Unit 2: Shapes, Intermolecular Forces & Structures

Delving deeper into the geometry and arrangement of chemical species.

2.7 Shapes of Molecules & VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the geometry of individual molecules from the number of electron pairs surrounding their central atoms. It states that electron pairs (both bonding and lone pairs) around a central atom will arrange themselves as far apart as possible to minimize repulsion, leading to specific molecular shapes and bond angles. Lone pairs repel more strongly than bonding pairs.

  • Linear: 2 electron domains (e.g., $CO_2$, BeCl$_2$). Bond angle: 180°.
  • Trigonal Planar: 3 electron domains (e.g., $BF_3$). Bond angle: 120°.
  • Tetrahedral: 4 electron domains (e.g., $CH_4$). Bond angle: 109.5°.
  • Trigonal Pyramidal: 4 electron domains (3 bonding, 1 lone pair) (e.g., $NH_3$). Bond angle: ~107° (lone pair repulsion).
  • Non-linear (Bent): 4 electron domains (2 bonding, 2 lone pairs) (e.g., $H_2O$). Bond angle: ~104.5° (two lone pair repulsions).
Solved Examples:
  1. Predict the shape of the $CO_2$ molecule and its bond angle.
    Solution: The central carbon atom has two double bonds to oxygen atoms and no lone pairs. This means there are two electron domains. According to VSEPR theory, these domains will repel each other to be as far apart as possible, resulting in a linear shape with a bond angle of 180°.
  2. Determine the shape and bond angle of $BF_3$.
    Solution: Boron is the central atom, forming three single bonds to fluorine atoms. There are no lone pairs on boron. Thus, there are three electron domains. These domains arrange themselves into a trigonal planar shape with bond angles of 120° to minimize repulsion.
  3. Explain the shape and bond angle of methane ($CH_4$).
    Solution: The central carbon atom forms four single bonds to hydrogen atoms and has no lone pairs. This results in four electron domains. To minimize repulsion, these domains orient themselves towards the vertices of a tetrahedron, giving a tetrahedral shape with bond angles of 109.5°.
  4. Why is the bond angle in ammonia ($NH_3$) approximately 107° instead of 109.5°?
    Solution: Ammonia has a central nitrogen atom with three bonding pairs to hydrogen and one lone pair. This gives a total of four electron domains. While the electron geometry is tetrahedral, the lone pair exerts a stronger repulsion on the bonding pairs than bonding pairs repel each other. This increased repulsion compresses the H-N-H bond angles from the ideal 109.5° to approximately 107°, resulting in a trigonal pyramidal molecular shape.
  5. Describe the shape and bond angle of a water ($H_2O$) molecule.
    Solution: The central oxygen atom in water forms two single bonds with hydrogen atoms and has two lone pairs. This means there are four electron domains in total. The two lone pairs exert even greater repulsion than a single lone pair, pushing the bonding pairs closer together. This results in a non-linear (bent) shape with an approximate bond angle of 104.5°, which is smaller than the ideal tetrahedral angle.

2.8 Intermolecular Forces

Intermolecular forces (IMFs) are attractive forces that exist *between* molecules. These forces are much weaker than intramolecular bonds (ionic, covalent, metallic) that hold atoms together *within* a molecule, but they are crucial in determining the physical properties of substances such as melting points, boiling points, and solubility.

1. Van der Waals Forces

These are the weakest IMFs and include:

  • London Dispersion Forces (LDFs): Present in all molecules (polar and nonpolar). They arise from temporary, instantaneous dipoles created by the momentary uneven distribution of electron density. LDFs increase with increasing number of electrons and larger molecular surface area, leading to higher boiling points.
  • Dipole-Dipole Forces: Occur between polar molecules that have permanent dipoles due to differences in electronegativity between bonded atoms. These are stronger than LDFs for molecules of comparable size.
2. Hydrogen Bonding

This is a special and particularly strong type of dipole-dipole interaction. It occurs when a hydrogen atom is directly bonded to a highly electronegative atom (N, O, or F) in one molecule, creating a strong positive charge on the hydrogen. This positively charged hydrogen is then strongly attracted to a lone pair on another highly electronegative atom (N, O, or F) in an adjacent molecule. Hydrogen bonding significantly affects properties like the unusually high boiling point of water.

