Class 11 Chemistry – Chapter 3: Chemical Bonding (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Chapter 3 – Chemical Bonding according to the Federal Board (FBISE) syllabus. Students will find clear explanations of concepts such as ionic and covalent bonding, electronegativity, MOT VBT structures, VSEPR theory, hybridization, molecular geometry, and intermolecular forces.
For better understanding, visual learning, and exam preparation, we also provide video lectures, MCQs practice, test series, and live sessions on our official YouTube channel.
▶ Visit AmurChem YouTube Channel 💬 Join Our WhatsApp Channel3.1 Electronegativity: Key Notes
Definition: Electronegativity is the power of a covalently bonded atom to attract a shared pair of electrons towards itself in a molecule. It is a dimensionless property (no units) measured most commonly on the Pauling Scale.
3.1.1 Factors Affecting Electronegativity
- Size of Atom: Smaller atoms have a stronger pull on electrons because the nucleus is closer to the shared pair.
- Nuclear Charge: More protons in the nucleus increase the positive charge, enhancing the attraction for electrons.
- Screening/Shielding Effect: Inner shell electrons block the nucleus's pull. More shells mean lower electronegativity.
3.1.2 Trends in the Periodic Table
| Direction | Trend | Reasoning |
|---|---|---|
| Across a Period (Left to Right) | Increases | Nuclear charge increases and atomic size decreases; shielding effect remains constant. |
| Down a Group (Top to Bottom) | Decreases | Atomic size increases and shielding effect increases due to more shells, overcoming the increased nuclear charge. |
3.1.3 Nature of Bond vs. Electronegativity Difference ($\Delta EN$)
The difference in electronegativity between two atoms determines the type of chemical bond formed:
- Ionic Bond: $\Delta EN > 1.8$
- Polar Covalent Bond: $0.4 \leq \Delta EN \leq 1.8$
- Nonpolar Covalent Bond: $\Delta EN < 0.4$
Relevant Questions & Answers
Q1: Why does fluorine have a higher electronegativity than iodine?
A: Both are in the same group, but fluorine is at the top. As you move down to iodine, the number of shells increases (shielding effect), and the atomic size increases. This weakens the nucleus's attraction for shared electrons, making iodine less electronegative than fluorine.
Q2: Calculate the electronegativity difference for $HCl$ and identify its bond type. (Given: $H = 2.20$, $Cl = 3.16$)
A:
$$\Delta EN = 3.16 - 2.20 = 0.96$$
Since $0.96$ falls between $0.4$ and $1.8$, the bond is Polar Covalent.
Q3: Why is electronegativity considered a "dimensionless" property?
A: It is considered dimensionless because it is not a direct physical measurement (like mass or length), but rather a relative "tendency" or "power" of an atom compared to others on a specific scale.
Q4: Predict the bond type of $KCl$ if $K = 0.82$ and $Cl = 3.16$.
A:
$$\Delta EN = 3.16 - 0.82 = 2.34$$
Because $2.34 > 1.8$, the bond is Ionic.
3.1.4 Covalent Character in a Compound
Covalent character refers to the sharing of electrons within what might otherwise be considered an ionic bond.
- Polarising Power: Covalent character depends on the polarizing power of the cation, which is influenced by its oxidation state.
- Trend: Higher polarizing power in a cation leads to greater covalent character.
- Periodic Trend: Covalent character increases when moving from left to right across a period.
- Example: Aluminum chloride ($AlCl_3$) is more covalent than magnesium chloride ($MgCl_2$) because the $Al^{3+}$ ion has a higher oxidation state ($+3$) and higher polarizing power than $Mg^{2+}$ ($+2$), causing it to polarize the anion and share electrons.
3.2 Dipole Moment ($\mu$)
Dipole moment is the product of the magnitude of the charges ($q$) and the distance between them ($r$).
$$\mu = q \cdot r$$
- Units: The SI unit is the debye (D).
- Conversion: $1 \text{ Debye} = 3.335 \times 10^{-30} \text{ Cm}$ (Coulomb meter).
- Function: It determines the overall polarity of a compound based on electronegativity differences.
3.3 Polar and Nonpolar Covalent Bonds
- Polar Covalent Bond: Formed when atoms have different electronegativities; the more electronegative atom attracts shared electrons, gaining a partial negative charge ($\delta^-$), while the other gains a partial positive charge ($\delta^+$).
- Nonpolar Molecules: Symmetrical polyatomic molecules (Linear, Trigonal Planar, Tetrahedral) are often non-polar because bond dipoles cancel each other out.
- Example ($CO_2$): Although the $C=O$ bond is polar, the linear molecule $CO_2$ is non-polar because the dipole moments cancel ($0.7 - 0.7 = 0$).
Relevant Questions & Answers
Q1: Why does covalent character increase from left to right in a period?
A: As you move left to right, the oxidation state of the cation increases. This increases its polarizing power, allowing it to pull electron density from the anion and develop covalent character.
Q2: Is $Cl_2$ polar or non-polar? (Using values: $Cl = 3.0$)
A: It is non-polar. Since both atoms are the same, the electronegativity difference is zero, and the pair of electrons is shared equally.
