if the radius of the octahedral void is $ r $ and the radius of the atom is close-packing is $ R $ , derive the relation between $ r $ and $ R $ .
Answer
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Hint: Close packing in crystals refers to space efficient arrangement of constituent particles in a crystal lattice. To understand this packing more clearly we have to assume all particles are of the same spherical solid shape. So the unit cell of a lattice is a cubic shape. If you observe a three-dimensional structure of a crystal lattice you will observe the gaps in between the spheres. These are the voids. So when the tetrahedral void of the first layer and the tetrahedral void of the second layer align together, they form an octahedral void.
Complete answer:
Three- dimensional close-packings of spheres have two kinds of voids: Such a void is called an octahedral void since the six spheres surrounding it lie at the corners of a regular octahedron. In a densely packed structure, voids refer to the empty space between constituent particles. When atoms are arranged in square close packing or hexagonal close packing, we see empty spaces between them in $ 2 - $ dimensional structures.
The total number of octahedral voids present in a cubic close packed structure is four. Besides the body centre, there is one octahedral void at the center of each of the twelve edges.
It is given that the radius of octahedral void is $ r $ and radius of the atoms in close packing is $ R $ .
Let the edge length be $ a $ .
In the right angle triangle $ ABC $
$ \Rightarrow AB = BC = a $
For diagonal $ AC $ ,
$ \Rightarrow AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{a^2} + {a^2}} = \sqrt {2a} $
$ \Rightarrow \dfrac{{AC}}{{AB}} = \dfrac{{\sqrt {2a} }}{a} $
And, $ AB = 2R $
$ \Rightarrow AC = R + 2r + R = 2R + 2r $
Therefore, $ \dfrac{{2R + 2r}}{{2R}} = \sqrt 2 $
$ \Rightarrow r = 0.414R $
Hence, the relation is $ \Rightarrow r = 0.414R $ .
Note:
The difference between octahedral and tetrahedral void is that a tetrahedral void is a simple triangular void in a crystal and is surrounded by four spheres arranged around it. On the other hand, an octahedral void is a double triangular void with one triangle vertex upwards and the other triangle vertex downwards and is surrounded by six spheres.
Complete answer:
Three- dimensional close-packings of spheres have two kinds of voids: Such a void is called an octahedral void since the six spheres surrounding it lie at the corners of a regular octahedron. In a densely packed structure, voids refer to the empty space between constituent particles. When atoms are arranged in square close packing or hexagonal close packing, we see empty spaces between them in $ 2 - $ dimensional structures.
The total number of octahedral voids present in a cubic close packed structure is four. Besides the body centre, there is one octahedral void at the center of each of the twelve edges.
It is given that the radius of octahedral void is $ r $ and radius of the atoms in close packing is $ R $ .
Let the edge length be $ a $ .
In the right angle triangle $ ABC $
$ \Rightarrow AB = BC = a $
For diagonal $ AC $ ,
$ \Rightarrow AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{a^2} + {a^2}} = \sqrt {2a} $
$ \Rightarrow \dfrac{{AC}}{{AB}} = \dfrac{{\sqrt {2a} }}{a} $
And, $ AB = 2R $
$ \Rightarrow AC = R + 2r + R = 2R + 2r $
Therefore, $ \dfrac{{2R + 2r}}{{2R}} = \sqrt 2 $
$ \Rightarrow r = 0.414R $
Hence, the relation is $ \Rightarrow r = 0.414R $ .
Note:
The difference between octahedral and tetrahedral void is that a tetrahedral void is a simple triangular void in a crystal and is surrounded by four spheres arranged around it. On the other hand, an octahedral void is a double triangular void with one triangle vertex upwards and the other triangle vertex downwards and is surrounded by six spheres.
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