Question

The edge length of the unit cell of a metal having a molecular weight of 75 g/mol is \[{\rm{5}}{{\rm{A}}^{\rm{o}}}\] which crystallises in a cubic lattice. If the density is 2 g/cc then find the radius of a metal atom. (\[{N_A} = 6 \times {10^{23}}\]). Answer in pm.
A. 217pm
B. 210pm
C. 220pm
D. 205pm

Answer 1

Hint: A lattice or a crystal lattice is a three-dimensional structure of ions, atoms, and molecules in space. For a simple cubic lattice, the relationship between edge length(a) and atomic radius(r) is \[r = \frac{a}{2}\]. For a face-centred unit cell and body-centred unit cell, \[r = \frac{a}{{2\surd 2}}\] and \[r = \frac{{\surd 3}}{4}a\].

Formula Used:
The density of a unit cell=\[\frac{{z \times M}}{{{a^3} \times {N_A}}}\]; where
z = no.of particles per unit cell
a = edge length of the unit cell.
\[{N_A} = 6 \times {10^{23}}\] = Avogadro's number
M = Atomic mass of the element

Complete Step by Step Solution:
In this question, the following are the given values:-
The edge length of the unit cell = \[{\rm{5}}{{\rm{A}}^{\rm{o}}}\]=\[5 \times {10^{ - 8}}cm\;\]
The molecular weight of the unit cell = 75g/mol.
Density = 2 g/cc.
\[{N_A} = 6 \times {10^{23}}\]
We have to find the radius of the metal.
We have to first find out the number of particles per unit cell. Then we will be able to know which unit cell it is.
The density of a unit cell = \[\frac{{z \times M}}{{{a^3} \times {N_A}}}\]
Putting the given values we get,
\[2g/cc = \frac{{z \times 75g/mol}}{{{{\left( {5 \times {{10}^{ - 8}}cm} \right)}^3} \times 6 \times {{10}^{23}}mo{l^{ - 1}}}}\]
\[ \Rightarrow 2g/cc = \frac{{z \times 75g}}{{\frac{{750}}{{10}}cc}}{\rm{ }}\]
\[ \Rightarrow z = 2\]
So, the unit cell is body-centred.
We know that the relationship between edge length(a) and atomic radius(r) for a body-centred unit cell is \[r = \frac{{\surd 3}}{4}a\].
So, \[r = \frac{{\surd 3}}{4} \times 5 \times {10^{ - 8}}cm\]
\[r = 2.16 \times {10^{ - 8}}cm = 216 \times {10^{ - 10}}cm\]
We know that \[1pm = {10^{ - 10}}cm\].
So, the radius of the unit cell is 216 pm
So, option B is correct.

Note: While attempting the question, one must mention the correct unit in each step of the calculation. In this question, z or no.of particles per unit cell does not have any unit. So, the value of z does not have any unit. The radius of the unit cell must be indicated at pm. One must remember that \[1pm = {10^{ - 10}}cm\]and \[{\rm{1}}{{\rm{A}}^o}{\rm{ = 5 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{cm}}\]. The density is given in g/cubic cm so the edge length was converted to centimetres.