Question
Question: \[{\text{N}}{{\text{H}}_4}{\text{Cl}}\]crystallizes in a body-centred cubic type lattice with a unit...
NH4Clcrystallizes in a body-centred cubic type lattice with a unit cell edge length of 387 pm. The distance (in pm) between the oppositely charged ions in the lattice is:
A.335.1
B.83.77
C.274.46
D.137.23
Solution
For this question we need to use the relation between the edge length and the radius of the atom in a body centered unit cell. In a body centered, atoms are present in the body center along with the corners.
Complete step by step solution:
The atoms in a body centered cube touch each other at the body diagonal. The length of a body diagonal in a cube is 3 a here ‘a’ is the edge length of the cube.
Now on a body diagonal there is one complete atom and two half atoms lies. Let us suppose the diameter of the atom is d so the length of body diagonal will be 2d+d+2d=2d
Comparing the body diagonal length in terms of edge length and diameter, we will get:
⇒ 3 a=2 d
Rearranging the equation we will get:
\Rightarrow$$$\dfrac{{\sqrt 3 {\text{ a}}}}{2} = {\text{d}}$$
We will now substitute the value of edge length given to us,
\Rightarrow$d=23× 387;pm2=335.15 pm
Hence the correct option is A.
Additional information:
A body centered unit cell is the unit cell that contains 8 atoms on the corner and 1 atom in the body center. One atom at the corner is shared by 8 other atoms. So the contribution of one atom at a corner is one eighth. The atom present in the body center has the full contribution that is 1.
So the total number of atom will be calculated by multiplying the number of atoms present in one unit cell with their contribution as:
81×8+1×1=2
Note: A unit cell is the building block of a crustal. A crystal is formed by repeating lattice and a lattice is formed by repeating unit cells in different directions. There are various types of unit cell such as simple cubic unit cell, face centered unit cell, body centered unit cell, edge centered unit cell etc.