Q3: Explain why $CBr_4$ (tetrahedral) is non-polar despite having polar $C-Br$ bonds.
A: $CBr_4$ is a symmetrical tetrahedral molecule. In such symmetrical shapes, the individual bond dipoles exert equal and opposite effects, effectively canceling each other out and resulting in zero net dipole moment.
Q4: Define a Debye in terms of Coulomb meters.
A: One Debye is equal to $3.335 \times 10^{-30} \text{ Cm}$.
3.4 Bond Energy (Bond Enthalpy)
Definition: The energy required to break one mole of bonds of the same type in a substance is known as bond energy. It is measured in $$kJmol^{-1}$$.
Factors Affecting Bond Energy
- Electronegativity Difference: Greater difference leads to more polar bonds and higher bond energy.
- Size of Atom: Smaller atoms generally form stronger bonds.
- Bond Order: The number of chemical bonds between a pair of atoms.
- Bond Length: Shorter bonds are typically stronger and require more energy to break.
Data Analysis: Hydrogen Halides (HX)
| Bond | Bond Energy ($$kJmol^{-1}$$) | Electronegativity Difference |
|---|---|---|
| H—F | 562 | 1.8 |
| H—Cl | 431 | 0.9 |
| H—Br | 366 | 0.7 |
| H—I | 299 | 0.4 |
Conclusion: As the electronegativity difference decreases from HF to HI, the bond energy also decreases.
Bond Length and Halogen Trends
Bond Length: The distance between the nuclei of two covalently bonded atoms, measured in Angstroms ($$Ã…$$) or picometers ($$pm$$). $$1 pm = 10^{-12} m$$.
Data Analysis: Halogen Molecules ($$X_2$$)
| Bond | Bond Energy ($$kJmol^{-1}$$) | Bond Length ($$pm$$) |
|---|---|---|
| F—F | 158 | 149 |
| Cl—Cl | 242 | |
| Br—Br | 193 | 228 |
| I—I | 151 | 266 |
The Fluorine Anomaly: Although bond energy generally decreases as atomic radius increases (from Cl to I), Fluorine ($$F_2$$) has a lower bond energy than Chlorine ($$Cl_2$$). This is due to the extremely small size of fluorine atoms, which causes their lone pairs to repel each other strongly, weakening the bond.
Review Questions and Answers
Q1: Why is the bond energy of $$F_2$$ lower than that of $$Cl_2$$?
A: Due to the small size of Fluorine atoms, the lone pairs of electrons on each atom are very close together and exert a strong repulsive force on each other. This repulsion makes it easier to break the bond, resulting in lower bond energy compared to Chlorine.
Q2: How does bond energy relate to the reactivity of a molecule?
A: Higher bond energy means a stronger, more stable bond, making the molecule less reactive. For example, alkyl fluorides are less reactive than alkyl chlorides because the C—F bond energy is greater than the C—Cl bond energy.
Q3: What is the relationship between bond length and bond energy?
A: Generally, they are inversely proportional. Smaller atoms have shorter bond lengths, which leads to stronger attraction between nuclei and shared electrons, resulting in higher bond energy.
Q4: Why does HF have the highest bond energy among hydrogen halides?
A: HF has the greatest electronegativity difference ($$1.8$$) between the bonded atoms, making the bond highly polar and strong. Additionally, Fluorine is the smallest halogen, resulting in a very short bond length.
Valence Shell Electron Pair Repulsion (VSEPR) Theory
Core Principles
VSEPR theory is used to predict the 3D molecular geometry and bond angles of a molecule based on the number of valence electron pairs (both bonding and lone pairs) surrounding a central atom.
- Repulsion: Electrons are negatively charged and repel each other. Electron pairs stay as far apart as possible to minimize these repulsive forces.
- Repulsion Order: Lone pairs ($LP$) occupy more space than bonding pairs ($BP$). The strength of repulsion follows this order:
$$LP-LP > LP-BP > BP-BP$$
Molecular Geometries
| Electron Pairs | Type (VSEPR) | Shape | Bond Angle | Example |
|---|---|---|---|---|
| 2 BP, 0 LP | $AX_2$ | Linear | 180° | $BeF_2$ |
| 3 BP, 0 LP | $AX_3$ | Trigonal Planar | 120° | $BCl_3$, $CO_3^{2-}$ |
| 2 BP, 1 LP | $AX_2E$ | Bent / V-shaped | < 120° | $SO_2$ |
| 4 BP, 0 LP | $AX_4$ | Tetrahedral | 109.5° | $CH_4$ |
| 3 BP, 1 LP | $AX_3E$ | Trigonal Pyramidal | < 109.5° (approx. 107°) | $NH_3$ |
| 2 BP, 2 LP | $AX_2E_2$ | Bent / V-shaped | < 107° (approx. 104.5°) | $H_2O$ |
| 5 BP, 0 LP | $AX_5$ | Trigonal Bipyramidal | 90° and 120° | $PCl_5$ |
| 6 BP, 0 LP | $AX_6$ | Octahedral | 90° | $SF_6$ |
Questions and Answers
Q1: Why is the bond angle in $H_2O$ (104.5°) smaller than in $NH_3$ (107°)?
