Chapter 3. Intermolecular Forces and Properties 

3.1 Intermolecular Forces

3.1.1

3.1.2

3.1.3

3.1.4

3.1.5

The boiling point of pentane (309 K) is higher than that of 2,2-dimethylpropane (283 K) because the linear structure of pentane allows for stronger London dispersion forces due to greater molecular surface area. In contrast, the compact, branched structure of 2,2-dimethylpropane reduces surface contact and weakens dispersion forces.

Detailed explanation here.

Intermolecular Forces:

  • CS2​: London dispersion forces.

  • COS: London dispersion forces and dipole-dipole interactions.

Detailed explanation here.

Stronger Intermolecular Forces:

  • CS2​ experiences stronger overall intermolecular forces, as indicated by its higher boiling point. The large sulfur atoms in CS2​ contribute to very strong London dispersion forces, which outweigh the dipole-dipole interactions present in COS.

Detailed explanation here.

The ion most likely to have the strongest interaction with nearby water molecules is S2-, due to its higher charge and relatively small ionic radius. According to Coulomb’s Law, force of attraction (F) depends on both the charge (q) and the radius (r): For S2−, q = −2, and r = 170 pm, yielding a stronger interaction compared to the other ions listed.

Detailed explanation here.

Compound 1 (CH3-O-CH3​) has weaker dispersion forces and no hydrogen bonding, resulting in a lower boiling point (249 K).

Compound 2 (CH3CH2OH) forms hydrogen bonds due to the -OH group, resulting in a higher boiling point (351 K).

Detailed explanation here.

On the provided structure:

  • Draw one dashed line from the oxygen atom of thymine’s carbonyl group to the hydrogen atom in adenine’s −NH group.

  • Draw another dashed line from the hydrogen atom in thymine’s −NH group to the nitrogen atom in adenine.

These dashed lines represent the two hydrogen bonds that stabilize the thymine-adenine base pair in the DNA double helix.

Detailed explanation here.

3.2  Properties of Solids

3.2.1

3.2.2

3.2.3

3.2.4

3.2.5

Part A:

  • Water has the highest boiling point due to strong hydrogen bonding, which requires more energy to break.

  • It also has the lowest vapor pressure because strong intermolecular forces prevent many molecules from escaping into the gas phase.

Detailed explanation here.

Part B:

  • Methanol has a higher boiling point (65°C) than acetone (56°C) because it exhibits hydrogen bonding, which is stronger than acetone's dipole-dipole interactions.

  • Acetone's lack of hydrogen bonding results in a lower boiling point.

Detailed explanation here.

Part C:

  • Water is predicted to have the highest melting point due to its extensive hydrogen bonding network, which forms a rigid lattice in the solid state. This structure requires significant energy to break, more so than the other compounds.

Detailed explanation here.

Water molecules, being polar with a partial negative charge on oxygen and partial positive charges on hydrogen, surround the ions. For sodium ions (Na⁺), water molecules orient with their oxygen ends pointing towards the positive ion, forming ion-dipole interactions. Conversely, for chloride ions (Cl⁻), water molecules orient with their hydrogen ends pointing towards the negative ion, also forming ion-dipole interactions.

Detailed explanation here.

Ion-dipole interactions between water molecules and ions are stronger than hydrogen bonds in pure water. These stronger interactions require more energy to overcome for vaporization. Consequently, the boiling point of the NaCl solution is elevated compared to pure water because more heat is needed to boil the solution.

Detailed explanation here.

Molecular solids: Solid X, Solid Z.

Detailed explanation here.

Ionic solids: Solid W.

Detailed explanation here.

Metallic solids: Solid Y.

Detailed explanation here.

Melting point comparison: Solid W > Solid X due to the stronger ionic interactions in Solid W compared to the weaker intermolecular forces in Solid X.

Detailed explanation here.

Both hexagonal boron nitride (h-BN) and cubic boron nitride (c-BN) represent covalent network solids (also sometimes referred to as atomic network solids).

Detailed explanation here.

The primary differences in melting points and mechanical properties between h-BN and c-BN arise from their structural differences:

  • h-BN (Hexagonal): Layered structure with strong covalent bonds within layers and weak van der Waals forces between layers. This leads to softness and a high melting point.

