Question

The radii of \[N{{a}^{+}}\]and \[C{{l}^{-}}\]are 95 pm and 181 pm respectively. The edge length of unit cell is:
A.276 pm
B.138 pm
C.552 pm
D.415 pm

Answer 1

Hint: To answer this question, we should know NaCl is a crystal structure with a face centered cubic. We should note that the total number of ions present in one unit cell of sodium chloride lattice is 8. It includes 4 (\[N{{a}^{+}}\]) ions and 4 \[\left( C{{l}^{-}} \right)\]ions.

Step by step answer:
We should first explain NaCl unit cell structure that is face centred cubic. As we know that the smallest repeating unit of the crystal lattice is the unit cell, the building block of NaCl crystal. We should note that in an FCC unit cell, it contains atoms at all the corners of the crystal lattice and at the center of all the faces of the cube. The atom present at the face-center is shared between 2 adjacent unit cells and only half of each atom belongs to an individual cell. In FCC unit cell atoms are present in all the corners of the crystal lattice Also; we should note that there is an atom present at the centre of every face of the cube.
Now, we will calculate the edge length of unit cell:
Give that radius of ion of sodium (\[N{{a}^{+}}\]) is= 95 pm
Chlorine\[\left( C{{l}^{-}} \right)\] ion radius is= 181 pm
We have to find the edge length:
We should know that in FCC:
\[{{r}^{+}}+{{r}^{-}}=\dfrac{a}{2}\]
In this (a) is the edge length.
\[2({{r}^{+}}+{{r}^{-}})=a\]
Now, we will put the radius of sodium ion in place of r+ and radius of chlorine in place of r-.
\[2({{r}^{+}}\left( N{{a}^{+}} \right)+{{r}^{-}}\left( C{{l}^{-}} \right))pm=a\]
Now, we will put the values in it.
\[\begin{align}
  & 2(95+181)pm=a \\
 & a=552pm \\
\end{align}\]
So, from the above calculation we can say that the radius of the edge of the unit cell of NaCl is 552 pm. Hence, the correct option is C.
Note:
We should know that there are three types of unit cell. They are:
* Simple Cubic Unit Cell
* Body-centred Cubic Unit Cell
* Face centered cubic unit cell
We should know that in a simple cubic cell, the atoms are present only at the corners. Every atom at the corner is shared among 8 adjacent unit cells. There are 4 unit cells in the same layer and 4 in the upper (or lower) layer.
In the body centred cubic cell, we should know that in BCC, atoms are present at each corner of the cube and an atom at the centre of the structure.