How to calculate radius ratio of square Voids?

The square voids are found in the planes of the close packed structures, whenever three spheres are in contact in such a fashion. The coordination number of square void is four.

Related Questions:

How to calculate radius ratio triangular voids?

How to calculate packing fraction or packing efficiency of two dimensional (2D) hexagonal packing solid atoms?

How to calculate radius ratio triangular voids?

The triangular voids are found in the planes of the close packed structures, whenever three spheres are in contact in such a fashion. Coordination number of triangular void is three. Here three atoms A, B, and C form triangular voids in which equilateral triangle form having angle 60⁰

Where, R= Radius of the (Atoms) sphere, r = maximum radius of a sphere that can be placed inside the void or radius of void.

How to calculate packing fraction or packing efficiency of two dimensional (2D) hexagonal packing solid atoms?

 Hexagonal close packing:

The particles in every next row are placed in the depressions between the particles of the first row. The particles in the third row will be vertically aligned with those in the first row. This type of packing gives a hexagonal pattern and is called hexagonal close packing

Packing efficiency calculation:

(1) One sphere will be in constant contact with 6 other spheres hence coordination number is 6

(2) Side of hexagon is a and a=2r where r is the radius of atom (sphere) .

(3) Area of hexagon unit cell is 6*area of six equilateral triangles 

       (4) Area of atoms in the hexagon unit 

      Packing fraction (PE) , fraction of area occupied by spheres 

HEXAGONAL CLOSE PACKING (HCP):

Step-(1) In order to develop three dimensional close packing take a 2D hexagonal close packing sheet as  first layer (A- layer).

Step-(2) Another 2D hexagonal close sheet (B-layer) is taken and it is just over the depression (Pit) of the first layer (A) .When the second layer is placed in such a way that its spheres find place in the ‘b’ voids of the first layer, the ‘c’ voids will be left unoccupied. Since under this arrangement no sphere can be placed in them, (c voids), i.e. only half (50%) the triangular voids in the first layer are occupied by spheres of the second layer (i.e. either b or c)

Step-(3) There are two alternative ways in which spheres in the third layer can be arranged over the second layer

(1)  When a third layer is placed over the second layer in such a way that the spheres cover the tetrahedral or ‘a’ voids; a three dimensional closest packing is obtained where the spheres in every third or alternate layers are vertically aligned (i.e. the third layer is directly above the first, the fourth above the second layer and so on) calling the first layer A and second layer as layer B, the arrangement is called ABAB …………. pattern or hexagonal close packing (HCP) as it has hexagonal symmetry.   

ANALYSIS OF HCP UNIT CELL:

(1) Number of effective atoms in HCP unit cell (Z):

Lattice point:  corner- total 12 carbon contribute 1/6 to the unit cell

Lattice point:   face- total face 2 contribute ½

Lattice point: body centre- total atom 3 (100% contribution)

(2) Radius of atom in HCP unit cell:

Let the edge of hexagonal base =( a) And the height of hexagon =( h) And radius of sphere =( r)

(3) Area of hexagon:

 Area of hexagonal can be divided into six equilateral triangles with side 2r

(4) Height of HCP unit cell:

(5)  Volume of hexagon = area of base x height

(6)  Volume of spheres:

(7) Packing efficiency: Percentage of space occupied by sphere.   

(8) Voids %: 100 - PE= 26 %

(9) Coordination Numbers:  12 (each spheres touches 6 spheres in its layer,3 above and 3 below).

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FACE CENTRED UNIT CELL (FCC/CCP):

CCP or FCC has two lattice point corner as well as face centred:

Suppose ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube As there are 4 sphere in FCC unit cell

(1) Relation between radius (r) and side (a)

In FCC, the corner spheres are in touch with the face centred sphere. Therefore, face diagonal AD is equal to four times the radius of sphere AC = 4r

But from the right angled triangle ABC:

(2)  Effective no. of atoms per unit cell (Z):

(3) Volume of four spheres (atoms):

(4) Volume of unit cube:

(5) Packing efficiency (PE): Percentage of space occupied by sphere

(6) Percentage Voids: 100- PE= 26 %

(7) Density of FCC(CCP):

(8) Coordination Numbers:

(9) LOCATION OF VOIDS: FCC/CCP UNIT CELL:

(A) Tetrahedral voids: The FCC/CCP unit cell has eight tetrahedral voids per unit cell. Just below every corner of the unit cell, there is one tetrahedral void. As there are eight corners, there are eight tetrahedral voids.