  • c-BN (Cubic): Three-dimensional network structure with strong covalent bonds extending in all directions. This leads to extreme hardness and a slightly lower (but still very high) melting point.

Detailed explanation here.

Mg atoms are smaller than Zn atoms, so they interfere more with the displacement of Al atoms in the alloy lattice.

This interference strengthens the Al/Mg alloy compared to the Al/Zn alloy.

Detailed explanation here.

3.3  Solids, Liquids, and Gases

3.3.1

3.3.2

3.3.3

3.3.4

No problem. XXX

The best explanation is (B) are relatively close together.

Detailed explanation here.

No problem. XXX

  1. The average spacing between gas particles in the larger balloon is greater than in the smaller balloon because the gas particles occupy a larger space while maintaining the same number of particles.

  2. If the smaller balloon’s volume doubles, the frequency of collisions with the walls decreases because the gas particles are more spread out, reducing the likelihood of collisions with the walls per unit time.

Detailed explanation here.

3.4  Ideal Gas Law

3.4.1

3.4.2

3.4.3

Mass of dry air = 1.18 g

Detailed explanation here.

Mass of container = 498.82 g

Detailed explanation here.

Mass of unknown gas. = 2.68 g

Detailed explanation here.

Calculate molar mass (M) = 65.6 g/mol (approximately)

Detailed explanation here.

Occurence 1: Yes, it could cause an error. If some dry air remained, the measured mass of the unknown gas would be higher than the actual mass. This would lead to an overestimation of the moles of gas present, which would result in an underestimation of the molar mass of the unknown gas.

Occurence 2: Yes, it could cause an error. If the actual pressure was lower than 760 torr, the calculated number of moles of the unknown gas would be lower than the actual number of moles. This would lead to an overestimation of the molar mass of the unknown gas.

Detailed explanation here.

PART A. total pressure is approximately 8.56 atm

Detailed explanation here.

(i) The mole fraction of N2(g) in the cylinder ≈ 0.4998

(II) partial pressure of N2 = 4.02 atm.

Detailed explanation here.

PART A. P₁V₁ = P₂V₂ or PV = k (where k is a constant)

Detailed explanation here.

PART B. The final pressure (P₂) is 1.5 atm.

Detailed explanation here.

PART C. Draw a new hyperbolic curve above the original curve. The new curve should maintain the same general shape but be shifted upwards. This indicates that at any volume, the pressure is higher when the temperature is increased.

Detailed explanation here.

3.5  Kinetic Molecular Theory

3.5.1

3.5.2

3.5.3

3.5.4

The Kinetic Molecular Theory (KMT) states that temperature is proportional to the average kinetic energy of particles.

At higher temperatures, particles move faster, resulting in a broader distribution of speeds and a peak shifted to the right. Conversely, at lower temperatures, particles have less energy and slower speeds, creating a narrower distribution with the peak shifted to the left.

If the gas is replaced with argon (Ar), the distribution will shift to the left for all temperatures because argon is heavier than helium. Heavier particles move slower at the same temperature because kinetic energy depends on both mass and speed ($KE = \frac{1}{2}mv^2$).

Detailed explanation here.

The ranking of the molar masses of substances A, B, and C, from highest to lowest, is: A > B > C. The lowest peak speed correlated with highest molar mass.

Detailed explanation here.

Since all balloons are at the same temperature (25 °C), each gas has the same average kinetic energy. The identity or molar mass of the gas does not change that fact—temperature alone sets the average kinetic energy for an ideal gas.

Detailed explanation here.

PART A.

When the temperature increases from 300 K to 400 K, the following changes occur:

  1. Faster Particles:

    • Temperature is a measure of the average kinetic energy of gas particles. As temperature increases, the particles gain more energy and move faster.

    • This causes the entire speed distribution to shift toward higher speeds, meaning more particles are moving quickly.

  2. Wider Range of Speeds:

    • At higher temperatures, the particles have a broader range of kinetic energies. Some particles move much faster than others, and the overall distribution flattens as high-speed particles become more common.

Detailed explanation here.

