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Chapter 10: Problem 61

The radius of gold is \(144 \mathrm{pm},\) and the density is 19.32 $\mathrm{g} / \mathrm{cm}^{3}$ . Does elemental gold have a face-centered cubic structure or a body-centered cubic structure?

Short Answer

Expert verified
Elemental gold has a face-centered cubic (FCC) structure. This is determined by comparing the calculated atomic radii for FCC and BCC structures, \(r_{FCC}\) = \(1.36 \times 10^{-8}\) cm, and \(r_{BCC}\) = \(1.05 \times 10^{-8}\) cm, with the given gold radius, \(1.44 \times 10^{-8}\) cm. The atomic radius for the FCC structure is closer to the given gold radius, indicating that gold has an FCC structure.

Step by step solution

01

Calculate atomic packing factors for FCC and BCC.

The atomic packing factor is a ratio that describes how much space is filled by atoms in a given atomic structure. An FCC structure has an atomic packing factor of 0.74, while a BCC structure has an atomic packing factor of 0.68.
02

Determine the mass of one gold atom.

First, we need to find the molar mass of gold (Au). From the periodic table, the molar mass of gold is approximately 197 g/mol. To find the mass of one gold atom, we divide the molar mass by Avogadro's number: \(Mass_{Au} = \frac{197 \,\text{g/mol}}{6.022 \times 10^{23} \,\text{atoms/mol}}\) \(Mass_{Au} \approx 3.27 \times 10^{-22} \,\text{g}\)
03

Determine the volume of one gold atom.

Next, we need to find the volume of one gold atom in the unit cell. The volume is determined by the following equation: \(Volume_{Au} = \frac{Mass_{Au}}{Density_{Au}}\) Where \(Density_{Au}\) is given as 19.32 g/cm³. Substitute the values: \(Volume_{Au} = \frac{3.27 \times 10^{-22} \,\text{g}}{19.32 \,\text{g/cm}^{3}}\) \(Volume_{Au} \approx 1.69 \times 10^{-23} \,\text{cm}^{3}\)
04

Calculate the volume of the unit cell for FCC and BCC structures.

To determine the volume of the unit cell, we need to use the atomic packing factors for FCC and BCC structures. To calculate the volume of the unit cell for FCC and BCC structures, we'll use the following formulas: \(Volume_{FCC} = \frac {Volume_{Au}} {APF_{FCC}}\) \(Volume_{BCC} = \frac {Volume_{Au}} {APF_{BCC}}\) Substitute the values and calculate: \(Volume_{FCC} = \frac {1.69 \times 10^{-23} \,\text{cm}^{3}}{0.74}\) \(Volume_{FCC} \approx 2.28 \times 10^{-23} \,\text{cm}^{3}\) \(Volume_{BCC} = \frac {1.69 \times 10^{-23} \,\text{cm}^{3}}{0.68}\) \(Volume_{BCC} \approx 2.49 \times 10^{-23} \,\text{cm}^{3}\)
05

Calculate the lattice constant for FCC and BCC structures.

Finally, we'll determine the lattice constant 'a' for both structures using the following formulas for FCC and BCC structures: \(a_{FCC} = \sqrt[3]{\frac{4*Volume_{FCC}}{\sqrt{2}}}\) \(a_{BCC} = \sqrt[3]{8*Volume_{BCC}}\) Substitute the values and calculate: \(a_{FCC} = \sqrt[3]{\frac{4*2.28 \times 10^{-23} \,\text{cm}^{3}}{\sqrt{2}}}\) \(a_{FCC} \approx 3.84 \times 10^{-8} \,\text{cm}\) \(a_{BCC} = \sqrt[3]{8*2.49 \times 10^{-23} \,\text{cm}^{3}}\) \(a_{BCC} \approx 4.19 \times 10^{-8} \,\text{cm}\)
06

Calculate the atomic radii for FCC and BCC structures.

