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Which member in each pair has the larger dispersion forces: (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{~S},(\mathbf{b}) \mathrm{CO}_{2}\) or \(\mathrm{CO}, \mathrm{C}\) (c) \(\mathrm{SiH}_{4}\) or \(\mathrm{GeH}_{4} ?\)

Short Answer

Expert verified
In each pair, the members with the larger dispersion forces are (a) Hâ‚‚S, due to its larger size and greater polarizability, (b) COâ‚‚, because of its higher electron count and increased polarizability, and (c) GeHâ‚„, as it has more electrons leading to stronger charge fluctuations and dispersion forces.

Step by step solution

01

Pair (a): Hâ‚‚O and Hâ‚‚S

Let us first analyze the properties of Hâ‚‚O and Hâ‚‚S. Hâ‚‚S is a larger molecule with more electrons, making it more polarizable compared to Hâ‚‚O. As a result, Hâ‚‚S will be more prone to charge fluctuations that induce dispersion forces. Therefore, Hâ‚‚S has larger dispersion forces than Hâ‚‚O.
02

Pair (b): COâ‚‚ and CO

Now, let's compare COâ‚‚ and CO. CO has 10 electrons, while COâ‚‚ has 22 electrons. Though COâ‚‚ is a linear and non-polar molecule, it has a higher electron count than CO, making it more polarizable. The increased electron count results in stronger charge fluctuations and subsequently larger dispersion forces. Therefore, COâ‚‚ has larger dispersion forces than CO.
03

Pair (c): SiHâ‚„ and GeHâ‚„

Finally, let us analyze and compare SiHâ‚„ and GeHâ‚„. SiHâ‚„ has 20 electrons, while GeHâ‚„ has 32 electrons. Both SiHâ‚„ and GeHâ‚„ have similar structures, but GeHâ‚„ contains more electrons due to germanium's higher atomic number. The increased electron count leads to larger charge fluctuations and, as a result, stronger dispersion forces. Therefore, GeHâ‚„ has larger dispersion forces than SiHâ‚„. In summary, the members with the larger dispersion forces in each pair are (a) Hâ‚‚S, (b) COâ‚‚, and (c) GeHâ‚„.

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Key Concepts

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

Polarizability
Polarizability is a key factor when discussing dispersion forces within molecules. It refers to the ability of a molecule to become polarized — which means that electrons within the molecule can be distorted by an external electric field, leading to temporary dipoles. These dipoles then affect how molecules interact with each other.

- **High Polarizability**: More easily allows for the creation of temporary dipoles, resulting in stronger dispersion forces. - **Factors Affecting Polarizability**: Larger molecules with more electrons are generally more polarizable, as the outer electrons are further away from the nucleus and are less tightly held, making them easier to distort.

In the exercise, we see that Hâ‚‚S, COâ‚‚, and GeHâ‚„ have more electrons or a larger atomic structure compared to their counterparts, making them more polarizable. As polarizability increases, so do the dispersion forces.
Molecular Size
The size of a molecule significantly influences its dispersion forces. Generally, the larger the molecule, the greater the dispersion forces it can experience. This is because larger molecules have more electron cloud volume, increasing the possibility of induced dipole interactions.

- **Larger Molecules**: Have increased surface area for interactions, allowing for stronger force generation between molecules. - **Example in Exercise**: Hâ‚‚S is larger than Hâ‚‚O, COâ‚‚ is larger than CO, and GeHâ‚„ is larger than SiHâ‚„, meaning they can exert stronger dispersion forces due to their greater size.

Understanding molecular size helps explain why larger molecules usually lead to stronger dispersion forces, helping to predict interactions between non-polar molecules.
Electron Count
Electron count plays a crucial role in determining the strength of dispersion forces. Molecules with a higher number of electrons can experience greater charge fluctuations, resulting in stronger temporary dipoles and thus stronger dispersion forces.

