Question

Sodium metal crystallizes in a body-centered cubic lattice with a unit cell edge of $4.29{{A}^{0}}$. The radius (in ${{A}^{0}}$) of sodium atom is approximately:
(a) 1.86
(b) 3.22
(c) 5.72
(d) 0.93

Answer 1

We know that the crystal lattice of a compound is generated by the complete three-dimensional rotation of a small portion of a specific arrangement of atoms.
To solve this question we relate the terms radius and the edge length of the unit cell and the equation obtained is as follows:
$r=\dfrac{\sqrt{3}a}{4}$

Complete step-by-step answer: In the question, it is asked that what will be the radius of the sodium atom, if the sodium metal crystallizes in a body-centered cubic lattice with an edge length$4.29{{A}^{0}}$.
Before going onto the solution of this question, first, let's briefly about the crystal lattice and the various arrangements of atoms in the crystal lattice.
We are very familiar that many of the ionic compounds have crystalline structures and the crystal lattice for a compound is generated when a small part of the atoms which is arranged in a specific fashion undergoes a complete three-dimensional rotation in their plane.
There are mainly three types of arrangement of atoms in the unit cell. And unit cell the basic entity of the crystalline structure.
The three arrangements of atoms in the unit cell are, simple cubic lattice, face-centered cubic lattice, and body-centered cubic lattice.
In a simple cubic lattice, the eight corners of the cube are occupied by the atoms.
In the body-centered cubic lattice, the eight corners are occupied with the atoms in addition to that one more atom is present wholly in the body center of the cube.
In the face-centered cubic lattice, in addition to the atoms on the eight corners atoms are present on the face of the cube.
Now we got a basic idea of the arrangements.
Now we have to know the relation between the radius of the atom and the edge length in a BCC lattice.
The equation is already deduced and known for us.
The equation relating these two parameters is, $r=\dfrac{\sqrt{3}a}{4}$, where r is the radius and a is the edge length.
$r=\dfrac{\sqrt{3}\times 4.29}{4}=1.8576\approx 1.86{{A}^{0}}$

Hence the correct answer for the question is option (a).

Note: For doing these types of problems we should know the type of arrangement of the atoms and also we should know the relation between various parameters like radius and edge length for each lattice.
If the given case was for simple cubic lattice the, $a=2r$.
If the given case of the arrangement of atoms was FCC lattice then the equation must have been, $a=2\sqrt{2}r$