Question
Question: The edge length of the unit cell of KCl(fcc) is \( 6.28\,\mathop A\limits^o \) . Assuming anion-cati...
The edge length of the unit cell of KCl(fcc) is 6.28Ao . Assuming anion-cation contact along the cell edge, calculate the radius in ( A0 ) of the potassium ion.
(A) 0.9
(B) 1.3
(C) 1.8
(D) None of the above
Solution
The chemical name of KCl is potassium chloride. It is a metal halide salt formed by the potassium cation and chloride anion. It is a white crystalline powder that has an FCC lattice. It is soluble in water and is used in industries for various purposes.
Complete step by step solution:
Here, we are given an FCC lattice. In a Face-centred cubic unit cell, atoms are placed at each corner and the centre of all the faces of the cell. In FCC there are total 8 corners and 6 face centre. So the total number of atoms in an FCC structure is 4 . The relation between the lattice parameter and the radius is given by the formula:
a=24r
Where a= edge or lattice parameter
r= radius of the lattice
Now, we are given the edge length =6.28Ao and . Now, we will calculate the radius of Cl− ion.
We know that the ratio of the radius of the cation to anion in an FCC lattice is 0.731 .
Also, the sum of the radius of cation and anion in FCC is given as equal to half of its edge length. It is represented as:
r++r−=2a
Where, r+= cation and r−= anion. In the given salt the cation is K+ and the anion is Cl− .
Putting all the values in the given relation we get:
rK++rCl−=2a
rK++1.8=26.28
rK++1.8=3.14=1.34
rK+=1.34A0
Hence the radius of the Potassium ion will be 1.34A0 .
Therefore, option (B) is correct.
Note:
Every crystal lattice is formed from a unit cell. A unit cell is the basic entity of a lattice. A unit cell is the smallest repeating unit of the cell which when repeated over and over gives the crystal lattice. A lattice constant or parameter is used to define the physical dimension of unit cells in a crystal lattice.