with a net dipole moment. This polarity allows ammonia to dissolve well in water and form hydrogen bonds

Figure 7.10: Trigonal pyramidal shape of NH₃

4. Sulphur Dioxide (SO2)

The bonds in SO₂are polar because oxygen pulls electrons more strongly than sulphur. The molecule has a bent shape with bond angles around 119°, caused by a lone pair of electrons on sulphur. Because of this shape, the bond pulls don’t cancel out, making SO₂a polar molecule. This polarity affects how SO₂behaves and reacts with other substances.

Figure 7.11: Bent shape of SO₂ Sigma Bonds (σ- Bonds) Sigma bonds are formed by the head-on (axial) overlap of atomic orbitals, resulting in electron density concentrated along the line connecting the nuclei of the bonded atoms. This type of overlap allows for maximum orbital interaction, making σ-bonds stronger than π\piπ-bonds.

Types of Overlap in Sigma Bonds

1. s−s Overlap: Two s-orbitals overlap directly.

Example: H₂molecule, where two hydrogen atoms form a sigma bond.

2. s−p Overlap: An s-orbital overlaps with a p-orbital.

Example: In CH₄, the carbon atom’s sp³-hybrid orbitals overlap with the s-orbitals hydrogen atoms.

3. p−p Overlap: Two p-orbitals overlap along their axes.

Example: In F₂, two fluorine atoms form a sigma bond through p−p overlap.

Sigma bonds are always present in single bonds and in the first bond of double or triple bonds.

Figure 7.12: Three types of sigma bonds between s-s, s-p, and p-p atomic orbitals Characteristics of Sigma Bonds

1. Sigma (σ) bonds are formed by the direct, head-on (axial) overlap of atomic orbitals.

2. They are always the first bond formed between two atoms and are found in all single bonds.

3. Sigma bonds are generally stronger than pi (π\piπ) bonds due to the greater overlap of orbitals.

4. The electron density in a sigma bond is concentrated along the internuclear axis (the line connecting the nuclei of the bonded atoms).

5. Sigma bonds have cylindrical symmetry around the bond axis, meaning the electron density is evenly distributed in a circular fashion.

6. Atoms connected by a sigma bond can freely rotate around the bond axis without breaking the bond.

7. They can be formed through s−s, s−p, or p−p orbital overlaps, and also through hybrid orbital overlaps (e.g., sp³−s).

Pi Bonds (π Bonds)

A pi (π) bond is a type of covalent bond formed by the sideways (lateral) overlap of atomic orbitals. It typically occurs in addition to a sigma (σ) bond in double or triple bonds.

Characteristics of Pi Bonds

1. Formed by the sideways overlap of unhybridized p-orbitals.

2. Found in double bonds (1 π-bond and 1 σ-bond) and triple bonds (2 π-bonds and 1 σ-bond).

3. The electron density in a π-bond is concentrated above and below the bond axis, not along it.

4. Weaker than sigma bonds due to the less effective overlap of orbitals.

5. π-bonds restrict rotation around the bond axis because breaking the sideways overlap requires energy.

Figure 7.13: pi bond between p-p atomic orbitals Differences Between Sigma (σ) and Pi (π) Bonds

Table 7.1: Differences Between σ and π Bonds Sigma (σ) Bond Pi (π) Bond Formed by the head-on (axial) overlap of atomic orbitals.

Formed by the sideways (lateral) overlap of p-orbitals.

Electron density is concentrated along the internuclear axis.

Electron density is above and below the internuclear axis.

Sigma (σ) Bond Pi (π) Bond

Stronger due to greater orbital overlap.

Weaker due to less effective orbital overlap.

Allows free rotation of bonded atoms around the bond axis.

Restricts rotation due to the rigidity of the bond.

Found in single, double, and triple bonds (always one σ).

Found only in double and triple bonds (accompanies a σ).

Can involve s−s, s−p, or p−p overlaps.

Involves p−p overlaps only.

Activity 7.1 Exploring Molecular Shapes, Polarity, and Bonding

Materials needed: A table of electronegativity values, clay or playdough, toothpicks a list of element pairs (e.g., H-F, C-H, Na-Cl, CH₄, NH₃, H₂O, CO₂, SO₂) Steps

1. Calculate electronegativity differences for each pair.

H-F, C-H, Na-Cl, CH₄, NH₃, H₂O

Classify the bonds as: Nonpolar covalent, polar covalent and ionic

2. a. Use VSEPR theory to predict shapes and bond angles.

H-F, C-H, Na-Cl, CH₄, NH₃, H₂O

b. Build 3D models representing molecular shapes (linear, trigonal planar, tetrahedral).