A: Water has two lone pairs on the central oxygen atom, whereas ammonia has only one lone pair on the nitrogen. Since lone pair-lone pair ($LP-LP$) repulsion is stronger than lone pair-bonding pair ($LP-BP$) repulsion, the two lone pairs in $H_2O$ push the O-H bonds closer together, resulting in a smaller bond angle.
Q2: What is meant by an "expanded octet"? Give an example.
A: An expanded octet occurs when a central atom is surrounded by more than eight valence electrons. This is common in elements from Period 3 and below. An example is $SF_6$, where sulfur is surrounded by six bonding pairs, totaling twelve valence electrons.
Q3: Explain the geometry of $PCl_5$.
A: $PCl_5$ has five bonding pairs and no lone pairs ($AX_5$). Its shape is Trigonal Bipyramidal. Three chlorine atoms lie in a horizontal plane with 120° angles (equatorial), and two chlorine atoms are positioned vertically above and below this plane at 90° (axial).
Q4: How does the "space" occupied by a lone pair affect molecular shape?
A: Lone pairs are more concentrated and closer to the nucleus of the central atom than bonding pairs. Their electron clouds are wider, causing more repulsion against neighboring pairs. This "extra space" taken by lone pairs forces the bonding pairs closer together, deviating from ideal geometric angles.
Study Notes: Valence Bond Theory (VBT) & Chemical Bonding
1. Core Principles of Valence Bond Theory
Valence Bond Theory explains how covalent bonds are formed through the interaction of atomic orbitals. Key postulates include:
- Orbital Overlap: A covalent bond forms when half-filled atomic orbitals from two different atoms overlap. The orbitals share a common region in space.
- Electron Spin: The electrons in the overlapping orbitals must have opposite spins to pair up.
- Identity Retention: Even after overlapping, the atomic orbitals largely retain their individual character.
- Bond Strength: The strength of the bond is directly proportional to the extent of the overlap. Greater overlap leads to a more stable, lower-energy state.
- Bond Axis: The imaginary line joining the nuclei of the two bonded atoms is called the bond axis.
2. Types of Covalent Bonds
Based on the method of overlapping, there are two primary types of bonds:
A. Sigma Bond ($\sigma$)
Formed by the head-on (axial) overlapping of atomic orbitals along the bond axis. It is the strongest type of covalent bond and is present in all single bonds.
- s-s overlap: Two s-orbitals overlap.
- s-p overlap: An s-orbital overlaps with a p-orbital.
- p-p overlap: Two p-orbitals overlap end-to-end.
B. Pi Bond ($\pi$)
Formed by the lateral (sideways) overlapping of atomic orbitals. This occurs only when a sigma bond has already been established between two atoms (forming double or triple bonds).
3. Practice Exercise: Molecular Geometry & Structures
Questions & Answers
| Molecule | Question | Answer/Prediction |
|---|---|---|
| $CCl_{4}$ | Predict the shape | Tetrahedral (4 bond pairs, 0 lone pairs) |
| $BeCl_{2}$ | Predict the shape | Linear (2 bond pairs, 0 lone pairs) |
| $PCl_{3}$ | Predict the shape | Trigonal Pyramidal (3 bond pairs, 1 lone pair) |
| $H_{2}S$ | Predict the shape | Bent / V-shaped (2 bond pairs, 2 lone pairs) |
| $PH_{3}$ | Predict the shape | Trigonal Pyramidal (3 bond pairs, 1 lone pair) |
| $SeF_{6}$ | Predict the shape | Octahedral (6 bond pairs, 0 lone pairs) |
| $PCl_{5}$ | Draw the shape | Trigonal Bipyramidal |
Note: For dot-and-cross diagrams, ensure the central atom's valence electrons are represented by dots/crosses and paired with the electrons from the surrounding atoms to achieve stable configurations (Octet or Expanded Octet).
Study Notes: Sigma ($\sigma$) and Pi ($\pi$) Bonds
1. Nature and Strength of Bonds
- Sigma Bond ($\sigma$): This bond is stronger than a pi bond. The electron density is concentrated directly between the line joining the two nuclei.
- Pi Bond ($\pi$): This bond is weaker and more diffuse because the electron density is located above and below the plane of the nuclei. Less energy is required to break this bond.
- Multiple Bonds:
- A double bond consists of one sigma ($\sigma$) bond and one pi ($\pi$) bond.
- A triple bond consists of one sigma ($\sigma$) bond and two pi ($\pi$) bonds.
2. Applications of Valence Bond Theory (VBT)
VBT explains the formation of various molecules through orbital overlapping:
A. Single Bond Formation ($\sigma$ bonds)
- $H_{2}$ Molecule: Formed by the overlap of two half-filled $1s^{1}$ orbitals.
- $F_{2}$ Molecule: Formed by the head-on overlap of two half-filled $2p_{z}$ orbitals.
- HF Molecule: Formed by the overlap of a half-filled $1s$ orbital of Hydrogen and a half-filled $2p_{z}$ orbital of Fluorine.