(B) Octahedral voids: In an FCC/CCP unit cell, there are four octahedral voids. They are present at all the edge centres and at the body centre. The contribution of the edge centre is 1/4

Hence, total number of octahedral voids:

In CCP/FCC:

 Rank (Z) = 4,

 Number of tetrahedral voids = 8 and

 Number of tetrahedral voids = 2 × Z

Number of tetrahedral voids in close packing = 2 × eff. no. of spheres.

Hence, there are two Tetrahedral Voids per sphere in closed packing arrangements.

In CCP/FCC:

 Z = 4

Number of octahedral voids = 4 

Number of octahedral voids = Z

There is exactly one OV per sphere in close packing.

BODY CENTERED CUBIC CELL (BCC):

(1) A more efficiently packed cubic structure is the "body-centered cubic" (bcc).

(2) The first layer of a square array is expanded slightly in all directions. Then, the second layer is shifted so its spheres nestle in the spaces of the first layer.

(3) This repeating order of the layers is often symbolized as "ABA...".The considerable space shown between the spheres is misleading: spheres are closely packed in bcc solids and touch along the body diagonal.

(4) The packing efficiency of the bcc structure is about 68%. The coordination number for an atom in the bcc structure is eight.

(5) Both SCC and BCC are  not a closed packing in 3D.

(6) BCC crystal lattice is basically fusion of two simple unit cell in such that coner of one simple unit cell become body centred atom of another simple unit cell.

(7) Radius of atom and Packing efficiency in BBC unit cell:

Consider ‘r’ is  the radius of sphere and ‘a’ be the edge length of the cube and the sphere at the centre touches the sphere at the corner. Therefore body diagonal

Packing efficiency (PE): Percentage of space occupied by spheres

No. of spheres in bcc = 2

Volume of 2 spheres = 2´4/3pr³

Empty space (% Voids) = 100- 68= 32 %

(8)  Density of BCC unit cell:

(9) Coordination Numbers:

(1) The nearest neighbour distance is half of body diagonal (a) root 3/2 (along body diagonal) therefore coordination number for a given atom in BCC unit cell is 8.

(2) The next nearest neighbour are 6 at distance (a) Lattice parameter.

(3) 3^rd neighbour (Next to Next nearest neighbour) are (12) at distance (a root 2) ( All corner along face diagonal in x,y and Z plane).

ILLUSTRATIVE EXAMPLE (1): An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element ?

SOLUTION:

ILLUSTRATIVE EXAMPLE (1): How many 'nearest' and 'next nearest' neighbours respectively does potassium have in b.c.c. lattice

SOLUTION: 8 and 6

SIMPLE CUBIC CELL(SCC):

(1) 2D square close packing sheets are involved to generate simple cubic cell as well as body centred cell. In which each corner atom is touching potion with its adjacent corner atom.

(2) Take two 2D square close packing sheet and Placing a second square packing layer (sheet) directly over a first square packing layer forms a "simple cubic" structure.

(3) The simple “cube” appearance of the resulting unit cell is the basis for the name of this three dimensional structure.

(4) This packing arrangement is often symbolized as "AA...", the letters refer to the repeating order of the layers, starting with the bottom layer.

(5) The coordination number of each lattice point is six. This becomes apparent when inspecting part of an adjacent unit cell.

(6) The unit cell contain eight corner spheres, however, the total number of spheres within the unit cell is 1 (only 1/8th of each sphere is actually inside the unit cell). The remaining 7/8ths of each corner sphere resides in 7 adjacent unit cells.

(7) PACKING EFFICIENCY):

In simple cubic unit cell:

(1) Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.

(2) As sphere are touching each other therefore a = 2r

(3) No. of spheres per unit cell = 8*1/8=1

(4) Volume of the sphere = 4/3(pi) r³

(5) Volume of the cube = a³= (2r)³ = 8r³

(6)  Packing efficiency (space occupied):

(7) Density of simple unit cell:

(8) Coordination Number:

(1) The nearest neighbour distance is just the lattice parameter (a) therefore coordination number for a given atom in SCC unit cell is 6 (six).

(2) The next nearest neighbour are 12 at distance a/root 2 (each face diagonal in x ,y and Z plane).

(3) 3^rd neighbour (Next to Next nearest neighbour) are (8) at distance a root 3 (each corner along body diagonal.