3. Predicting bond polarity

a. Identify bond polarities based on electronegativity differences:

CH₄,CO₂, SO₂

b. Predict molecular shapes using VSEPR theory.

c. Determine if the molecule is polar or nonpolar.

4. Analyse how molecular shape influences overall polarity.

Give examples of real-life implications

Activity 7.2 Exploring Molecular Orbitals and Bond Formation

Materials needed: Diagrams of s- and p-orbitals.

Steps

1. Brainstorm what is meant by sigma and pi bonds. Make notes for each of the definitions.

2. Draw diagrams to illustrate the overlaps. Make sure to included s-p, p-p and and s-s.

3. Use examples to explain sigma bonds (e.g., single bonds in H₂) and pi bonds (e.g., double bond in C₂H₄). Consider why each bond would apply to each scenario that you choose.

4. Compare sigma and pi bonds, ensure that you include mention of how the electrons are contained within these bond structures.

Hybridisation is a key concept in organic chemistry that explains the bonding and geometry of carbon-containing compounds. Carbon atoms form stable covalent bonds with other carbon atoms and elements, creating a wide variety of molecular structures. Hybridisation involves mixing atomic orbitals to create new hybrid orbitals, enabling carbon to form these diverse structures.

The three main types of hybridisation in organic compounds are:

• sp³: Tetrahedral geometry

• sp²: Trigonal planar geometry

• sp: Linear geometry.

Each type determines the molecular geometry and bonding patterns, influencing the stability and reactivity of organic molecules. Hybridisation is essential for understanding organic bonding.

Types of Hybridisation and Bonding

1. sp³ Hybridisation sp³ hybridisation occurs in alkanes, where each carbon atom forms four sigma (σ) bonds, resulting in a tetrahedral geometry with bond angles of approximately 109.5°.

Examples:

Methane (CH₄) Carbon forms four sp³-hybrid orbitals, each overlapping with hydrogen’s 1s-orbital to form four σ-bonds.

Ethane (C₂H₆) Each carbon atom starts with 1s²2s²2p², having only two unpaired electrons in the 2p-orbitals.

One electron from the 2s-orbital moves to the 2p_(z)-orbital, creating four unpaired electrons with the configuration 1s²2s¹2p³.

The 2s-orbital mixes with the 2pₓ, 2p_(y), and 2p_(z)-orbitals, forming four equivalent sp³-hybrid orbitals, each with 75% p character and 25% s character.

Each carbon uses one sp³-orbital to overlap with the other carbon’s sp³-orbital, forming a single σ-bond.

The remaining three sp³-orbitals on each carbon overlap with hydrogen’s 1s¹- orbitals, forming six σ-bonds (three per carbon).

The four sp³-orbitals arrange in a tetrahedral shape to minimise repulsion, with bond angles close to 109.5°.

2. sp² Hybridisation sp² hybridisation occurs in alkenes like ethene (C₂H₄).

Each carbon atom forms three sigma (σ) bonds and one pi (π) bond.

The sp² hybrid orbitals lie in a plane with bond angles of approximately 120°.

The unhybridised 2p_(z)orbital forms the π-bond.

Each carbon atom has the electron configuration 1s²2s²2p².

There are two unpaired electrons in the 2p orbitals—insufficient for forming four bonds.

One electron from the 2s orbital is promoted to the empty 2p orbital.

New configuration: 1s²2s¹2pₓ ¹2p_(y) ¹2p_(z) ¹.

This results in four unpaired electrons.

The 2s orbital mixes with two 2p orbitals (2pₓand 2p_(y)).

Forms three equivalent sp² hybrid orbitals.

One 2p_(z)orbital remains unhybridised.

Each sp² hybrid orbital has 33% s character and 67% p character.

The three sp² hybrid orbitals arrange themselves 120° apart in a plane to minimize electron repulsion.

Each carbon atom uses its three sp² hybrid orbitals to form σ-bonds:

Two σ-bonds with hydrogen atoms.

One σ-bond with the other carbon atom.

Pi (π) Bond Formation: The unhybridised 2p_(z)orbitals on each carbon overlap side-by-side. This forms a π-bond above and below the plane of the molecule.

Resulting double bond: The double bond between the carbon atoms consists of:

One σ-bond (from sp²-sp² overlap).

One π-bond (from 2p_(z)overlap).

3. sp Hybridisation sp hybridisation occurs in alkynes like ethyne (C₂H₂).

Each carbon atom forms two sigma (σ) bonds and two pi (π) bonds.

The sp hybrid orbitals are linearly arranged with bond angles of 180°.

Electron Configuration and Hybridisation:

Each carbon atom starts with 1s²2s²2pₓ ¹2p_(y) ¹.

