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Chapter 12: Problem 76

Germanium has a cubic unit cell with a side edge of \(565 \mathrm{pm} .\) The density of germanium is \(5.36 \mathrm{g} / \mathrm{cm}^{3}\) What is the crystal system adopted by germanium?

Short Answer

Expert verified
The crystal system adopted by Germanium is Diamond Cubic.

Step by step solution

01

Calculate the number of atoms in a unit cell

Germanium has a diamond cubic crystal structure. Therefore, there are 8 atoms in a unit cell.
02

Calculate the volume of the unit cell

The edge length is given in picometres (pm), we need to convert this first to centimetres (cm). \(1 \mathrm{pm} = 1 \times 10^{-10} \mathrm{cm}\) So, the edge length is \(565 \times 10^{-10} \mathrm{cm}\). The volume (V) of the cube would be \(V = a^3 = (565 \times 10^{-10})^3 \mathrm{cm}^3\)
03

Calculate the Theoretical density (ρ)

The theoretical density can be calculated as \(\rho = \frac{n \times m}{V \times N_A}\) where \(n\) is the number of atoms in the unit cell, \(m\) is the molar mass of germanium, \(V\) is the volume of unit cell, \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \, \text{atoms/mol}\)). Substituting the respective values we get the theoretical density of germanium. If this matches the given density, then the crystal structure assumed (i.e., diamond cubic in this case) is correct.

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Most popular questions from this chapter

Because solid \(p\) -dichlorobenzene, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) sublimes rather easily, it has been used as a moth repellent. From the data given, estimate the sublimation pressure of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}(\mathrm{s})\) at \(25^{\circ} \mathrm{C} .\) For \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2} ; \mathrm{mp}=\) \(53.1^{\circ} \mathrm{C} ;\) vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)\) at \(54.8^{\circ} \mathrm{C}\) is \(10.0 \mathrm{mmHg} ; \Delta H_{\text {fus }}=17.88 \mathrm{kJ} \mathrm{mol}^{-1} ; \Delta H_{\text {vap }}=\) \(72.22 \mathrm{k}] \mathrm{mol}^{-1}\)

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