B. Multiple Bond Formation
- $O_{2}$ Molecule (Double Bond): Each Oxygen atom has two half-filled orbitals ($2p_{y}$ and $2p_{z}$). One bond is a sigma bond (head-on overlap), and the other is a pi bond (parallel/lateral overlap).
- $N_{2}$ Molecule (Triple Bond): Each Nitrogen atom has three half-filled $p$ orbitals. One sigma bond is formed by head-on overlap of $2p_{x}$ orbitals, while two pi bonds are formed by the parallel overlap of $2p_{y}$ and $2p_{z}$ orbitals.
3. Relevant Questions & Answers
Q1: Why is a sigma bond stronger than a pi bond?
A: A sigma bond is stronger because the orbital overlap occurs directly along the internuclear axis, concentrating electron density between the nuclei. In contrast, pi bonds result from lateral overlap, making them more diffuse and easier to break.
Q2: Describe the bonding in a Nitrogen ($N_{2}$) molecule according to VBT.
A: Nitrogen has the electronic configuration $1s^{2} 2s^{2} 2p^{3}$. The three half-filled $2p$ orbitals overlap to form three bonds: one sigma bond (head-on) and two pi bonds (lateral), resulting in a triple bond.
Q3: What types of bonds are present in Ethene ($C_{2}H_{4}$)?
A: In Ethene, the carbon-carbon double bond consists of one sigma bond and one pi bond. The carbon-hydrogen bonds are all sigma bonds.
Study Notes: Applications of VBT & Hybridization
1. Bond Strength and Structures
- Bond Composition:
- Double bonds consist of one sigma ($\sigma$) and one pi ($\pi$) bond.
- Triple bonds consist of one sigma ($\sigma$) and two pi ($\pi$) bonds.
- Relative Strength: Sigma bonds are stronger than pi bonds because electron density is concentrated between the nuclei, whereas in pi bonds, it is diffuse and located above/below the plane.
2. Formation of Specific Molecules
- Hydrogen ($H_{2}$): Formed by the overlap of two $1s^{1}$ orbitals, creating a $\sigma$ bond.
- Fluorine ($F_{2}$): Formed by head-on overlap of half-filled $2p_{z}$ orbitals.
- Hydrogen Fluoride (HF): Formed by the overlap of the $1s$ orbital of H and the $2p_{z}$ orbital of F.
- Oxygen ($O_{2}$): Involves one head-on overlap ($\sigma$ bond) and one parallel overlap ($\pi$ bond).
- Nitrogen ($N_{2}$): Contains a triple bond consisting of one head-on $\sigma$ overlap and two lateral $\pi$ overlaps.
3. Concept of Hybridization
Hybridization is the mixing of atomic orbitals of different energies and shapes to produce a new set of equivalent hybrid orbitals.
Types of Hybridization:
- sp Hybridization: Mixing of one $s$ and one $p$ orbital to form two linear hybrid orbitals (e.g., $BeCl_{2}$, Acetylene $C_{2}H_{2}$).
- $sp^{2}$ Hybridization: Mixing of one $s$ and two $p$ orbitals to form three trigonal planar hybrid orbitals (e.g., $BCl_{3}$).
4. Relevant Questions & Answers
Q1: Why is Acetylene ($C_{2}H_{2}$) linear in shape?
A: Each carbon atom undergoes $sp$ hybridization, mixing its $2s$ and one $2p$ orbital to form two $sp$ hybrid orbitals arranged linearly at a $180^{\circ}$ angle.
Q2: Compare the electronic configuration of Beryllium in ground vs. excited states for $sp$ hybridization.
A: In the ground state, Be is $1s^{2} 2s^{2}$ (no unpaired electrons). In the excited state, one $2s$ electron is promoted to an empty $2p$ orbital, resulting in $1s^{2} 2s^{1} 2p^{1}$, allowing for two hybrid bonds.
Q3: Describe the bonding in Boron Trichloride ($BCl_{3}$).
A: Boron undergoes $sp^{2}$ hybridization using one $2s$ and two $2p$ orbitals to form three half-filled hybrid orbitals oriented in a trigonal planar symmetry.
Study Notes: Applications of Valence Bond Theory and Hybridization
1. Sigma ($\sigma$) and Pi ($\pi$) Bonds
- Bond Nature: When a double bond exists between two atoms, one is a sigma bond and the other is a pi bond. A triple bond consists of one sigma and two pi bonds.
- Strength: Sigma bonds are stronger than pi bonds. In a sigma bond, electron density is concentrated between the nuclei along the bond axis. In a pi bond, electron density is located above and below the plane of the nuclei, making it more diffuse and easier to break.
2. Orbital Overlapping in Molecules
- Hydrogen ($H_2$): Formed by the overlap of two $1s^1$ orbitals, creating a sigma bond where electron density is between the nuclei.
- Fluorine ($F_2$): Formed by the head-on overlap of half-filled $2p_z$ orbitals, resulting in a sigma bond.
- Hydrogen Fluoride (HF): Formed by the overlap of the half-filled $1s$ orbital of H with the half-filled $2p_z$ orbital of F.
