Chemical Bonding and Molecular Structure: Answers

1. VSEPR Theory and Geometry

a) Explain the VSEPR theory briefly.

The Valence Shell Electron Pair Repulsion (VSEPR) theory states that the shape of a molecule is determined by repulsion between electron pairs (bonding and lone pairs) around a central atom. Electron pairs arrange themselves as far apart as possible to minimize repulsion, determining the geometry of the molecule.

b) Compare the bond angles of NH₃, H₂O, and CH₄ with explanation.

• CH₄ (Methane): Tetrahedral geometry, bond angle ≈ 109.5°. No lone pairs on C, so perfect geometry.
• NH₃ (Ammonia): Trigonal pyramidal geometry, bond angle ≈ 107°. One lone pair on N pushes bonds closer together.
• H₂O (Water): Bent geometry, bond angle ≈ 104.5°. Two lone pairs on O push the H atoms closer together.

c) Determine the geometry of PCl₅.

The geometry of PCl₅ is trigonal bipyramidal. It has five bonding pairs of electrons around the central phosphorus atom—three in the equatorial plane and two in the axial positions.

2. Dipole Moment

a) Define dipole moment. Discuss its applications.

Dipole moment (μ) is the product of the magnitude of the charge (Q) and the distance (d) between centers of positive and negative charge: μ = Q × d. It is a vector quantity measured in Debye (D).

• Applications:
  • Distinguishing polar and nonpolar molecules
  • Predicting molecular shape
  • Determining ionic character of bonds
  • Explaining physical properties like solubility

b) Comment on the dipole moments of CCl₄, BCl₃, and H₂O.

• CCl₄ and BCl₃ are non-polar as their dipole moments cancel due to symmetrical geometry (tetrahedral and trigonal planar, respectively), so μ = 0.
• H₂O has a bent shape, so the bond moments do not cancel. Water has a high dipole moment (μ ≠ 0).

3. Lattice Energy

a) What is lattice energy? Define it.

Lattice energy is the energy required to separate 1 mole of an ionic solid into its gaseous ions, or equivalently, the energy released when gaseous ions form one mole of a crystalline lattice. It reflects the strength of ionic bonding and is measured in kJ/mol.

b) Compare the lattice energies of NaCl, KCl, and RbCl.

The order of lattice energy is: NaCl > KCl > RbCl, because smaller cations (Na⁺ < K⁺ < Rb⁺) create stronger electrostatic attractions, resulting in greater lattice energies.

c) Illustrate the Born-Haber cycle for the calculation of lattice energy.

The Born-Haber cycle is a thermochemical cycle that analyzes the steps in forming an ionic compound:

1. Sublimation of the metal (solid to gas)
2. Ionization of the gaseous metal atom
3. Dissociation of non-metal molecules (if needed)
4. Addition of electrons (electron affinity) to the non-metal atom
5. Formation of the ionic crystal (lattice energy)

Lattice energy can be calculated using Hess’s Law, rearranging the cycle:
Lattice Energy = [Sum of other steps] – [Enthalpy of formation]

4. Molecular Orbitals

a) Draw the molecular orbital (MO) diagram of Be₂.

MO diagram for Be₂ involves 4 electrons (2 from each Be atom): σ1s², σ1s*², σ2s², σ2s*². All bonding and antibonding pairs cancel, making bond order 0. Be₂ is not stable under normal conditions.

b) Compare the bond orders of O₂, O₂⁺, and O₂^–.

• O₂: Bond order = 2
• O₂⁺: Bond order = 2.5
• O₂^–: Bond order = 1.5

c) Discuss whether He₂ exists or not, using MO theory.

He₂ does not exist because its bond order is zero: both bonding and antibonding orbitals are equally filled, so there is no net bond formation.

5. Resonance in Carbonate Ion

Draw the resonating structures of the carbonate ion (CO₃²⁻).

CO₃²⁻ has three resonance structures, each with a double bond between carbon and one oxygen atom, and two single-bonded oxygens each carrying a negative charge. The double bond location rotates among the oxygens, resulting in equivalent bond lengths due to delocalization.

6. Hybridization and Ethylene

a) Define hybridization. Give reasons for hybridization.

Hybridization is the mixing of atomic orbitals to form new, equivalent hybrid orbitals suitable for bonding. This explains observed molecular shapes and bond angles, which cannot be justified by unhybridized atomic orbitals alone.

b) Draw and explain the orbital diagram of ethylene.

In ethylene (C₂H₄), each carbon undergoes sp² hybridization, forming three sigma bonds (two with hydrogen, one with carbon). The unhybridized p orbital on each carbon overlaps sideways to create a pi bond, resulting in a double bond between the carbons.

7. Fajan’s Rule of Covalency and Comparison

a) State Fajan’s rule of covalency.

Fajan’s Rule predicts that covalent character increases when the cation is small and highly charged, or the anion is large and highly polarizable. Greater polarizing power of the cation and polarizability of the anion lead to greater covalency.

b) Compare covalency of LiF, MgO and AlCl₃.

• LiF: low covalency
• MgO: moderate covalency
• AlCl₃: greatest covalency (Al³⁺ is small and highly charged, Cl^– is large and polarizable; thus, AlCl₃ is covalent in nature)

8. Formal Charge and Ozone

Define formal charge. Calculate formal charge of each oxygen in O₃.

Formal charge for an atom = (valence electrons) – (nonbonding electrons) – (bonding electrons/2).
In ozone (O₃):

• Central oxygen: +1
• Double-bonded oxygen: 0
• Single-bonded oxygen: –1