Question

$ CuCl $ has a zinc blende structure. Its density is $ 3.4{\text{g c}}{{\text{m}}^{ - 3}} $ . The length of its unit cell edge in pm is:

Answer 1

Hint: To solve this question firstly we will find the type of cubic unit cell means we will find that the given compound is a face centered cubic unit cell or body centered cubic unit cell or simple unit cell.
After finding the type of cubic unit cell we will calculate the number of atoms in the cubic unit cell. Finally we will apply the density formula and we will get the length of the cubic unit cell.

Complete step by step solution:
As in the given question we have $ CuCl $ has a zinc blende structure .
And also we all know have studied that zinc blenders have FCC type of cubic unit cell.
Means $ CuCl $ also has FCC type of cubic unit cell.
FCC means face centered cubic unit cell.
Means, all faces of FCC unit cells have part of an atom in the center of faces.
 $ \therefore $ we all know that in the FCC type close packing atom contribute its half for each face:
So the total number of atoms will be 1/2 times the number of faces in the unit cell .
Now, we all know that a cube has 6 faces.
So the total number of atoms in FCC packing will be : $ \dfrac{1}{2} \times 6 = 3 + \dfrac{1}{8} $ atoms in the 8 corners of the cubic unit cell.
So the total number of atoms in FCC will be 4.
Now we have calculated the total number of atoms in the FCC close packing.
Now we will apply the density formula for cubic unit cell:
Density formula ( $ \rho $ )= $ \dfrac{{ZM}}{{{N_0}{a^3}}} $
Where, Z is the number of atoms in the cubic unit cell which we have calculated =3
M is molar mass of the given compound $ CuCl $
So to calculate the mass of $ CuCl $ we will add molar masses of Cu and Cl.
As we all know that molar mass of Cu is 63.5 and molar mass of Cl is 35.5
So the total mass of $ CuCl $ is 99.
 $ {N_0} $ is a constant known as Avogadro's constant and its value is $ 6.023 \times {10^{23}} $
So, after putting the all value of values in the above density formula for cubic unit cell:
 $ \rho = \dfrac{{ZM}}{{{N_0}{a^3}}} $
On substituting the values we get
 $
3.4 = \dfrac{{4 \times 99}}{{6.023 \times {{10}^{23}} \times {a^3}}} \\
\\
 $
 $
{a^3} = \dfrac{{3 \times 99}}{{6.023 \times {{10}^{23}} \times 3.4}} \\
{a^3} = \dfrac{{396}}{{20.4748}} \times {10^{ - 23}} \\
\\
 $
On simplifying we get
 $
{a^3} = 19.3453835 \times {10^{ - 23}} \\
{a^3} = 193453835 \times \times {10^{ - 30}} \\
a = 578 \times {10^{ - 10}}cm \\
a = 578pm \\
 $
Hence, the length of the edge of the cubic unit cell will be 578 picometer.

Note:
Face centered cubic unit cell(FCC): In this unit cell corners as well as the face centered are occupied .
 $ \therefore $ face centered contributes 1/2 particle to one cubic cell.
And corners contribute 1/8 particle to one cubic unit cell.
There are 6 faces and 8 corners in FCC
So total number of atoms will be: $ 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 4 $
Hence, a face centered cubic unit cell has four constituents.