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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

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 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.

Sketch a generic phase diagram for a substance that has a more dense solid phase than a liquid phase. Label all regions, lines, and points.

Refer to Figure \(11.27(\mathrm{~b})\), and describe the phase changes (and the temperatures at which they occur) when \(\mathrm{CO}_{2}\) is heated from \(-80{ }^{\circ} \mathrm{C}\) to \(-20{ }^{\circ} \mathrm{C}\) at \((\mathrm{a})\) a constant pressure of \(3 \mathrm{~atm}\), (b) a constant pressure of \(6 \mathrm{~atm}\).

Compounds like \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) are known as chlorofluorocarbons, or CFCs. These compounds were once widely used as refrigerants but are now being replaced by compounds that are believed to be less harmful to the environment. The heat of vaporization of \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(289 \mathrm{~J} / \mathrm{g}\). What mass of this substance must evaporate to freeze \(200 \mathrm{~g}\) of water initially at \(15^{\circ} \mathrm{C} ?\) (The heat of fusion of water is \(334 \mathrm{~J} / \mathrm{g} ;\) the specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K}\).)

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