Using the lattice constants, we can now calculate the atomic radii for FCC and BCC structures using the following formulas: \(r_{FCC} = \frac{a_{FCC}}{2\sqrt{2}}\) \(r_{BCC} = \frac{a_{BCC}}{4}\) Substitute the values and calculate: \(r_{FCC} = \frac{3.84 \times 10^{-8} \,\text{cm}}{2\sqrt{2}}\) \(r_{FCC} \approx 1.36 \times 10^{-8} \,\text{cm}\) \(r_{BCC} = \frac{4.19 \times 10^{-8} \,\text{cm}}{4}\) \(r_{BCC} \approx 1.05 \times 10^{-8} \,\text{cm}\) Now, convert the given radius of gold to centimeters: \(144 \,\text{pm} = 1.44 \times 10^{-8} \,\text{cm}\)
07

Compare the calculated atomic radii with the given radius of gold.

We compare the calculated atomic radii with the given radius and see which one is closer: \(r_{FCC} = 1.36 \times 10^{-8} \,\text{cm}\) Given radius: \(1.44 \times 10^{-8} \,\text{cm}\) \(r_{BCC} = 1.05 \times 10^{-8} \,\text{cm}\) From the comparison, we can see that the atomic radius for the FCC structure is closer to the given value of the golad radius. Therefore, elemental gold has a face-centered cubic structure.

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Most popular questions from this chapter

Rationalize the difference in boiling points for each of the following pairs of substances: $$\begin{array}{rr}{\text { a. Ar }} & {-186^{\circ} \mathrm{C}} \\\ {\mathrm{HCl}} & {-85^{\circ} \mathrm{C}}\end{array}$$ $$\begin{array}{rr}{\text { b. } \mathrm{HF}} & {20^{\circ} \mathrm{C}} \\\ {\mathrm{HCl}} & {-85^{\circ} \mathrm{C}}\end{array}$$ $$\begin{array}{cc}{\text { c. } \mathrm{HCl}} & {-85^{\circ} \mathrm{C}} \\\ {\mathrm{LiCl}} & {1360^{\circ} \mathrm{C}}\end{array}$$ $$\begin{array}{ccc}{\text { d. } n \text { -pentane }} & {\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}} & {36.2^{\circ} \mathrm{C}} \\ {n \text { -hexane }} & {\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}} & {69^{\circ} \mathrm{C}}\end{array}$$

What is the formula for the compound that crystallizes with a cubic closest packed array of sulfur ions, and that contains zinc ions in \(\frac{1}{8}\) of the tetrahedral holes and aluminum ions in \(\frac{1}{2}\) of the octahedral holes?

Which of the following statements is (are) true? a. LiF will have a higher vapor pressure at \(25^{\circ} \mathrm{C}\) than \(\mathrm{H}_{2} \mathrm{S}\) . b. HF will have a lower vapor pressure at \(-50^{\circ} \mathrm{C}\) than \(\mathrm{HBr}\) . c. \(\mathrm{Cl}_{2}\) will have a higher boiling point than Ar. d. HCl is more soluble in water than in \(\mathrm{CCl}_{4}\) e. \(\mathrm{MgO}\) will have a higher vapor pressure at \(25^{\circ} \mathrm{C}\) than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) .

Which of the following statements about intermolecular forces is(are) true? a. London dispersion forces are the only type of intermolecular force that nonpolar molecules exhibit. b. Molecules that have only London dispersion forces will always be gases at room temperature \(\left(25^{\circ} \mathrm{C}\right)\) c. The hydrogen-bonding forces in \(\mathrm{NH}_{3}\) are stronger than those in \(\mathrm{H}_{2} \mathrm{O}\) . d. The molecules in \(\mathrm{SO}_{2}(g)\) exhibit dipole-dipole intermolecular interactions. e. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) has stronger London dispersion forces than does \(\mathrm{CH}_{4}\) .

Identify the most important types of interparticle forces present in the solids of each of the following substances a. Ar b. \(\mathrm{HCl}\)l c. \(\mathrm{HF}\) d. \(\mathrm{CaCl}_{2}\) e. \(C \mathrm{H}_{4}\) f. \(C O\) g. \(N a N O_{3}\)
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