- Oxygen ($O_2$): Contains one sigma bond (head-on overlap of $p$ orbitals) and one pi bond (parallel overlap of $p$ orbitals).
- Nitrogen ($N_2$): Formed by three bonds: one sigma bond (head-on $2p_x$ overlap) and two pi bonds (parallel $2p_y$ and $2p_z$ overlaps).
3. Hybridization
Hybridization is the intermixing of atomic orbitals of different energies and shapes to produce a new set of equivalent "hybrid" orbitals.
A. $sp$ Hybridization
- Definition: Mixing of one $s$ and one $p$ orbital to form two $sp$ hybrid orbitals arranged linearly ($180^\circ$ angle).
- Beryllium Chloride ($BeCl_2$): Be promotes a $2s$ electron to $2p$ and hybridizes to form a linear molecule.
- Acetylene ($C_2H_2$): Each Carbon undergoes $sp$ hybridization to form one sigma bond with H and one sigma bond with the other Carbon. The two unhybridized $p$ orbitals on each C form two pi bonds, resulting in a triple bond.
B. $sp^2$ Hybridization
- Definition: Mixing of one $s$ and two $p$ orbitals to form three hybrid orbitals oriented in a trigonal planar symmetry.
- Boron Trichloride ($BCl_3$): Boron hybridizes its $2s$ and two $2p$ orbitals to form three bonds with Chlorine.
C. $sp^3$ Hybridization
- Definition: Mixing of one $s$ and three $p$ orbitals to produce four hybrid orbitals in a tetrahedral geometry.
- Methane ($CH_4$): Carbon forms four $\sigma$ $sp^3$-$s$ bonds with Hydrogen. Bond angle: $109.5^\circ$.
- Ammonia ($NH_3$): Nitrogen undergoes $sp^3$ hybridization. It has three bond pairs and one lone pair. Repulsion from the lone pair reduces the bond angle to $107^\circ$, resulting in a trigonal pyramidal shape.
- Water ($H_2O$): Oxygen undergoes $sp^3$ hybridization with two bond pairs and two lone pairs. Repulsion from two lone pairs reduces the bond angle to $104.5^\circ$, giving it a bent or V-shape.
Relevant Questions and Answers
Q1: Why is the bond angle in $NH_3$ ($107^\circ$) less than the ideal tetrahedral angle ($109.5^\circ$)?A: According to VBT and hybridization, Ammonia has one lone pair of electrons. This lone pair exerts more repulsion on the bond pairs than the bond pairs exert on each other, causing the bond angles to contract.
Q2: Describe the bonding in Acetylene ($C_2H_2$) based on hybridization.A: Each carbon atom undergoes $sp$ hybridization, creating two hybrid orbitals at $180^\circ$. One hybrid orbital forms a $\sigma$ bond with a Hydrogen $1s$ orbital, and the other forms a $\sigma$ bond with the second Carbon. The two remaining unhybridized $p$ orbitals on each Carbon atom overlap laterally to form two $\pi$ bonds, resulting in a C-C triple bond.
Q3: What is the electronic configuration of a Carbon atom in its excited state for $sp^3$ hybridization?A: In the excited state, Carbon's electronic configuration is $1s^2 2s^1 2p_x^1 2p_y^1 2p_z^1$.
Q4: How many sigma and pi bonds are found in a molecule of Ethene?A: Ethene contains five sigma bonds (one C-C and four C-H) and one pi bond (part of the C=C double bond).
Valence Bond Theory (VBT)
Proposed by Heitler and London (1927) and developed by Pauling, VBT explains bond energies, lengths, and molecular shapes.
- Orbital Overlap: Covalent bonds form when partially filled atomic orbitals of two atoms overlap, sharing a common region in space while retaining their identity.
- Electron Spin: Electrons in overlapping orbitals must have opposite spins.
- Bond Count: The number of bonds equals the number of unpaired electrons in the atom's outermost shell.
- Bond Axis: Overlapping orbitals must have the same symmetry relative to the bond axis (the line joining the nuclei).
- Energy and Strength: Energy is released during overlapping; greater overlap results in more energy released and a stronger bond.
Types of Covalent Bonds
Sigma Bond ($\sigma$)
Formed by the head-on overlapping of atomic orbitals. It occurs between $s-s$, $s-p$, and $p-p$ (axial) orbitals.
Pi Bond ($\pi$)
Formed by the parallel or lateral overlap of half-filled $p$ orbitals.
- Single Bond: Always a sigma bond.
- Double Bond: One sigma ($\sigma$) and one pi ($\pi$) bond.
- Triple Bond: One sigma ($\sigma$) and two pi ($\pi$) bonds.
- Relative Strength: Sigma bonds are stronger than pi bonds because electron density is concentrated between the nuclei, whereas pi bond density is diffuse and located above/below the plane.
Hybridization
The mixing of atomic orbitals of different energies and shapes to produce a new set of equivalent "hybrid" orbitals.
Types of Hybridization
- $sp$ Hybridization: Mixing one $s$ and one $p$ orbital to form two linear orbitals ($180^\circ$ angle). Example: $BeCl_2$, $C_2H_2$.
