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Which member of the following pairs has the larger London dispersion forces: (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{~S}\), (b) \(\mathrm{CO}_{2}\) or \(\mathrm{CO}\), (c) \(\mathrm{SiH}_{4}\) or \(\mathrm{GeH}_{4} ?\)

Short Answer

Expert verified
The molecules with larger London dispersion forces are \(H_2S\) (pair 1), \(CO_2\) (pair 2), and \(GeH_4\) (pair 3).

Step by step solution

01

Pair 1: H2O vs. H2S

First, we will compare the electron count for each molecule. Hydrogen has atomic number 1, oxygen has atomic number 8, and sulfur has atomic number 16. The electron counts for H2O and H2S will be: - H2O: 2(H) + 1(O) = 10 electrons - H2S: 2(H) + 1(S) = 18 electrons H2S has more electrons compared to H2O. Also, sulfur is larger in size compared to oxygen. Therefore, H2S will have larger London dispersion forces than H2O.
02

Pair 2: CO2 vs. CO

In this pair, we will compare the electron count for each molecule. Carbon has atomic number 6 and oxygen has atomic number 8. The electron counts for CO2 and CO will be: - CO2: 1(C) + 2(O) = 22 electrons - CO: 1(C) + 1(O) = 14 electrons CO2 has more electrons compared to CO. Also, CO2 is larger in size and has a linear shape. Therefore, CO2 will have larger London dispersion forces than CO.
03

Pair 3: SiH4 vs. GeH4

For this pair, we will again compare the electron count for each molecule. Silicon has atomic number 14, and germanium has atomic number 32. The electron counts for SiH4 and GeH4 will be: - SiH4: 1(Si) + 4(H) = 18 electrons - GeH4: 1(Ge) + 4(H) = 36 electrons GeH4 has more electrons compared to SiH4. Also, germanium is larger in size compared to silicon. Therefore, GeH4 will have larger London dispersion forces than SiH4. Thus, the molecules with larger London dispersion forces are H2S (pair 1), CO2 (pair 2), and GeH4 (pair 3).

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Key Concepts

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

Electron Count
In the world of chemistry, the number of electrons in a molecule plays a crucial role in determining the strength of its London dispersion forces. This is because London dispersion forces are a type of intermolecular force that arises from temporary shifts in electron density. These shifts create fleeting dipoles, which then induce dipoles in neighboring molecules, leading to weak attractions.
Increasing the electron count results in greater likelihood of these temporay fluctuations. More electrons mean larger temporary dipole moments and, consequently, stronger dispersion forces.
Let's take a look at our examples:
  • For \(H_2O\), there are 10 electrons, while \(H_2S\) has 18 electrons. \(H_2S\) has a higher electron count, leading to larger dispersion forces.
  • Comparing \(CO_2\) (22 electrons) and \(CO\) (14 electrons), \(CO_2\) wins with its greater electron count.
  • In the case of \(SiH_4\) versus \(GeH_4\), \(GeH_4\) has 36, vis-a-vis \(SiH_4\)'s 18 electrons. This higher electron count in \(GeH_4\) again points to stronger dispersion forces.
Molecular Size
Molecular size is another key factor when considering London dispersion forces. As a general rule, larger molecules exhibit stronger London dispersion forces. This is because larger molecules have a bigger surface area allowing for more contact between molecules, increasing the chances of the formation of these temporary dipoles.
Larger molecules also tend to have more electrons, contributing further to stronger dispersion forces, as discussed earlier.
In case studies:
  • In the \(H_2O\) vs. \(H_2S\) comparison, \(H_2S\), which contains sulfur, is larger in size than \(H_2O\) with oxygen.
  • Comparing \(CO_2\) and \(CO\), \(CO_2\) is not only a larger molecule but also structured in a linear way that maximizes its size.
  • Between \(SiH_4\) and \(GeH_4\), germanium in \(GeH_4\) is larger than silicon in \(SiH_4\).
Intermolecular Forces
Intermolecular forces are the adhesive forces that hold molecules together. Among these, London dispersion forces are particularly notable because they act between all particles, regardless of polarity. They are the weakest, yet they are the only type of intermolecular force present in nonpolar compounds.
London dispersion forces arise from temporary dipoles that occur due to fluctuations in electron cloud density. These forces increase with higher electron count and larger molecular size.
Examples include:
  • Water \(H_2O\) versus hydrogen sulfide \(H_2S\): even though \(H_2O\) can form hydrogen bonds, \(H_2S\) has greater London dispersion forces due to more electrons.
  • Carbon dioxide \(CO_2\) has greater dispersion forces than carbon monoxide \(CO\) because of its larger size and higher electron count.
  • For \(SiH_4\) compared to \(GeH_4\), the intermolecular forces are enhanced in \(GeH_4\) thanks to its greater number of electrons and bigger size.
Chemical Bonding
Chemical bonding refers to the attractions that hold atoms together within a molecule. In the context of London dispersion forces, chemical bonding isn't directly involved, yet orbitals and structure influence how electron density can change.
While chemical bonds are strong and hold a molecule together, the weak London dispersion forces are more about interactions between molecules rather than within.
For example:
  • The covalent bonds in \(H_2O\) between hydrogen and oxygen are strong, yet the London dispersion forces when comparing with \(H_2S\) are large due to \(H_2S\)’s higher electron count.
  • In \(CO_2\) and \(CO\), the types of bonding within the molecules differ, with \(CO_2\) having two double bonds, yet dispersion forces consider outside molecular interactions and electron number.
  • For \(SiH_4\) and \(GeH_4\), both have covalent bonds within, but the increased dispersion force in \(GeH_4\) is primarily due to its electron count and molecular size.

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Most popular questions from this chapter

Amorphous silica has a density of about \(2.2 \mathrm{~g} / \mathrm{cm}^{3}\), whereas the density of crystalline quartz is \(2.65 \mathrm{~g} / \mathrm{cm}^{3}\). Account for this difference in densities.

Clausthalite is a mineral composed of lead selenide (PbSe). The mineral adopts a NaCl-type structure. The density of PbSe at \(25^{\circ} \mathrm{C}\) is \(8.27 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the length of an edge of the PbSe unit cell.

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\), the major substance in antifreeze, has a normal boiling point of \(198^{\circ} \mathrm{C}\). By comparison, ethyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) boils at \(78^{\circ} \mathrm{C}\) at atmospheric pressure. Ethylene glycol dimethyl ether \(\left(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}\right)\) has a normal boiling point of \(83^{\circ} \mathrm{C}\), and ethyl methyl ether \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{3}\right)\) has a normal boiling point of \(11^{\circ} \mathrm{C}\). (a) Explain why replacement of a hydrogen on the oxygen by \(\mathrm{CH}_{3}\) generally results in a lower boiling point. (b) What are the major factors responsible for the difference in boiling points of the two ethers?

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the ClausiusClapeyron equation, Equation 11.1, derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a major component of which is octane, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}\) \(\mathrm{C} \mathrm{H}_{2} \mathrm{C} \mathrm{H}_{3} .\) Octane has a vapor pressure of \(13.95\) torr at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(144.78\) torr at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. (c) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise \(11.86 .\) (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

(a) What is the significance of the critical pressure of a substance? (b) What happens to the critical temperature of a series of compounds as the force of attraction between molecules increases? (c) Which of the substances listed in Table \(11.5\) can be liquefied at the temperature of liquid nitrogen \(\left(-196^{\circ} \mathrm{C}\right)\) ?

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