Question

Chromium metal crystallizes with a body-centred cubic lattice. The length of the unit cell edge is found to be 287 pm. Calculate the atomic radius. What would be the density of chromium in g/$c{m^3}$ ? (atomic mass of Cr = 52.99)

Answer 1

The atomic radius is the half of distance between two adjacent particles in a cubic unit cell. It can be calculated in a body centred cubic unit cell by the formula -
‘r’ = $\dfrac{{\sqrt 3 \times a}}{4}$
Where ‘r’ is the radius of the atom
‘a’ is the edge length of the unit cell
The density is the ratio of mass of unit cell to the volume of unit cell. It can be calculated as-
Density (D) = $\dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$
Where ‘Z’ = Number of atoms in unit cell
For BCC type of unit cell, Z = 2
M = Molar mass of the element
${N_A}$= Avogadro’s constant = 6.023$\times {10^{23}}$
‘a’ = Edge length of unit cell

Complete answer:
Let us write what is given to us and what we need to find out.
Given :
Cubic lattice = Body centred
Length of unit cell (a) = 287 pm = 287$\times {10^{ - 10}}$cm
Atomic mass of Cr = 52.99
To find :
Atomic radius
Density of chromium ( g/$c{m^3}$)
We know that the atomic radius in a body centred cubic unit cell is given by the formula -
‘r’ = $\dfrac{{\sqrt 3 \times a}}{4}$
Where ‘r’ is the radius of the atom
‘a’ is the edge length of the unit cell
So, filling the value of ‘a’, we can find radius as-
‘r’ = $\dfrac{{\sqrt 3 \times 287}}{4}$
‘r’ = $\dfrac{{1.732 \times 287}}{4}$
‘r’ = 124.47 pm
We have the formula for finding the density of solid crystals as -
Density (D) = $\dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$
Where ‘Z’ = Number of atoms in unit cell
For BCC type of unit cell, Z = 2
M = Molar mass of the element
${N_A}$= Avogadro’s constant = 6.023$\times {10^{23}}$
‘a’ = Edge length of unit cell
Substituting the values for all; we get -
Density (D) = $\dfrac{{2 \times 52.99}}{{6.023 \times {{10}^{23}} \times {{(287 \times {{10}^{ - 10}})}^3}}}$
Density = 7.44 g/$c{m^3}$

Note:
The formula derived for density is standard for all types of unit cell. It is derived as-
We have density = $\dfrac{{Mass{\text{ of unit cell}}}}{{Volume{\text{ of unit cell}}}}$
Mass of a unit cell = Number of atoms in unit cell (Z) $\times$Mass of one atom (m)
Mass of one atom = $\dfrac{{Atomic{\text{ mass(M)}}}}{{{N_A}}}$
Volume of unit cell = ${a^3}$
Thus, combining these, we get above formula.