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Chapter 11: Problem 20

True or false: (a) Molecules containing polar bonds must be polar molecules and have dipole-dipole forces. (b) For the halogen gases, the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, the more polar bonds there are in a molecule, the stronger the dipole-dipole interaction. \((\mathbf{d})\) All other factors being the same, total attractive forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The more electronegative the atom, the more polarizable it is.

Short Answer

Expert verified
a) False, b) False, c) False, d) True, e) False.

Step by step solution

01

Polar Bonds and Polar Molecules

Examine whether molecules with polar bonds are necessarily polar molecules with dipole-dipole forces. Molecules can have polar bonds but be nonpolar if the molecular geometry causes the dipoles to cancel, such as in carbon dioxide. Therefore, a molecule with polar bonds is not necessarily a polar molecule or possess dipole-dipole forces. **Answer:** False.
02

Dispersion Forces and Boiling Points of Halogens

Determine the relationship between dispersion forces and boiling points in halogen gases down the periodic table. As we go from fluorine to iodine, the size and mass of the molecules increase, increasing dispersion forces. Higher dispersion forces correlate with higher boiling points. The statement that dispersion forces decrease is incorrect. **Answer:** False.
03

More Polar Bonds and Dipole-Dipole Interactions

Evaluate if more polar bonds lead to stronger dipole-dipole interactions. While more polar bonds could contribute more dipole moments, the overall molecular shape and arrangement are what determine the dipole-dipole interactions. Thus, this statement isn't necessarily true. **Answer:** False.
04

Shape Impact on Attractive Forces

Assess how molecular shape affects total attractive forces. Linear molecules can pack closer together, increasing the contact area for intermolecular forces compared to spherical molecules. As such, linear molecules generally exhibit stronger total attractive forces assuming all else is equal. **Answer:** True.
05

Electronegativity and Polarizability

Consider the link between electronegativity and polarizability. More electronegative atoms hold onto their electrons more tightly, making them less polarizable. Hence, more electronegative atoms are generally less polarizable. **Answer:** False.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Bonds
Polar bonds occur when there is a difference in electronegativity between two atoms in a molecule. This difference causes an uneven distribution of electron density, leading to a partial positive charge on one atom and a partial negative charge on the other. We refer to the separation of charges as a dipole moment.
An important aspect to keep in mind is that having polar bonds doesn't automatically make a molecule polar.
  • Molecular symmetry can cause dipole moments to cancel each other out.
  • For example, carbon dioxide (CO鈧�) has polar bonds but is nonpolar due to its linear shape.
In such cases, the overall molecule lacks a net dipole moment, meaning it will not demonstrate dipole-dipole interactions.
Therefore, while polar bonds are essential for forming polar molecules, other factors like molecular geometry play a critical role.
Dispersion Forces
Dispersion forces, often called London dispersion forces, are weak intermolecular forces that arise from temporary dipoles in molecules. These forces are present in all molecular interactions but tend to be the sole intermolecular forces in nonpolar molecules.
Interestingly, as you move down the halogens in the periodic table from fluorine to iodine:
  • The size and molecular weight increase, enhancing the strength of the dispersion forces.
  • This increase results in higher boiling points with heavier halogens as they offer more electrons that can shift to create temporary dipoles.
Thus, contrary to some misconceptions, the increase in dispersion forces leads to increased boiling points down the group rather than a decrease.
Induced Dipole
An induced dipole occurs when a nonpolar molecule temporarily becomes polar due to the influence of nearby charged particles or polar molecules. This happens when an external electric field distorts the electron cloud of the nonpolar molecule, creating a temporary dipole.
This concept is crucial in understanding dispersion forces and their interaction with polar and nonpolar substances. Induced dipoles are key players when:
  • Two nonpolar substances interact, creating temporary alignment and attraction.
  • Nonpolar molecules are near polar ones, enhancing the strength of the overall interaction due to mutual attraction.
While typically weaker than permanent dipole-dipole interactions, induced dipole forces are essential in the behavior of gases and liquids, especially under varying temperature and pressure conditions.

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Most popular questions from this chapter

Rationalize the difference in boiling points in each pair: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\left(-23^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \quad\left(78^{\circ} \mathrm{C}\right)\), (b) \(\mathrm{CO}_{2}\left(-78.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CS}_{2}\left(46.2^{\circ} \mathrm{C}\right),(\mathbf{c}) \mathrm{CH}_{3} \mathrm{COCH}_{3}\left(50.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{COOH}\left(101^{\circ} \mathrm{C}\right)\).

A watch with a liquid crystal display (LCD) does not function properly when it is exposed to low temperatures during a trip to Antarctica. Explain why the LCD might not function well at low temperature.

(a) What is the relationship between surface tension and temperature? (b) What is the relationship between viscosity and temperature? (c) Why do substances with high surface tension also tend to have high viscosities?

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) and pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) are both liquids at room temperature and room pressure, and have about the same molecular weight. (a) One of these liquids is much more viscous than the other. Which one do you predict is more viscous? (b) One of these liquids has a much lower normal boiling point \(\left(36.1^{\circ} \mathrm{C}\right)\) compared to the other one \(\left(198^{\circ} \mathrm{C}\right) .\) Which liquid has the lower normal boiling point? (c) One of these liquids is the major component in antifreeze in automobile engines. Which liquid would you expect to be used as antifreeze? (d) One of these liquids is used as a "blowing agent" in the manufacture of polystyrene foam because it is so volatile. Which liquid would you expect to be used as a blowing agent?

Name the phase transition in each of the following situations and indicate whether it is exothermic or endothermic: (a) Ice-cream melts at room temperature. (b) Potato slices become crisp when fried. \((\mathbf{c})\) Droplets of water appear on the lid of a food container containing freshly baked bread. (d) You can see your own breath on a cold day.
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