Q12.133CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Diamond has a face-centered cubic unit cell, with four more C atoms in tetrahedral holes within the cell. Densities of diamonds vary from   to   because C atoms are missing from some holes.

1. Calculate the unit-cell edge length of the densest diamond.
2. Assuming the cell dimensions are fixed, how many C atoms are in the unit cell of the diamond with the lowest density?

 

Step-by-Step Solution

Answer

Answer

 

1. The unit-cell edge length of the densest diamond having C-atom in three-dimensional structure is  .$0.613\times {10}^{3}cm$
2. The number of C-atom in the unit cell of the diamond with the lowest density is 8 C-atoms.

Step 1: Definition.

Density can be defined as the ratio of mass to volume. Density is inversely proportional to the volume of the unit cell as the mass of the atom is constant.

 

Density equation for the unit cell:

 $\mathbf{\rho }\mathbf{=}\frac{\mathbf{Z}\mathbf{\times }\mathbf{M}}{\mathbf{Na}\mathbf{\times }{\mathbf{a}}^{\mathbf{3}}}$

 

 

Here,

ρ is Density,  

Z is the Number of atoms in contact in the particular structure

M is Molar mass

  is Avogadro number

a is the atomic radius.

 

For FCC, the value of   is the number of atoms in contact in a particular structure. 

 

Step 2: Subpart (a) The unit-cell edge length of the densest diamond.

The largest density of the unit cell,  $\rho =3.52 g/c{m}^3$.

Avogadro number,  ${\mathbf{N}}_{\mathbf{a}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$ atoms/molecules.

Atomic radius,  a=?

The molar mass of C-atom or molecule,  $\text{m = 12 g/mole}$.

 

As,

 $\mathbf{\rho }\mathbf{=}\frac{\mathbf{Z}\mathbf{\times }\mathbf{M}}{\mathbf{Na}\mathbf{\times }{\mathbf{a}}^{\mathbf{3}}}$

 

Therefore,

 

$$\begin{gathered} \mathbf{3}\mathbf{.}\mathbf{52}\mathbf{=}\frac{\mathbf{4}\mathbf{\times }\mathbf{12}}{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{\times }{\mathbf{a}}^{\mathbf{3}}} \\ {\mathbf{a}}^{\mathbf{3}}\mathbf{=}\frac{\mathbf{48}}{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{52}} \\ \mathbf{a}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{24}}{\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{3}}} \\ \mathbf{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{613}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{cm} \end{gathered}$$

Step 3: Subpart (b) The number of C atoms present in the unit cell of the diamond with the lowest density.

The diamond lattice is structured in FCC geometry.

Diamond unit cell has C-atom present at 8 corners, 6 face centers, and 4 internal.

 $$\begin{gathered} \text{Number of C - atom = 8×}\frac{1}{8}\text{+6×}\frac{1}{2}\text{+4} \\ \text{Number of C - atom = 8} \end{gathered}$$

 

 

The number of C-atom in the unit cell of the diamond with the lowest density is 8 C-atoms.