- **Higher Electron Counts**: Lead to larger and more frequent oscillations in electron density, enhancing the temporary polarity of molecules. - **Specific Pair Analyses**: - **COâ‚‚ vs. CO**: COâ‚‚ has more electrons than CO, creating larger induced dipoles despite COâ‚‚ being non-polar. - **GeHâ‚„ vs. SiHâ‚„**: GeHâ‚„'s greater number of electrons compared to SiHâ‚„ supports more significant charge fluctuations. - **Hâ‚‚S vs. Hâ‚‚O**: Hâ‚‚S boasts more electrons, leading to stronger dispersion forces.

A comprehensive understanding of electron count helps predict which molecules will exhibit stronger dispersion forces, critical for understanding molecular interactions in chemistry.

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Most popular questions from this chapter

Liquid butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is stored in cylinders to be used as a fuel. The normal boiling point of butane is listed as \(-0.5^{\circ} \mathrm{C}\). (a) Suppose the tank is standing in the sun and reaches a temperature of \(35^{\circ} \mathrm{C}\). Would you expect the pressure in the tank to be greater or less than atmospheric pressure? How does the pressure within the tank depend on how much liquid butane is in it? (b) Suppose the valve to the tank is opened and a few liters of butane are allowed to escape rapidly. What do you expect would happen to the temperature of the remaining liquid butane in the tank? Explain. (c) How much heat must be added to vaporize \(250 \mathrm{~g}\) of butane if its heat of vaporization is \(21.3 \mathrm{~kJ} / \mathrm{mol} ?\) What volume does this much butane occupy at 755 torr and \(35^{\circ} \mathrm{C} ?\)

The table shown here lists the molar heats of vaporization for several organic compounds. Use specific examples from this list to illustrate how the heat of vaporization varies with (a) molar mass, (b) molecular shape, (c) molecular polarity, (d) hydrogen-bonding interactions. Explain these comparisons in terms of the nature of the intermolecular forces at work. (You may find it helpful to draw out the structural formula for each compound.) $$ \begin{array}{ll} \text { Compound } & \begin{array}{l} \text { Heat of Vaporization } \\ \mathbf{( k J / m o l )} \end{array} \\ \hline \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3} & 19.0 \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} & 27.6 \\ \mathrm{CH}_{3} \mathrm{CHBrCH}_{3} & 31.8 \\ \mathrm{CH}_{3} \mathrm{COCH}_{3} & 32.0 \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} & 33.6 \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 47.3 \end{array} $$

Rationalize the difference in boiling points in each pair: (a) \(\mathrm{HF}\left(20^{\circ} \mathrm{C}\right)\) and \(\mathrm{HCl}\left(-85^{\circ} \mathrm{C}\right),(\mathbf{b}) \mathrm{CHCl}_{3}\left(61{ }^{\circ} \mathrm{C}\right)\) and \(\mathrm{CHBr}_{3}\) \(\left(150^{\circ} \mathrm{C}\right),(\mathrm{c}) \mathrm{Br}_{2}\left(59^{\circ} \mathrm{C}\right)\) and \(\mathrm{ICl}\left(97^{\circ} \mathrm{C}\right)\)

(a) When you exercise vigorously, you sweat. How does this help your body cool? (b) A flask of water is connected to a vacuum pump. A few moments after the pump is turned on, the water begins to boil. After a few minutes, the water begins to freeze. Explain why these processes occur.

Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) is widely used as an industrial solvent. (a) Draw the Lewis structure for the acetone molecule and predict the geometry around each carbon atom. (b) Is the acetone molecule polar or nonpolar? (c) What kinds of intermolecular attractive forces exist between acetone molecules? (d) 1-Propanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH},\) has a molecular weight that is very similar to that of acetone, yet acetone boils at \(56.5^{\circ} \mathrm{C}\) and 1-propanol boils at \(97.2^{\circ} \mathrm{C}\). Explain the difference.

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