PART B. As the temperature increases, the fraction of particles with speeds exceeding 1000 m/s will increase significantly.

Detailed explanation here.

PART C. At 300 K, the Maxwell-Boltzmann distribution for H2​ would be shifted towards much higher speeds compared to the distribution for O2​. The H2​ distribution curve would be broader and flatter, while the O2​ distribution curve would be narrower and taller and peaked at lower speeds.

Detailed explanation here.

3.6  Deviations from Ideal Gas Law

3.6.1

PART A. Hydrogen (H2​) behaves more ideally because its molecules are smaller, lighter, and have weaker intermolecular forces (London dispersion only).

Detailed explanation here.

PART B. Hydrogen (H2​) would show slight deviations because its weak intermolecular forces make it less likely to condense or interact strongly. Carbon dioxide (CO2​) would deviate significantly because its stronger intermolecular forces become more influential as the particles lose kinetic energy, increasing attractions and reducing the pressure compared to ideal predictions.

Detailed explanation here.

PART C. Both gases deviate from the ideal gas law because particles are forced closer together, making intermolecular forces significant. Carbon dioxide (CO2​) deviates more because its larger size and stronger intermolecular forces lead to greater particle interactions, reducing the measured pressure compared to the ideal prediction.

Detailed explanation here.

3.7  Solutions and Mixtures

3.7.1

3.7.2

(1) A cup of black coffee - Homogeneous mixture; Black coffee is a solution where water is the solvent and compounds from the coffee beans (like caffeine) are dissolved uniformly.

(2) A salad with lettuce, tomatoes, and cucumbers - Heterogeneous mixture; The components (lettuce, tomatoes, cucumbers) are not uniformly distributed and can be easily distinguished and separated physically.

(3) Seawater - Homogeneous mixture; Seawater is a solution of water (solvent) and dissolved salts (solutes) that is uniform throughout.

(4) A gold ring (14-karat gold) - Homogeneous mixture (alloy); gold is an alloy, which is a uniform mixture of gold and other metals (such as silver or copper). The composition is consistent throughout the ring.

(5) A bottle of soda with visible bubbles - Heterogeneous mixture; While soda in its sealed, undisturbed state might appear homogeneous, the presence of visible bubbles of carbon dioxide gas in the liquid phase makes it a heterogeneous mixture because the gas and liquid phases are distinct and not uniformly mixed.

Detailed explanation here.

PART A. The molarity of the solution is 0.856 M

Detailed explanation here.

PART B. 0.0856 mol of NaCl are present in 100.0 mL of solution.

Detailed explanation here.

PART C. 5.00 g of NaCl are present in 100.0 mL of solution.

Detailed explanation here.

3.8  Representations of Solutions

3.8.1

PART A. The ions present after K2SO4 dissolves are potassium ions (K+) and sulfate ions (SO4^2-). Potassium ions have a +1 charge, and sulfate ions have a -2 charge.

Detailed explanation here.

PART B. For every one sulfate ion (SO4^2-) in the solution, there will be two potassium ions (K+). Therefore, if you were to illustrate this with 15 water molecules, you would show twice as many K+ ions as SO4^2- ions. For example, you could represent it with 2 SO4^2- ions and 4 K+ ions alongside 15 water molecules to maintain the 2:1 ratio of K+ to SO4^2-.

Detailed explanation here.

PART C. The interactions between water molecules and the ions are ion-dipole interactions. Water molecules, being polar, surround the ions. The positive potassium ions (K+) are attracted to the negative (oxygen) end of water molecules, while the negative sulfate ions (SO4^2-) are attracted to the positive (hydrogen) end of water molecules. These ion-dipole forces are the primary forces present between these particles.

Detailed explanation here.

3.9  Separation of Solutions and Mixtures Chromatography

3.9.1

PART A. Water has a higher boiling point than ethanol because its intermolecular forces (hydrogen bonding) are stronger than those in ethanol. Water forms extensive hydrogen bonds due to its smaller size and higher polarity, leading to a more cohesive liquid structure.

Detailed explanation here.