- $sp^2$ Hybridization: Mixing one $s$ and two $p$ orbitals to form three trigonal planar orbitals. Example: $BCl_3$.
- $sp^3$ Hybridization: Mixing one $s$ and three $p$ orbitals to produce four tetrahedral orbitals.
Molecular Examples ($sp^3$)
| Molecule | Geometry | Bond Angle | Notes |
|---|---|---|---|
| Methane ($CH_4$) | Tetrahedral | $109.5^\circ$ | Four $\sigma$ $sp^3-s$ bonds. |
| Ammonia ($NH_3$) | Trigonal Pyramidal | $107^\circ$ | Reduced angle due to lone pair repulsion. |
| Water ($H_2O$) | Bent / V-shape | $104.5^\circ$ | Reduced angle due to two lone pairs. |
Co-ordinate Covalent Bond
Formed when one atom provides both electrons for a shared pair.
- Donor: Atom with a lone pair.
- Acceptor: Electron-deficient atom with an empty orbital.
- Example: Formation of Hydronium ion ($H_3O^+$) from water and a proton.
Intermolecular Forces
Forces present between covalent molecules that explain physical properties like boiling points.
- Permanent Dipole-Dipole: Between polar molecules with electronegativity differences.
- Instantaneous Dipole-Induced Dipole (Van der Waals): Between nonpolar molecules due to temporary uneven electron distribution. Strength increases with molecular size and mass.
Review Questions & Answers
Q1: Why is the boiling point of Propane ($C_3H_8$) higher than Ethane ($C_2H_6$)?A: Propane has a greater molecular mass ($44$ vs $30$), leading to stronger Van der Waals forces.
Q2: Predict the shape of Selenium Hexafluoride ($SeF_6$).A: Octahedral.
Q3: Describe the bonding in Nitrogen ($N_2$).A: It contains a triple bond formed by one head-on $2p_x$ overlap ($\sigma$ bond) and two lateral $2p_y$ and $2p_z$ overlaps (two $\pi$ bonds).
Q4: What is the bond angle in Beryllium Chloride ($BeCl_2$)?A: $180^\circ$ due to linear $sp$ hybridization.
Valence Bond Theory and Intermolecular Forces
1. Valence Bond Theory (VBT)
Proposed by Heitler and London (1927) and developed by Pauling, VBT explains bond energies, lengths, and molecular shapes.
- Overlap of Orbitals: Covalent bonds form when partially filled atomic orbitals of two atoms overlap.
- Identity Retention: Overlapping orbitals retain their individual identity.
- Electron Spin: Electrons in overlapping orbitals must have opposite spins.
- Bond Strength: Greater overlap results in more energy released and a stronger bond.
2. Types of Covalent Bonds
Sigma Bond ($\sigma$)
Formed by the head-on overlap of atomic orbitals. It is the bond present in single covalent bonds.
- Types include s-s, s-p, and p-p axial overlap.
Pi Bond ($\pi$)
Formed by the lateral (sideways) overlap of half-filled p-orbitals.
- Strength: $\sigma$ bonds are stronger than $\pi$ bonds because $\sigma$ bond electron density is concentrated between nuclei, while $\pi$ bond density is diffuse and located above and below the nuclear plane.
- Multiple Bonds: A double bond contains one $\sigma$ and one $\pi$ bond; a triple bond contains one $\sigma$ and two $\pi$ bonds.
3. Hybridization
The concept of mixing atomic orbitals of different energies and shapes to produce new, equivalent "hybrid" orbitals.
- sp Hybridization: One s and one p orbital mix to form two linear orbitals ($180^\circ$ angle). Example: $BeCl_2$.
- $sp^2$ Hybridization: One s and two p orbitals mix to form three trigonal planar orbitals ($120^\circ$ angle). Example: $BCl_3$.
- $sp^3$ Hybridization: One s and three p orbitals mix to form four tetrahedral orbitals.
- Methane ($CH_4$): Tetrahedral shape, $109.5^\circ$ bond angle.
- Ammonia ($NH_3$): Trigonal pyramidal shape, $107^\circ$ angle due to lone pair repulsion.
- Water ($H_2O$): Bent/Angular shape, $104.5^\circ$ angle due to repulsion from two lone pairs.
4. Intermolecular Forces
Forces present between covalent molecules that determine physical properties.
- Permanent Dipole-Dipole: Occur between polar molecules with electronegativity differences.
- Instantaneous Dipole-Induced Dipole (Van der Waals): Present in non-polar molecules due to temporary uneven electron distribution. Strength increases with molecular mass and size.
- Hydrogen Bonding: A strong attraction between a hydrogen atom bonded to a highly electronegative element (F, N, O) and the lone pair of another such atom.
- Ice is less dense than water because hydrogen bonding creates an open, rigid lattice in the solid state.
Relevant Questions and Answers
Q1: Why is a sigma bond stronger than a pi bond?A: In a sigma bond, the electron density is concentrated directly between the two nuclei along the bond axis. In a pi bond, the electron density is located above and below the plane of the nuclei, making the bond more diffuse and requiring less energy to break.