Solved Examples:
  1. Explain why iodine ($I_2$) is a solid at room temperature, while bromine ($Br_2$) is a liquid and chlorine ($Cl_2$) is a gas.
    Solution: All three are nonpolar molecules, so only London Dispersion Forces (LDFs) are present. The strength of LDFs increases with increasing number of electrons (and thus increasing molecular size/mass). Iodine has the most electrons, leading to the strongest LDFs, requiring more energy to overcome them, hence it's a solid. Bromine has fewer electrons than iodine but more than chlorine, resulting in intermediate LDFs (liquid). Chlorine has the fewest electrons and weakest LDFs (gas).
  2. Why does water ($H_2O$) have a significantly higher boiling point ($100^\circ C$) than hydrogen sulfide ($H_2S$, $-60^\circ C$), despite sulfur being heavier than oxygen?
    Solution: Water molecules exhibit strong hydrogen bonding because hydrogen is directly bonded to the highly electronegative oxygen atom. Hydrogen sulfide, while polar, does not form hydrogen bonds as sulfur is less electronegative than oxygen. The dominant intermolecular forces in $H_2S$ are dipole-dipole interactions and London Dispersion Forces. The much stronger hydrogen bonds in water require significantly more energy to overcome during boiling, leading to its exceptionally high boiling point.
  3. Explain why ethanol ($CH_3CH_2OH$) is highly soluble in water, but hexane ($C_6H_{14}$) is not.
    Solution: Ethanol has an -OH group, which enables it to form hydrogen bonds with water molecules. Since water also forms hydrogen bonds, these strong intermolecular attractions allow ethanol to dissolve readily ("like dissolves like"). Hexane is a nonpolar hydrocarbon that only experiences weak London Dispersion Forces. It cannot form hydrogen bonds with water, and the energy required to break the strong hydrogen bonds in water to accommodate the hexane molecules is not compensated by favorable interactions, making it largely insoluble.
  4. Compare the boiling points of methane ($CH_4$) and ammonia ($NH_3$), and explain the difference.
    Solution: Methane is a nonpolar molecule and only experiences weak London Dispersion Forces, resulting in a very low boiling point ($-162^\circ C$). Ammonia is a polar molecule with N-H bonds, allowing it to form hydrogen bonds, which are much stronger intermolecular forces. More energy is required to overcome these hydrogen bonds, giving $NH_3$ a significantly higher boiling point ($-33^\circ C$) than $CH_4$.
  5. Why does hydrogen chloride ($HCl$) have a higher boiling point than fluorine ($F_2$), even though $F_2$ has a slightly larger molar mass?
    Solution: $HCl$ is a polar molecule due to the electronegativity difference between H and Cl, experiencing dipole-dipole interactions in addition to London Dispersion Forces. $F_2$ is a nonpolar molecule and only has London Dispersion Forces. Despite $F_2$'s slightly larger molar mass (which means slightly stronger LDFs), the presence of permanent dipole-dipole forces in $HCl$ makes its intermolecular attractions stronger overall, leading to a higher boiling point for $HCl$ ($-85^\circ C$) compared to $F_2$ ($-188^\circ C$).

2.9 Effect of Structure & Intermolecular Forces on Solubility

Solubility is the ability of a solute to dissolve in a solvent to form a homogeneous solution. The principle "like dissolves like" is a fundamental rule in predicting solubility, meaning substances with similar types and strengths of intermolecular forces are likely to be soluble in each other.

