Question

If the effective radius of the iron ${\rm{(Fe)}}$ atom is 141.4 pm which has a rock salt type structure, then its density (in ${\rm{g c}}{{\rm{m}}^{{\rm{ - 3}}}}$) will be:
${\rm{[Fe = 56 amu]}}$
A. 5.83
B. 2.63
C. 1.08
D. None of the above

Answer 1

A rock salt type structure corresponds to the FCC unit cell. Using the relation between the radius of the atom and edge length of the unit cell calculate the edge length of the unit cell. Using the edge length of the unit cell calculate the volume of the unit cell. Using the given atomic mass of iron atoms calculate the mass of the number of atoms present in the unit cell. Finally using mass and volume calculate the density.

Formula Used:
$r = \dfrac{{\sqrt 2 a}}{4}\\\$
${\rm{Volume of cube = }}{a^3}\\\$
${\rm{Density = }}\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}$

Complete answer:
We have given that iron has a rock salt type structure. A rock salt type structure corresponds to the face centered cubic (FCC) unit cell so we can say that iron has FCC structure.
Using the given radius of the iron atom we can calculate the edge length of FCC unit cell as follows:
The mathematical expression related to the radius and edge length is:
$r = \dfrac{{\sqrt 2 a}}{4}$
Here,
$r$ = radius of atom
$a$= edge length of the unit cell
Radius of iron atom = 141.4 pm = $1.414 \times {10^{ - 8}}{\rm{cm}}$ ( As ${\rm{1pm = 1}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 10}}}}{\rm{ cm}}$)
Now, substitute $1.414 \times {10^{ - 8}}{\rm{cm}}$ for the radius of the atom and calculate the edge length.
$\Rightarrow 1.414 \times {10^{ - 8}}{\rm{cm}} = \dfrac{{\sqrt 2 a}}{4}$
$\Rightarrow a = 4.0 \times {10^{ - 8}}{\rm{cm}}$
Now, using the edge length of the unit cell we can calculate the volume of the unit cell as follows:
${\rm{Volume of cube = }}{a^3}$
$\Rightarrow {\rm{Volume of cube = (}}4.0 \times {10^{ - 8}}{\rm{cm}}{)^3} = {\rm{6}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}{\rm{c}}{{\rm{m}}^{\rm{3}}}$
Now, we can calculate the mass of atoms present in the unit cell as follows:
As iron atoms have FCC structure so we can say that there are 4 atoms per unit cell.
We have given an atomic mass of ${\rm{Fe}}$ as 56 amu. Atomic mass corresponds to the mass of Avogadro number of atoms (${\rm{6}}{\rm{.02}} \times {\rm{1}}{{\rm{0}}^{{\rm{23}}}}$).
So we can calculate the mass of 4 atoms as follows:
$\Rightarrow \dfrac{{4{\rm{ atoms }} \times {\rm{ 56 amu}}}}{{{\rm{6}}{\rm{.02}} \times {\rm{1}}{{\rm{0}}^{{\rm{23}}}}{\rm{atoms}}}} = 3.72 \times {10^{ - 22}}{\rm{ g}}$

Now, using calculate volume and mass we can calculate the density as follows:
${\rm{Density = }}\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}$
Substitute $3.72 \times {10^{ - 22}}{\rm{ g}}$ for mass and ${\rm{6}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}{\rm{c}}{{\rm{m}}^{\rm{3}}}$ for volume and calculate the density.
$\Rightarrow {\rm{Density = }}\dfrac{{3.72 \times {{10}^{ - 22}}{\rm{ g}}}}{{{\rm{6}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}{\rm{c}}{{\rm{m}}^{\rm{3}}}}} = {\rm{ 5}}{\rm{.83g/c}}{{\rm{m}}^{\rm{3}}}$
Thus, the density of the iron atom is ${\rm{5}}{\rm{.83g/c}}{{\rm{m}}^{\rm{3}}}$.

**Hence, the correct option is (A) 5.83

Note:**
Different unit cells have a different number of atoms so determining the correct type of unit cell is the crucial step in this type of problem. Also the relation between atomic radius and edge length varies with the type of unit cell.