Q2: Predict the shape of $BeCl_2$ and explain its bonding.A: $BeCl_2$ has a linear shape with a bond angle of $180^\circ$. The Beryllium atom undergoes $sp$ hybridization by mixing one $2s$ and one $2p$ orbital to form two equivalent hybrid orbitals.
Q3: Explain the trend in boiling points of Alkanes (Methane $-164^\circ C$ to Butane $-1^\circ C$).A: As the molecular size and mass increase from Methane to Butane, the strength of the Van der Waals (instantaneous dipole-induced dipole) forces increases. Stronger intermolecular forces require more energy to overcome, resulting in higher boiling points.
Q4: Why does ice float on water?A: In the solid state (ice), water molecules form a three-dimensional hydrogen-bonded network that creates a "more open" rigid lattice. This causes water in the solid state to occupy more space, making ice less dense than liquid water.
Q5: Describe the composition of the triple bond in a Nitrogen ($N_2$) molecule.A: According to VBT, the triple bond in $N_2$ consists of one sigma ($\sigma$) bond formed by head-on overlap and two pi ($\pi$) bonds formed by parallel/lateral overlap of p-orbitals.
Study Notes: Chemical Bonding and Valence Bond Theory
1. Valence Bond Theory (VBT)
Proposed by Heitler and London (1927) and developed by Pauling, this theory explains bond energies, lengths, and shapes of covalent molecules.
- Orbital Overlap: Covalent bonds form by the overlapping of partially filled atomic orbitals. These orbitals share a common region in space but retain their individual identity.
- Electron Spin: Electrons in the overlapping orbitals must have opposite spins.
- Bond Strength: The greater the overlap between orbitals, the more energy is released and the stronger the resulting bond.
- Bond Axis: The imaginary line joining the nuclei of two bonded atoms is known as the bond axis.
2. Types of Covalent Bonds
There are two main types of covalent bonds determined by the method of orbital overlapping.
A. Sigma Bond ($\sigma$)
A sigma bond is formed by the head-on overlapping of atomic orbitals.
- It can occur between s-s, s-p, or p-p orbitals.
- It is stronger than a pi bond because electron density is concentrated directly between the nuclei.
B. Pi Bond ($\pi$)
A pi bond is formed by the lateral or sideways overlapping of half-filled p-orbitals.
- Electron density is located above and below the plane of the nuclei, making it more diffuse and weaker than a sigma bond.
- A double bond consists of one $\sigma$ and one $\pi$ bond.
- A triple bond consists of one $\sigma$ and two $\pi$ bonds.
3. Concept Assessment Exercise 3.2
Below are predictions for molecular shapes based on valence shell electron pair repulsion and VBT principles.
| Molecule | Formula | Predicted Shape |
|---|---|---|
| Tetrachloromethane | $CCl_{4}$ | Tetrahedral |
| Beryllium chloride | $BeCl_{2}$ | Linear |
| Phosphorus (III) chloride | $PCl_{3}$ | Trigonal Pyramidal |
| Hydrogen sulphide | $H_{2}S$ | Bent / V-shaped |
| Phosphine | $PH_{3}$ | Trigonal Pyramidal |
| Selenium hexafluoride | $SeF_{6}$ | Octahedral |
Relevant Questions & Answers
Q1: What is the primary difference between a sigma bond and a pi bond in terms of strength?
A: A sigma bond is stronger than a pi bond. This is because the orbital overlap in a sigma bond occurs directly along the bond axis between the nuclei, whereas pi bonds result from lateral overlap, making the electron density more diffuse and easier to break.
Q2: How many and what types of bonds are present in a molecule of Ethene ($C_{2}H_{4}$)?
A: In Ethene, there is a double bond between the two carbon atoms, which consists of one sigma ($\sigma$) bond and one pi ($\pi$) bond. All carbon-hydrogen bonds are sigma bonds.
Q3: According to VBT, what determines the number of bonds an atom can form?
A: The number of bonds formed is equal to the number of unpaired electrons present in the outermost shell of the atom.
Q4: Describe the shape of Phosphorus (V) chloride ($PCl_{5}$).
A: $PCl_{5}$ has a trigonal bipyramidal shape.
Study Notes: Valence Bond Theory and Covalent Bonding
1. Valence Bond Theory (VBT)
Valence Bond Theory was proposed by Heitler and London in 1927 and later developed by Pauling. It provides a framework for understanding bond energies, bond lengths, and the shapes of covalent molecules.
- Orbital Overlap: Covalent bonds are formed when a partially filled atomic orbital of one atom overlaps with a partially filled atomic orbital of another atom.
- Identity: Overlapping atomic orbitals retain their individual identity during the process.
- Electron Spin: Electrons within the overlapping orbitals must have opposite spins.
- Bond Count: The number of bonds an atom forms is equal to the number of unpaired electrons in its outermost shell.
- Symmetry: To form a bond, the overlapping orbitals must have the same symmetry with respect to the bond axis (the line joining the nuclei of the two bonded atoms).
- Energy: Energy is released during the overlapping of orbitals. The greater the overlap, the more energy is released and the stronger the bond becomes.
