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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) H₂S, due to its larger size and greater polarizability, (b) CO₂, because of its higher electron count and increased polarizability, and (c) GeH₄, as it has more electrons leading to stronger charge fluctuations and dispersion forces.

Step by step solution

01

Pair (a): H₂O and H₂S

Let us first analyze the properties of H₂O and H₂S. H₂S is a larger molecule with more electrons, making it more polarizable compared to H₂O. As a result, H₂S will be more prone to charge fluctuations that induce dispersion forces. Therefore, H₂S has larger dispersion forces than H₂O.
02

Pair (b): CO₂ and CO

Now, let's compare CO₂ and CO. CO has 10 electrons, while CO₂ has 22 electrons. Though CO₂ 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, CO₂ has larger dispersion forces than CO.
03

Pair (c): SiH₄ and GeH₄

Finally, let us analyze and compare SiH₄ and GeH₄. SiH₄ has 20 electrons, while GeH₄ has 32 electrons. Both SiH₄ and GeH₄ have similar structures, but GeH₄ 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, GeH₄ has larger dispersion forces than SiH₄. In summary, the members with the larger dispersion forces in each pair are (a) H₂S, (b) CO₂, and (c) GeH₄.

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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 H₂S, CO₂, and GeH₄ 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**: H₂S is larger than H₂O, CO₂ is larger than CO, and GeH₄ is larger than SiH₄, 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**: - **CO₂ vs. CO**: CO₂ has more electrons than CO, creating larger induced dipoles despite CO₂ being non-polar. - **GeH₄ vs. SiH₄**: GeH₄'s greater number of electrons compared to SiH₄ supports more significant charge fluctuations. - **H₂S vs. H₂O**: H₂S 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

The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is a \(-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melting points of the salts that form between the 1 -alkyl-3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{m.p.}=40^{\circ} \mathrm{C}\right)\) \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=10^{\circ} \mathrm{C}\right)\) and \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=-61^{\circ} \mathrm{C}\right) . \mathrm{Why}\) does the melting point decrease as the length of alkyl group increases?

The following quote about ammonia \(\left(\mathrm{NH}_{3}\right)\) is from a textbook of inorganic chemistry: "It is estimated that \(26 \%\) of the hydrogen bonding in \(\mathrm{NH}_{3}\) breaks down on melting, \(7 \%\) on warming from the melting to the boiling point, and the final \(67 \%\) on transfer to the gas phase at the boiling point." From the standpoint of the kinetic energy of the molecules, explain (a) why there is a decrease of hydrogen-bonding energy on melting and (b) why most of the loss in hydrogen bonding occurs in the transition from the liquid to the vapor state.

The critical temperatures \((\mathrm{K})\) and pressures \((\mathrm{atm})\) of a series of halogenated methanes are as follows: $$ \begin{array}{lcccc} \text { Compound } & \mathbf{C C l}_{3} \mathbf{F} & \mathbf{C C l}_{2} \mathbf{F}_{2} & \mathbf{C C I F}_{3} & \mathbf{C F}_{4} \\ \hline \text { Critical temperature } & 471 & 385 & 302 & 227 \\ \text { Critical pressure } & 43.5 & 40.6 & 38.2 & 37.0 \end{array} $$ (a) List the intermolecular forces that occur for each compound. (b) Predict the order of increasing intermolecular attraction, from least to most, for this series of compounds. (c) Predict the critical temperature and pressure for \(\mathrm{CCl}_{4}\) based on the trends in this table. Look up the experimentally determined critical temperatures and pressures for \(\mathrm{CCl}_{4}\), using a source such as the CRC Handbook of Chemistry and Physics, and suggest a reason for any discrepancies.

At standard temperature and pressure the molar volume of \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) gases are \(22.06 \mathrm{~L}\) and \(22.40 \mathrm{~L},\) respectively (a) Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same? (b) \(\mathrm{On}\) cooling to \(160 \mathrm{~K}\), both substances form crystalline solids. Do you expect the molar volumes to decrease or increase on cooling to \(160 \mathrm{~K} ?\) (c) The densities of crystalline \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) at \(160 \mathrm{~K}\) are \(2.02 \mathrm{~g} / \mathrm{cm}^{3}\) and \(0.84 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. Calculate their molar volumes. (d) Are the molar volumes in the solid state as similar as they are in the gaseous state? Explain. (e) Would you expect the molar volumes in the liquid state to be closer to those in the solid or gaseous state?

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water? (The heat of vaporization of water in this temperature range is \(2.4 \mathrm{~kJ} / \mathrm{g}\). The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .)\)

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