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The elements xenon and gold both have solid-state structures with face- centered cubic unit cells, yet \(\mathrm{Xe}\) melts at \(-112^{\circ} \mathrm{C}\) and gold melts at \(1064^{\circ} \mathrm{C}\). Account for these greatly different melting points.

Short Answer

Expert verified
In summary, the difference in melting points between Xe and Au, both having face-centered cubic unit cell structures, can be attributed to the bond strength. Xe has weak London dispersion forces leading to a low melting point of \(-112^{\circ} \mathrm{C}\), whereas Au has strong metallic bonding resulting in a high melting point of \(1064^{\circ} \mathrm{C}\).

Step by step solution

01

Identify the type of bonding for both elements

While Xe is a noble gas with weak London dispersion forces, Au is a metal with metallic bonding. The differences in the type of bonding and the forces between atoms lead to different melting points.
02

Compare the strength of bonding in Xe and Au

In Xe's solid-state, the atoms are held together by weak London dispersion forces. These are temporary, induced dipoles between atoms from fluctuations in the electron distribution. Due to its full outer electron shell and large atomic size, dispersion forces in Xe are relatively weak. Contrastingly, Au has metallic bonding, which is strong due to the attraction between positive metal ions and delocalized electrons. Metallic bonding allows for electrons to move freely, creating a strong bond between atoms. In Au's case, the strong metallic bonding leads to a high melting point.
03

Relating the bond strength to melting points

The melting point of a substance depends on its bond strength. The stronger the bonds between atoms, the more energy (heat) is required to break those bonds and turn the solid into a liquid. In Xe, since it's only held together by weak London dispersion forces, it requires a lesser amount of energy to overcome these weak forces, and hence it melts at a lower temperature (-112°C). In contrast, Au has strong metallic bonding requiring a higher amount of energy to break, which results in a much higher melting point (1064°C).
04

Conclusion:

In summary, the greatly different melting points between Xe and Au are due to the differences in the bond strength. Xe's weak London dispersion forces result in a correspondingly low melting point, whereas Au's strong metallic bonding leads to a high melting point. Despite both elements having face-centered cubic unit cell structures, their melting points are influenced primarily by their bond strength.

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Key Concepts

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

Face-Centered Cubic Unit Cell
The face-centered cubic unit cell is a type of crystal structure that many elements can adopt in their solid state. Essentially, it includes one atom at each of the corners of a cube and one atom positioned on the center of each face of the cube. This arrangement forms a highly symmetrical and dense packing of atoms. Although both xenon and gold have this type of arrangement, the face-centered cubic unit cell does not account for their melting points. That's because it is the nature of the bonding between the atoms, not the unit cell itself, that heavily influences melting points. Thus, xenon and gold may share the same crystal structure but still have vastly different melting temperatures due to their distinct types of bonding.
Metallic Bonding
Metallic bonding is a unique form of chemical bonding that holds metal atoms together. It comes from the attraction between positively charged metal ions and a sea of delocalized electrons that move freely throughout the metal. This free flow of electrons is what imparts properties typical of metals, such as electrical conductivity and malleability. The strength of metallic bonds depends on several factors, including the number of delocalized electrons and the charge of the metal ion. Because these bonds are quite strong, metals like gold require significant energy to break these bonds, thus explaining its high melting point of 1064°C. The robust nature of metallic bonding ensures that the atoms are tightly bound, resulting in a resilient structure.
London Dispersion Forces
London dispersion forces are the weakest type of intermolecular forces and arise from temporary shifts in the electron cloud around atoms or molecules. As electrons move, they can create transient dipoles, inducing similar dipoles in adjacent particles, leading to a weak attraction. These forces are especially prominent in noble gases like xenon, where other stronger bonding interactions are absent. Because these forces are weak, they are easily overcome with little energy input, which explains why xenon has a low melting point of -112°C. Despite the face-centered cubic arrangement, the weak dispersion forces in xenon result in a significantly lower melting point compared to strong metallic bonds in other elements like gold. These forces tend to increase in strength with larger electron clouds, which is why xenon's dispersion forces are stronger than those in smaller noble gases, but still comparatively weak overall.

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Most popular questions from this chapter

Using the following list of normal boiling points for a series of hydrocarbons, estimate the normal boiling point for octane, \(\mathrm{C}_{8} \mathrm{H}_{18}:\) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8},-42.1{ }^{\circ} \mathrm{C}\right)\), bu- tane \(\left(\mathrm{C}_{4} \mathrm{H}_{10},-0.5^{\circ} \mathrm{C}\right)\), pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}, 36.1^{\circ} \mathrm{C}\right)\), hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}, 68.7^{\circ} \mathrm{C}\right)\), heptane \(\left(\mathrm{C}_{7} \mathrm{H}_{16}, 98.4{ }^{\circ} \mathrm{C}\right)\). Explain the trend in the boiling points.

Refer to Figure \(11.27(\mathrm{a})\), and describe all the phase changes that would occur in each of the following cases: (a) Water vapor originally at \(0.005 \mathrm{~atm}\) and \(-0.5^{\circ} \mathrm{C}\) is slowly compressed at constant temperature until the final pressure is \(20 \mathrm{~atm} .\) (b) Water originally at \(100.0^{\circ} \mathrm{C}\) and \(0.50 \mathrm{~atm}\) is cooled at constant pressure until the temperature is \(-10^{\circ} \mathrm{C}\).

Using information in Appendices \(\mathrm{B}\) and \(\mathrm{C}\), calculate the minimum number of grams of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\), that must be combusted to provide the energy necessary to convert \(5.50 \mathrm{~kg}\) of ice at \(-20^{\circ} \mathrm{C}\) to liquid water at \(75^{\circ} \mathrm{C}\).

The normal melting and boiling points of \(\mathrm{O}_{2}\) are \(-218{ }^{\circ} \mathrm{C}\) and \(-183{ }^{\circ} \mathrm{C}\) respectively. Its triple point is at \(-219^{\circ} \mathrm{C}\) and \(1.14\) torr, and its critical point is at \(-119^{\circ} \mathrm{C}\) and \(49.8\) atm. (a) Sketch the phase diagram for \(\mathrm{O}_{2}\), showing the four points given and indicating the area in which each phase is stable. (b) Will \(\mathrm{O}_{2}(s)\) float on \(\mathrm{O}_{2}(I)\) ? Explain. (c) As it is heated, will solid \(\mathrm{O}_{2}\) sublime or melt under a pressure of 1 atm?

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 \(\mathrm{GeH}_{4} ?\)

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