1. Polar Solutes in Polar Solvents
  • Mechanism: Polar solutes (which have permanent dipoles or can form hydrogen bonds) dissolve well in polar solvents (like water, ethanol) because favorable dipole-dipole interactions or hydrogen bonds can form between the solute and solvent molecules. The energy released from these new interactions compensates for the energy required to break the existing intermolecular forces in both the solute and solvent.
  • Examples: Ionic compounds (e.g., NaCl), alcohols (e.g., $CH_3OH$), and small organic acids are typically soluble in water.
2. Nonpolar Solutes in Nonpolar Solvents
  • Mechanism: Nonpolar solutes dissolve well in nonpolar solvents (like hexane, benzene) because both rely primarily on weak London Dispersion Forces. When mixed, new London Dispersion Forces can form between solute and solvent molecules, which are comparable in strength to the forces being broken.
  • Examples: Oils, fats, and hydrocarbons (e.g., $C_6H_{14}$) are soluble in nonpolar solvents.
3. Polar Solutes in Nonpolar Solvents (and vice versa)
  • Mechanism: These combinations generally exhibit low solubility. A polar substance would need to overcome its strong intermolecular forces (dipole-dipole or hydrogen bonding) to mix with a nonpolar solvent, but the nonpolar solvent can only offer weak London Dispersion Forces in return. The energy cost of breaking the strong interactions in the polar substance is not compensated by the weak interactions with the nonpolar solvent. Similarly, nonpolar substances struggle to break the strong attractions in polar solvents.
  • Examples: Oil does not dissolve in water; salt does not dissolve in hexane.
4. Effect of Molecular Size and Structure on Solubility (within similar types)
  • For Simple Molecular Structures:
    • Size: For a homologous series, as molecular size (and thus the number of electrons) increases, London Dispersion Forces become stronger. This can make larger molecules less soluble in polar solvents if their polar functional groups are 'diluted' by a large nonpolar chain (e.g., long-chain alcohols become less soluble in water). Conversely, larger nonpolar molecules might be more soluble in nonpolar solvents due to stronger LDFs.
    • Branching: Increased branching in molecules can sometimes slightly decrease solubility in certain solvents due to less efficient packing or reduced surface area for interaction.
  • For Ionic Compounds:
    • Lattice Energy vs. Hydration Energy: Solubility of ionic compounds in water depends on the balance between lattice energy (energy required to break apart the ionic lattice) and hydration energy (energy released when ions are solvated by water molecules). If hydration energy is greater than lattice energy, the compound is soluble. Factors like ion charge and size influence these energies.
Solved Examples:
  1. Why is glucose ($C_6H_{12}O_6$) highly soluble in water?
    Solution: Glucose is a polar molecule containing many -OH (hydroxyl) groups. These -OH groups allow glucose to form extensive hydrogen bonds with water molecules. Since water also forms hydrogen bonds, the strong, favorable interactions between glucose and water molecules lead to high solubility.
  2. Octane ($C_8H_{18}$) is a liquid at room temperature and is insoluble in water. Explain why.
    Solution: Octane is a nonpolar hydrocarbon. The only intermolecular forces present in octane are weak London Dispersion Forces. Water is a polar solvent with strong hydrogen bonds. Octane cannot form hydrogen bonds with water, and the weak LDFs it forms with water molecules are not strong enough to overcome the strong hydrogen bonds between water molecules, making it insoluble ("like dissolves like").
  3. Potassium iodide ($KI$) is very soluble in water, but iodine ($I_2$) is not. Explain this difference.
    Solution: $KI$ is an ionic compound. When dissolved in water, the strong electrostatic attractions between $K^+$ and $I^-$ ions are overcome by the favorable ion-dipole interactions with polar water molecules, leading to dissolution. Iodine ($I_2$) is a nonpolar simple molecular substance held together by weak London Dispersion Forces. While these LDFs can be overcome, the strong hydrogen bonds between water molecules are not sufficiently disrupted by the weak interactions with $I_2$, making $I_2$ largely insoluble in water.
  4. Why is propan-1-ol ($CH_3CH_2CH_2OH$) more soluble in water than hexan-1-ol ($CH_3(CH_2)_5OH$)?
    Solution: Both propan-1-ol and hexan-1-ol have an -OH group and can form hydrogen bonds with water. However, hexan-1-ol has a much longer nonpolar hydrocarbon chain ($C_6H_{13}$) compared to propan-1-ol ($C_3H_7$). As the nonpolar part of the molecule increases, the influence of the polar -OH group diminishes, and the molecule becomes more "hydrophobic" (water-fearing). The larger nonpolar chain in hexan-1-ol means that the overall molecule interacts less favorably with water, reducing its solubility.
  5. Suggest a suitable solvent for grease, a nonpolar substance.
    Solution: According to the "like dissolves like" principle, a nonpolar solvent would be suitable. Examples include hydrocarbons like hexane, benzene, or other organic solvents such as tetrachloromethane ($CCl_4$). These solvents interact with grease primarily through London Dispersion Forces, leading to dissolution.

2.10 Structures

The macroscopic properties of a substance are largely determined by its microscopic structure and the type of bonding present. In the solid state, particles (atoms, ions, or molecules) often arrange themselves in a regular, repeating pattern called a lattice. There are four main types of chemical structures:

1. Simple Molecular Structures
  • Bonding: Strong covalent bonds within molecules, but weak intermolecular forces between molecules.
  • Properties: Low melting/boiling points (weak IMFs easily overcome), often gases or liquids at room temp, do not conduct electricity (no free ions or delocalized electrons), generally soft.
  • Examples: $H_2O$, $CO_2$, $O_2$, $CH_4$, $I_2$.
2. Giant Covalent Structures (Macromolecular)
  • Bonding: Strong covalent bonds extend throughout the entire structure in a continuous, extensive network.
  • Properties: Very high melting/boiling points (strong covalent bonds require vast energy to break), generally hard and brittle, most do not conduct electricity (except graphite), insoluble in common solvents.
  • Examples: Diamond, graphite, silicon ($Si$), silicon dioxide ($SiO_2$).
3. Giant Ionic Structures
  • Bonding: Strong electrostatic forces of attraction between oppositely charged ions in a regular three-dimensional lattice.
  • Properties: High melting/boiling points (strong electrostatic forces), hard and brittle, conduct electricity only when molten or in aqueous solution (ions are mobile), soluble in polar solvents like water.
  • Examples: Sodium chloride ($NaCl$), magnesium oxide ($MgO$).
4. Giant Metallic Structures
  • Bonding: Strong electrostatic forces of attraction between a lattice of positive metal ions and a 'sea' of delocalized electrons.
  • Properties: High melting/boiling points (strong metallic bonds), good conductors of electricity (mobile delocalized electrons), good conductors of heat, malleable and ductile (layers of ions can slide), generally insoluble.
  • Examples: Copper ($Cu$), Iron ($Fe$), Magnesium ($Mg$), Sodium ($Na$).
Solved Examples:
  1. Sodium chloride ($NaCl$) has a very high melting point and conducts electricity when molten but not when solid. Explain these properties.
    Solution: $NaCl$ has a giant ionic structure with strong electrostatic forces between $Na^+$ and $Cl^-$ ions in a rigid lattice. Much energy is needed to overcome these forces, hence the high melting point. In solid state, ions are fixed, so no conductivity. When molten, ions become mobile and can carry charge, allowing conductivity.
  2. Graphite conducts electricity while diamond does not, even though both are carbon allotropes. Explain.
    Solution: Both are giant covalent structures. In diamond, all carbon valence electrons are in covalent bonds, so no free electrons for conduction. In graphite, each carbon atom forms three bonds, leaving one delocalized electron per atom in layers. These mobile electrons allow graphite to conduct electricity.
  3. Why are metals like copper malleable and ductile?
    Solution: Metals have a giant metallic structure: a lattice of positive ions in a 'sea' of delocalized electrons. When force is applied, layers of ions can slide past each other without breaking the metallic bond because the delocalized electrons maintain the attractive forces, allowing for malleability (sheets) and ductility (wires).
  4. Methane ($CH_4$) has a very low boiling point ($-162^\circ C$). Explain this property.
    Solution: Methane has a simple molecular structure. While there are strong covalent bonds within each molecule, only weak London Dispersion Forces exist between molecules. Little energy is required to overcome these weak intermolecular forces to separate the molecules, resulting in a very low boiling point.
  5. Silicon dioxide ($SiO_2$) has a very high melting point and is hard. Explain.
    Solution: Silicon dioxide has a giant covalent structure, forming an extensive 3D network where each silicon is bonded to four oxygen atoms. The entire structure is held by vast numbers of strong covalent bonds. A significant amount of energy is needed to break these strong bonds, leading to its very high melting point and hardness.

Knowledge Check (20 Questions)

Answer: Linear, 180°.

Answer: Lone pairs are held closer to the central atom's nucleus and occupy more space, leading to greater repulsion.

Answer: Bent (or non-linear).

Answer: $SO_3$. (It has 3 electron domains around S, no lone pairs).

Answer: ~107°.

Answer: London Dispersion Forces < Dipole-Dipole Forces < Hydrogen bonding.

Answer: London Dispersion Forces.

Answer: $HF$. (H bonded to F, O, or N).

Answer: Both are nonpolar with only LDFs. $Cl_2$ has more electrons than $F_2$, leading to stronger LDFs and thus a higher boiling point.

Answer: Simple molecular structure.

Answer: Diamond, Graphite, Silicon, Silicon dioxide ($SiO_2$).

Answer: Electrical conductivity, thermal conductivity, malleability, and ductility.

Answer: Strong electrostatic forces hold ions in fixed positions. A strong force can cause layers to shift, bringing like charges together and leading to repulsion and fracture.

Answer: No, because they have no free ions or delocalized electrons to carry charge.

Answer: Electrostatic attraction between a lattice of positive metal ions and a 'sea' of delocalized electrons.

Answer: $KCl$ is an ionic compound. In solid state, ions are fixed. In aqueous solution, ions dissociate and become mobile, carrying charge.

Answer: $H_2O$ (due to hydrogen bonding).

Answer: Bent (or non-linear).

Answer: Water, Ethanol, Bromine, etc.

Answer: Quartz has a giant covalent structure where all atoms are held by strong, extensive covalent bonds in a 3D network. A large amount of energy is required to break these bonds.
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