Chapter 9, Problem 59E

The radius of gold is 144 pm, and the density is 19.32 g/cm³. Does elemental gold have a face-centered cubic structure or a body-centered cubic structure?

Interpretation Introduction

Interpretation:

        The lattice structure of Gold has to be identified and justified.

Concept introduction:

        In packing of atoms in a crystal structure, the atoms are imagined as spheres. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing.

        In body-centered cubic unit cell, each of the six corners is occupied by every single atom. Center of the cube is occupied by one atom.

        Each atom in the corner is shared by eight unit cells and a single atom in the center of the cube remains unshared. Thus the number of atoms per unit cell in BCC unit cell is,

                  $\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{1}\text{atom}\text{at}\text{the}\text{center}\text{=}\text{1}\text{+}\text{1}\text{=}\text{2}\text{atoms}\\ \\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\frac{\text{4R}}{\sqrt{\text{3}}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\\ \end{array}$

         In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

         Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

                       $\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{6}\text{×}\frac{\text{1}}{\text{2}}\text{atoms}\text{in}\text{faces}\text{=}\text{1}\text{+}\text{3}\text{=}\text{4}\text{atoms}\\ \\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\\ \end{array}$

Answer to Problem 59E

Answer

               The lattice structure of Gold is identified as cubic close packing with face-centered cubic unit cell.

Explanation of Solution

Explanation

Calculate the density of Gold by assuming its structure as BCC.

$\begin{array}{l}\text{given}\text{data:}\text{R}\text{=}144\text{pm}\text{=}144\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{a}\text{=}\frac{\text{4R}}{\sqrt{\text{3}}}\\ \text{=}\frac{\text{4}\text{×}144\text{×}{\text{10}}^{\text{-12}}\text{m}}{\text{1}\text{.732}}\\ \text{=}332.6\text{pm}\\ \\ \text{volume of unit cell}\text{=}{\text{a}}^{\text{3}}\\ \text{=}{\text{(332}}^{\text{.6}}{\text{m}}^{\text{3}}\\ =3.68\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

$\begin{array}{l}\text{Average mass of one gold atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{gold}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{197\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}32.71\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{Each}\text{BCC}\text{unit cell contains}\text{2}\text{gold atoms}\text{. therefore,}\\ \text{Mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{2}\text{×}32.71\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}65.42\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

           $\begin{array}{l}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{65.42{\text{×10}}^{\text{-23}}\text{g}}{\text{3}\text{.68}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}}\\ =1\text{7}\text{.78}{\text{g/cm}}^{\text{3}}\end{array}$

             The atomic radius of Gold is given. The unit cell is assumed as that of body-centered cubic and its edge length is calculated. Accordingly, the volume, mass and density of BCC unit cell are calculated. The obtained value does not agree with the actual value of density of gold.

Calculate the density of gold by assuming its structure as FCC.

$\begin{array}{l}\text{given}\text{data:}\text{R}\text{=}\text{144pm}\text{=}\text{144×}{\text{10}}^{\text{-12}}\text{m}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{=}\text{2}\text{×}\text{144}\text{×}{\text{10}}^{\text{-12}}\text{m}\text{×}\text{1}\text{.414}\\ \text{=}\text{407}\text{.2}\text{pm}\\ \text{volume of unit cell}\text{=}{\text{a}}^{\text{3}}\\ \text{=}{\text{(407}}^{\text{.2}}{\text{m}}^{\text{3}}\\ \text{=}\text{6}\text{.75}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\\ \text{Average mass of one gold atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{gold}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{197}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{32}\text{.71}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \\ \text{Each}\text{FCC}\text{unit cell contains}\text{4}\text{gold atoms}\text{. therefore,}\\ \\ \text{Mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}4\text{×}\text{32}\text{.71}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}130.84\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

                                   $\begin{array}{l}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{\text{130}\text{.84}{\text{×10}}^{\text{-23}}\text{g}}{\text{6}\text{.75}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}}\\ \text{=}\text{19}\text{.38}{\text{g/cm}}^{\text{3}}\end{array}$

             The atomic radius of gold is given. The unit cell is assumed as that of face-centered cubic and its edge length is calculated. Accordingly, the volume, mass and density of FCC unit cell are calculated. The obtained value agrees well with the actual value of density of gold.

Conclusion

Conclusion

            The density of gold calculated by assuming its structure as FCC lattice is accurate. So, it is identified that gold has face-centered cubic structure and the same has been justified.