/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Which member of each of the foll... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which member of each of the following pairs of compounds has the higher boiling point? (a) \(\mathrm{O}_{2}\) or \(\mathrm{N}_{2}\) (b) \(\mathrm{SO}_{2}\) or \(\mathrm{CO}_{2}\) (c) HF or HI (d) \(\mathrm{SiH}_{4}\) or \(\mathrm{GeH}_{4}\)

Short Answer

Expert verified
(a) \(\mathrm{O}_{2}\), (b) \(\mathrm{SO}_{2}\), (c) HF, (d) \(\mathrm{GeH}_{4}\).

Step by step solution

01

Understanding Boiling Points

The boiling point of a compound is influenced by the strength and type of intermolecular forces present. Factors such as molecular size, shape, and the ability to form hydrogen bonds or other dipole interactions play a significant role.
02

Analyzing (a) \\mathrm{O}_{2} vs. \\mathrm{N}_{2}

Both \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) are diatomic non-polar molecules, and their only intermolecular forces are London dispersion forces. Larger electron clouds in \(\mathrm{O}_{2}\) make its dispersion forces slightly stronger than \(\mathrm{N}_{2}\), thus \(\mathrm{O}_{2}\) has a higher boiling point.
03

Analyzing (b) \\mathrm{SO}_{2} vs. \\mathrm{CO}_{2}

\(\mathrm{SO}_{2}\) is a bent polar molecule which can have dipole-dipole interactions, while \(\mathrm{CO}_{2}\) is a linear non-polar molecule and only has London dispersion forces. The stronger dipole-dipole interactions in \(\mathrm{SO}_{2}\) lead to a higher boiling point than \(\mathrm{CO}_{2}\).
04

Analyzing (c) HF vs. HI

Although both HF and HI are polar, HF can form hydrogen bonds, which are significantly stronger than the dipole-dipole attractions in HI. Consequently, HF has a much higher boiling point than HI.
05

Analyzing (d) \\mathrm{SiH}_{4} vs. \\mathrm{GeH}_{4}

Both \(\mathrm{SiH}_{4}\) and \(\mathrm{GeH}_{4}\) are similar molecules, but \(\mathrm{GeH}_{4}\) is larger and has more electrons, resulting in stronger London dispersion forces. Therefore, \(\mathrm{GeH}_{4}\) has a higher boiling point than \(\mathrm{SiH}_{4}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Boiling Point
Boiling points are essential when discussing the properties of substances. The boiling point of a compound is the temperature at which it transitions from a liquid to a gas. This is heavily influenced by the intermolecular forces acting within the substance. Stronger intermolecular forces result in a higher boiling point because it requires more energy to separate the molecules. Various factors, such as molecular size, shape, or the presence of specific types of intermolecular forces like hydrogen bonding, significantly affect the boiling point. Therefore, understanding the boiling point is crucial for predicting the behavior of substances under different temperatures.
Dipole-Dipole Interactions
Dipole-dipole interactions occur in polar molecules where there exists a permanent dipole moment. This happens when there is a significant difference in electronegativity between bonded atoms, causing a partial positive and a partial negative charge to develop at different ends of the molecule. Dipole-dipole interactions result from the attraction between the positively and negatively charged ends of different molecules. These interactions significantly increase the boiling point of substances because the attraction needs to be overcome to convert the substance from liquid to vapor. Therefore, molecules with significant dipole moments usually possess higher boiling points compared to those with weaker or no dipole interactions.
London Dispersion Forces
London Dispersion Forces are a type of van der Waals force that arise even in non-polar molecules. They are due to temporary, instantaneous dipoles that occur due to fluctuations in the electron cloud density. Heavier atoms or larger molecules tend to have stronger dispersion forces because they have larger electron clouds that are more easily polarizable. While these forces are typically weaker than other intermolecular forces like hydrogen bonds or permanent dipole interactions, they can still significantly influence boiling points. This is especially true for larger non-polar molecules or atoms, where dispersion forces become the primary intermolecular attraction.
Hydrogen Bonding
Hydrogen bonding is a special type of dipole-dipole interaction but considerably stronger. It occurs in molecules where hydrogen is directly bonded to highly electronegative atoms such as nitrogen, oxygen, or fluorine. This bond leads to a pronounced dipole where the hydrogen atom carries a partial positive charge due to the unequal sharing of electrons. This allows the hydrogen to form a strong attraction with lone pairs from other electronegative atoms in nearby molecules. This powerful interaction results in higher boiling points compared to other molecules that do not involve hydrogen bonding. Substances capable of hydrogen bonding often require significantly higher temperatures to overcome the intermolecular forces and transition into the gaseous state.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Acetaminophen is used in analgesics. (A model is in Organic: Alcohols.) (a) Draw the structure of acetaminophen. (b) Is the molecule capable of hydrogen bonding? If so, what are the sites of hydrogen bonding?

Assuming that in a simple cubic unit cell the spherical atoms or ions just touch along the cube's edges, calculate the percentage of empty space within the unit cell. (Recall that the volume of a sphere is \((4 / 3) \pi r^{3},\) where \(r\) is the radius of the sphere.) (IMAGE NOT COPY)

The specific heat capacity of silver is \(0.235 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\). Its melting point is \(962^{\circ} \mathrm{C},\) and its heat of fusion is \(11.3 \mathrm{kJ} / \mathrm{mol}\) What quantity of heat, in joules, is required to change \(5.00 \mathrm{g}\) of silver from a solid at \(25^{\circ} \mathrm{C}\) to a liquid at \(962^{\circ} \mathrm{C} ?\)

If your air conditioner is more than several years old, it may use the chlorofluorocarbon \(\mathrm{CCl}_{2} \mathrm{F}_{2}\) as the heat transfer fluid. The normal boiling point of \(\mathrm{CCl}_{2} \mathrm{F}_{2}\) is \(-29.8^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(20.11 \mathrm{kJ} / \mathrm{mol}\). The gas and the liquid have specific heats of \(117.2 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) and \(72.3 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K},\) respectively. How much heat is evolved when \(20.0 \mathrm{g}\) of \(\mathrm{CCl}_{2} \mathrm{F}_{2}\) is cooled from \(+40^{\circ} \mathrm{C}\) to \(-40^{\circ} \mathrm{C} ?\)

The density of copper metal is \(8.95 \mathrm{g} / \mathrm{cm}^{3} .\) If the radius of a copper atom is \(127.8 \mathrm{pm},\) is the copper unit cell simple cubic, body-centered cubic, or face-centered cubic?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.