Question

Calculate the density in $ gc{m^{ - 1}}$ of diamond from the fact that it has face-centered cubic structure with the two atoms per lattice point and unit cell edge length of $ 3.569{A^ \circ }$ .
(A) $ 7.5$
(B) $ 1.7$
(C) $ 3.5$
(D) None of the above

Answer 1

Hint: This question gives the knowledge about the fcc crystal lattice. Fcc crystal lattice is called as face-centered cubic crystal lattice. It contains six atoms at each face of the cube and eight atoms at the corners of the cube. Diamond experiences cubic crystal structure.

Formula used The formula used to determine the number of atoms is as follows:
$ \smallint = \dfrac{{zM}}{{{a^3}{N_A}}}$
Where $ \smallint $ is density, $ z$ is an effective number of atoms in a unit cell, $ M$ is molar mass, $ {N_A}$ is Avogadro’s number and $ a$ is the edge of the cube.

Complete step by step solution
Fcc crystal lattice contains six atoms at each face of the cube and eight atoms at the corners of the cube. The effective number of atoms in a fcc crystal unit cell is always $ 4$ . The coordination number of FCC lattice is $ 12$ . Diamond experiences cubic crystal structure. It is very expensive and is mainly used in making ornaments or jewelry.
The formula to determine the number of atoms is considered as follows:
$ \Rightarrow \smallint = \dfrac{{zM}}{{{a^3}{N_A}}}$
Substitute edge of the cube as $ 3.569{A^ \circ }$ or $ 3.569 \times {10^{ - 8}}$ and $ z$ as$ 8$ because two atoms are present at each lattice point in the above equation as follows:
$ \Rightarrow \smallint = \dfrac{{8 \times M}}{{{{\left( {3.569 \times {{10}^{ - 8}}} \right)}^3}{N_A}}}$
Substitute $ {N_A}$ as $ 6.023 \times {10^{23}}$ and $ M$ as$ 12$ in the above equation as follows:
$ \Rightarrow \smallint = \dfrac{{8 \times 12}}{{{{\left( {3.569 \times {{10}^{ - 8}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}$
On simplifying, we get
$ \Rightarrow \smallint = \dfrac{{96}}{{{{\left( {3.569} \right)}^3} \times {{10}^{ - 24}} \times 6.023 \times {{10}^{23}}}}$
On further simplifying, we get
$ \Rightarrow \smallint = 3.506$
Therefore, the density of diamond is $ 3.506gc{m^{ - 1}}$ .
Hence, option $ C$ is the correct option.

Note
Always remember that the effective number of atoms in a face centered cubic crystal unit cell is always $ 4$ . FCC crystal lattice contains six atoms at each face of the cube and eight atoms at the corners of the cube.