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Chemical Bonding and Structure - Intermolecular forces

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Intermolecular forces (IMFs) are the forces of attraction between molecules, which determine physical properties such as boiling point, melting point, and solubility. They are much weaker than intramolecular forces like covalent or ionic bonds.

London (Dispersion) Forces: The weakest IMF, present in all molecules. They arise from temporary dipoles caused by the movement of electrons. The strength of LDFs increases with increasing molar mass (MrM_r) and larger surface area (e.g., I2I_2 has stronger LDFs than F2F_2).

Dipole-Dipole Forces: Forces of attraction between the permanent dipoles of polar molecules. A molecule is polar if it has polar bonds (difference in electronegativity Δχ>0.4\Delta \chi > 0.4) and a non-symmetrical geometry (e.g., HClHCl, SO2SO_2).

Hydrogen Bonding: A specific, strong type of dipole-dipole attraction. It occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (NN, OO, or FF) and is attracted to the lone pair of an NN, OO, or FF atom in a neighboring molecule.

The relative strength of IMFs generally follows the order: London Dispersion<Dipole-Dipole<Hydrogen Bonding\text{London Dispersion} < \text{Dipole-Dipole} < \text{Hydrogen Bonding}. However, very large molecules with only LDFs can sometimes have higher boiling points than small molecules with hydrogen bonding.

Solubility: 'Like dissolves like.' Polar solutes (like C6H12O6C_6H_{12}O_6) dissolve in polar solvents (like H2OH_2O) due to favorable dipole-dipole or hydrogen bonding interactions. Non-polar solutes (like I2I_2) dissolve in non-polar solvents (like C6H14C_6H_{14}) via LDFs.

📐Formulae

μ=q×d\mu = q \times d

Δχ=χAχB\Delta \chi = |\chi_A - \chi_B|

Fq1q2r2F \propto \frac{q_1 q_2}{r^2}

💡Examples

Problem 1:

Explain why ethanol (CH3CH2OHCH_3CH_2OH) has a significantly higher boiling point (78C78^\circ C) than dimethyl ether (CH3OCH3CH_3OCH_3, 24C-24^\circ C), despite both having the same molecular formula C2H6OC_2H_6O.

Solution:

Ethanol contains an OH-OH group, which allows it to form strong intermolecular hydrogen bonds. Dimethyl ether is a polar molecule but lacks HH bonded directly to OO, so it only exhibits dipole-dipole forces and London dispersion forces.

Explanation:

Hydrogen bonds are significantly stronger than dipole-dipole interactions, requiring more energy (higher temperature) to overcome during the phase change from liquid to gas.

Problem 2:

Predict the trend in boiling points for the following noble gases: HeHe, NeNe, ArAr, and KrKr.

Solution:

He<Ne<Ar<KrHe < Ne < Ar < Kr

Explanation:

Noble gases are non-polar monatomic species that only experience London Dispersion Forces. As we move down Group 18, the number of electrons increases and the electron cloud becomes larger and more polarizable. This leads to stronger temporary dipoles and increased LDF strength, resulting in higher boiling points.

Problem 3:

Identify the dominant intermolecular force in NH3NH_3, H2SH_2S, and CF4CF_4.

Solution:

NH3NH_3: Hydrogen bonding; H2SH_2S: Dipole-dipole; CF4CF_4: London dispersion forces.

Explanation:

In NH3NH_3, HH is bonded to NN, enabling hydrogen bonding. H2SH_2S is a polar molecule (bent geometry) but SS is not electronegative enough for hydrogen bonding, so it uses dipole-dipole. CF4CF_4 is non-polar because its tetrahedral geometry causes the individual CFC-F bond dipoles to cancel out, leaving only LDFs.

Intermolecular forces - Revision Notes & Key Formulas | IB Grade 12 Chemistry