
The edge length of the unit cell of KCl(fcc) is $ 6.28\,\mathop A\limits^o $ . Assuming anion-cation contact along the cell edge, calculate the radius in ( $ \mathop A\limits^0 $ ) of the potassium ion.
(A) $ 0.9 $
(B) $ 1.3 $
(C) $ 1.8 $
(D) None of the above
Answer
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Hint: 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 = \dfrac{{4r}}{{\sqrt 2 }} $
Where $ a = $ edge or lattice parameter
$ r = $ radius of the lattice
Now, we are given the edge length $ = 6.28\,\mathop A\limits^o $ and . Now, we will calculate the radius of $ C{l^ - } $ 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^ - } = \dfrac{a}{2} $
Where, $ {r^ + } = $ cation and $ {r^ - } = $ anion. In the given salt the cation is $ {K^ + } $ and the anion is $ C{l^ - } $ .
Putting all the values in the given relation we get:
$ {r_{{K^ + }}} + {r_{C{l^ - }}} = \dfrac{a}{2} $
$ {r_{{K^ + }}} + 1.8 = \dfrac{{6.28}}{2} $
$ {r_{{K^ + }}} + 1.8 = 3.14 = 1.34 $
$ {r_{{K^ + }}} = 1.34\,\mathop A\limits^0 $
Hence the radius of the Potassium ion will be $ 1.34\,\mathop A\limits^0 $ .
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.
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 = \dfrac{{4r}}{{\sqrt 2 }} $
Where $ a = $ edge or lattice parameter
$ r = $ radius of the lattice
Now, we are given the edge length $ = 6.28\,\mathop A\limits^o $ and . Now, we will calculate the radius of $ C{l^ - } $ 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^ - } = \dfrac{a}{2} $
Where, $ {r^ + } = $ cation and $ {r^ - } = $ anion. In the given salt the cation is $ {K^ + } $ and the anion is $ C{l^ - } $ .
Putting all the values in the given relation we get:
$ {r_{{K^ + }}} + {r_{C{l^ - }}} = \dfrac{a}{2} $
$ {r_{{K^ + }}} + 1.8 = \dfrac{{6.28}}{2} $
$ {r_{{K^ + }}} + 1.8 = 3.14 = 1.34 $
$ {r_{{K^ + }}} = 1.34\,\mathop A\limits^0 $
Hence the radius of the Potassium ion will be $ 1.34\,\mathop A\limits^0 $ .
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.
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