2. Types of Covalent Bonds
There are two primary types of covalent bonds obtained through orbital overlapping:
A. Sigma Bond ($\sigma$)
A sigma bond is formed by the head-on overlapping of atomic orbitals. It is the bond present when there is a single bond between two atoms.
- s-s overlap: Two s-orbitals overlap.
- s-p overlap: An s-orbital overlaps with a p-orbital.
- p-p overlap: Two p-orbitals overlap axially.
B. Pi Bond ($\pi$)
Pi bonds are formed by lateral or sideways overlapping of orbitals. By overlapping additional orbitals beyond the first sigma bond, multiple bonds (double and triple) are formed.
3. Predicted Molecular Shapes
Based on the principles of bonding and electron pairs, the following shapes are predicted for common molecules:
| Molecule | Chemical Formula | Predicted Shape |
|---|---|---|
| Tetrachloromethane | $CCl_{4}$ | Tetrahedral |
| Beryllium chloride | $BeCl_{2}$ | Linear |
| Phosphorus (III) chloride | $PCl_{3}$ | Trigonal Pyramidal |
| Hydrogen sulphide | $H_{2}S$ | Bent / V-shaped |
| Phosphine | $PH_{3}$ | Trigonal Pyramidal |
| Selenium hexafluoride | $SeF_{6}$ | Octahedral |
| Phosphorus (V) chloride | $PCl_{5}$ | Trigonal Bipyramidal |
Relevant Questions & Answers
Q1: What is a "bond axis" in Valence Bond Theory?A: The bond axis is defined as the line joining the nuclei of two bonded atoms.
Q2: How does the extent of orbital overlap affect the strength of a covalent bond?
A: The greater the overlap between the orbitals, the greater the energy released and the stronger the resulting bond.
Q3: What are the three specific types of orbital overlaps that can result in a sigma bond?
A: Sigma bonds can be formed between s-s, s-p ($s-p_{x}$, $s-p_{y}$, $s-p_{z}$), and p-p ($p_{x}-p_{x}$) orbitals through head-on overlapping.
Q4: According to VBT, what condition must be met regarding electron spin for a bond to form?
A: The electrons of the overlapping orbitals must have opposite spin.
Study Notes: Valence Bond Theory (VBT) and Covalent Bonding
1. Introduction to Valence Bond Theory
Valence Bond Theory was proposed by Heitler and London (1927) and later developed by Pauling. It successfully explains bond energies, bond lengths, and the shapes of covalent molecules.
Main Points of VBT:
- A bond between two atoms is formed by the overlap of half-filled atomic orbitals of two atoms.
- These atomic orbitals retain their identity even after overlapping.
- Electrons of overlapping orbitals must have opposite spin.
- The number of bonds formed is equal to the number of unpaired electrons present in the outermost shell of the atom.
- Overlapping orbitals must have the same symmetry with respect to the bond axis (the line joining the nuclei of two bonded atoms).
- The greater the overlap between the orbitals, the greater the energy released and the stronger the bond formed.
2. Types of Covalent Bonds
Covalent bonds are classified based on the type of orbital overlapping.
A. Sigma Bond ($\sigma$)
A sigma bond is formed by the head-on overlapping of atomic orbitals. This is the bond present in every single covalent bond.
- s-s overlap: Two s-orbitals overlap.
- s-p overlap: An s-orbital overlaps with a p-orbital ($s-p_x, s-p_y, s-p_z$).
- p-p overlap: Two p-orbitals overlap axially ($p_x-p_x$).
B. Pi Bond ($\pi$)
Pi bonds are formed by the overlapping of additional orbitals beyond the initial sigma bond. These result in multiple bonds, such as double and triple bonds.
3. Predicted Molecular Shapes
Based on bonding principles, we can predict the geometry of various molecules:
| Molecule | Formula | Predicted Shape |
|---|---|---|
| Tetrachloromethane | $CCl_4$ | Tetrahedral |
| Beryllium chloride | $BeCl_2$ | Linear |
| Phosphorus (III) chloride | $PCl_3$ | Trigonal Pyramidal |
| Hydrogen sulphide | $H_2S$ | Bent / V-shaped |
| Phosphine | $PH_3$ | Trigonal Pyramidal |
| Selenium hexafluoride | $SeF_6$ | Octahedral |
| Phosphorus (V) chloride | $PCl_5$ | Trigonal Bipyramidal |
Relevant Questions & Answers
Q1: What is the "bond axis" according to Valence Bond Theory?
A: The bond axis is defined as the line joining the nuclei of two bonded atoms.
Q2: How does orbital overlap relate to bond strength?
A: The greater the overlap between the atomic orbitals, the greater the energy released and the stronger the resulting bond.
Q3: List the types of orbital overlaps that can result in a Sigma ($\sigma$) bond.
A: Sigma bonds can be formed through s-s overlap, s-p overlap, or p-p axial overlap.
Q4: What requirement does VBT place on the spin of electrons in a bond?
A: The electrons in the overlapping orbitals must have opposite spins.
Q5: Predict the shape of a Selenium hexafluoride ($SeF_6$) molecule.
A: The predicted shape for $SeF_6$ is Octahedral.