There are two unpaired electrons—insufficient for forming four bonds.

One electron from the 2s-orbital moves to the empty 2p_(z) ²-orbital.

New configuration: 1s²2s¹2pₓ ¹2p_(y) ¹2p_(z) ¹.

The 2s-orbital mixes with one 2pₓ-orbital to form two sp hybrid orbitals.

50% s character and 50% p character.

The remaining 2p_(y)-orbitals remain unhybridised.

The two sp hybrid orbitals arrange themselves 180° apart to minimize repulsion.

One sp orbital from each carbon overlaps to form a σ-bond between the carbons.

The other sp orbitals overlap with hydrogen’s 1s-orbitals to form two sigma bonds (one per carbon).

The unhybridised 2p_(y)and 2p_(z)-orbitals on each carbon overlap side-by-side to form two π-bonds.

The triple bond between the two carbon atoms consists of:

One σ-bond (head-on sp-sp overlap).

Two π-bonds (sideways overlap of 2p_(y)and 2p_(z)).

Activity 7.3 Building Carbon-Carbon Double and Triple Bonds

Material needed: Molecular model kits Organise yourselves into groups of no more than five for this activity.

Steps

1. Building a carbon-carbon double bond

a. Identify the following components:

2 carbon atoms, 4 hydrogen atoms, and 2 connectors for the double bond.

b. Connect two carbon atoms using two connectors to represent the double bond.

c. Attach two hydrogen atoms to each carbon atom using single connectors.

d. Observe the planar geometry with bond angles of 120°.

2. Building a Carbon-Carbon Triple Bond (10 minutes)

a. Identify the following components:

2 carbon atoms, 2 hydrogen atoms, and 3 connectors for the triple bond.

b. Connect two carbon atoms using three connectors to represent the triple bond.

c. Attach one hydrogen atom to each carbon atom using single connectors.

d. Observe the linear geometry with bond angles of 180°.

3. Present your model to the class, explaining:

a. How you built the double and triple bonds.

b. The bond types and geometry of each structure.

c. What is the difference between double and triple bonds?

d. How does geometry affect molecular properties?

Activity 7.4 Exploring sigma and pi bonds in unsaturated hydrocarbons Organise yourselves into groups of no more than five for this activity Materials needed: Molecular model kits Videos, charts, and pictures illustrating:

Formation of sigma (σ) and pi (π) bonds.

Orbital overlaps in double and triple bonds.

Examples of unsaturated hydrocarbons such as ethene (C₂H₄) and ethyne (C₂H₂).

Steps

1. Watch videos showing the formation of σ- and π-bonds. https://www.

youtube.com/watch?v= i2WY_JElXlU

a. Study charts and pictures, analyse diagrams, focusing on:

i. Sigma bonds.

ii. Pi bonds

b. Discuss your observations with the class.

2. Build the following molecular model of ethene and ethyne using the following steps:

Ethene (C₂H₄)

a. Two sigma bonds per carbon (one with hydrogen, one with the other carbon).

b. One pi bond between the carbons.

c. Observe planar geometry with 120° bond angles.

Ethyne (C₂H₂)

a. One sigma bond and two pi bonds between carbons.

b. One sigma bond between each carbon and hydrogen.

c. Observe linear geometry with 180° bond angles.

3. Using the model you have created, compare

a. the geometry of the molecules,

b. how this geometry is influenced by the bonding within the molecules.

Review Questions 7.1

1. What is a sigma bond, and how is it different from a pi bond?

2. Identify the shape and bond angle of CH₄.

3. Predict the shape of NH₃and explain how the lone pair affects the bond angles.

4. Distinguish between the types of bonds in C₂H₄(ethylene).

5. Predict the molecular shape and bond angles of SO₂using VSEPR theory and determine if it is polar or nonpolar.

6. Explain how sigma and pi bonds contribute to the rigidity of C₂H₄.

7. Analyse the bonding in C₂H₂(ethyne) and predict its geometry, bond angles, and the type of bonds present.

8. Compare the molecular shapes of BF₃, NH₃, and H₂O using VSEPR theory and discuss the role of lone pairs.

Review Questions 7.2

1. What is hybridisation?

2. Identify the three main types of hybridisation in organic compounds.

3. How does hybridisation explain the structure of methane (CH₄)?

4. Explain the difference between sigma (σ) and pi (π) bonds.

5. Apply hybridisation to explain the bonding in ethene (C₂H₄).

6. How does hybridisation explain the structure and bonding in ethyne (C₂H₂)?

7. Compare the geometry and bonding in methane (CH₄), ethene (C₂H₄), and ethyne (C₂H₂) based on hybridisation.

8. Evaluate how the presence of pi bonds affects the rigidity of molecules.