PART B. Ethanol has a lower boiling point, so at temperatures below 100∘ C, ethanol vaporizes more readily than water. The vapor above the mixture becomes enriched in ethanol because ethanol contributes more to the vapor pressure than water does.

Detailed explanation here.

PART C. Adding sodium chloride raises the boiling points of both ethanol and water (boiling point elevation), but the effect is more pronounced for water because it is more polar and dissolves more ions.

Detailed explanation here.

3.10  Solubility

3.10.1

PART A. Ethanol dissolves in water because both are polar and form hydrogen bonds. Hexane, being nonpolar, does not dissolve in polar water due to incompatible intermolecular forces. Conversely, nonpolar iodine dissolves in nonpolar hexane through London Dispersion Forces, but its interaction with polar water is weak, leading to slight solubility.

Detailed explanation here.

PART B. KCl will dissolve better in water because KCl is ionic, and water is polar, allowing for strong ion-dipole interactions that overcome KCl's ionic bonds. Hexane, being nonpolar, can only offer weak London Dispersion Forces, which are insufficient to dissolve KCl effectively.

Detailed explanation here.

3.11  Spectroscopy and the Electromagnetic Spectrum

3.11.1

PART A. Microwave radiation primarily affects molecular rotations, specifically the rotation of polar molecules like water.

Detailed explanation here.

PART B. Infrared (IR) radiation primarily affects molecular vibrations.

Detailed explanation here.

PART C. Ultraviolet (UV) and visible radiation absorption primarily causes electronic transitions in molecules like chlorophyll.

Detailed explanation here.

PART D. Different types of radiation affect molecules differently because they have different energies, and molecules have quantized energy levels associated with different types of motion and electronic states. The energy of the radiation must match the energy difference between molecular energy levels to be effectively absorbed and cause a change. Molecular structure determines what types of energy levels and transitions are available.

Detailed explanation here.

3.12  Photoelectric Effect

3.12.1

PART A. The frequency of light with a photon energy of 4.5 × 10⁻¹⁹ J is approximately 6.79 × 10¹⁴ Hz.

Detailed explanation here.

PART B. The wavelength of light with a photon energy of 4.5 × 10⁻¹⁹ J is approximately 442 nm.

Detailed explanation here.

PART C. This light belongs to the Ultraviolet/Visible region of the electromagnetic spectrum, specifically at the border between the ultraviolet and violet/blue visible light. For simplicity and common classifications, it can be broadly referred to as Visible or Violet/Blue Light.

Detailed explanation here.

PART D. Doubling the intensity of the light would double the number of ejected electrons. The energy (specifically, the maximum kinetic energy) of the ejected electrons would remain unchanged. This is because intensity increases the number of photons, but not the energy of each photon.

Detailed explanation here.

3.13  Beer-Lambert Law

3.13.1

3.13.2

PART A. The concentration of the dye in the solution is 2.5×10−3M or 0.0025M.

Detailed explanation here.

PART B. The new absorbance of the solution with a path length of 4.00 cm is 1.500.

Detailed explanation here.

PART C. The absorbance of this new solution with doubled concentration at a path length of 2.00 cm is predicted to be 1.500.

Detailed explanation here.

PART A. An appropriate wavelength to gather absorption data for Solution X would be around 575 nm to 600 nm.

Detailed explanation here.

PART B. The color of light that is absorbed is yellow-green to yellow.

Detailed explanation here.

PART C. The predicted color of this solution is violet to blue-violet.

Detailed explanation here.

PART D.

Differences: The new solution (unknown concentration - blue line in the second graph) has a lower absorbance across all wavelengths compared to the first solution (1.0M - red line in the second graph). Specifically, the peak absorbance of the new solution is lower. This indicates that the concentration of solute X in the new solution is lower than in the first solution.

Similarities: The shape of the absorption spectrum and the wavelength of maximum absorbance are the same for both solutions. Both spectra exhibit a similar peak in the yellow-green to yellow region (around 575-600 nm). This similarity in the spectrum's shape and peak position indicates that both solutions contain the same solute, X. The molar absorptivity (ε), which is related to the shape of the spectrum and peak wavelength, is a property of the substance itself and remains constant regardless of concentration.

Detailed explanation here.