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.

RADIUS RATIO RULES IN IONIC SOLIDS:

The structure of many ionic solids can be accounted by considering the relative sizes of the cation and anion, and their relative numbers. By simple calculations, we can work out as how

Many ions of a given size can be in contact with a smaller ion. Thus, we can predict the coordination number from the relative size of the ions.

Following conditions must be satisfied simultaneously during the stacking of ions of different sizes in an ionic crystal:

(1) An anion and a cation are assumed to be hard spheres always touching each other.

(2) Anions generally will not touch but may be close enough to be in contact with one another in a limiting situation.

(3) A cation should surround itself with as many anions as possible. Each ion tends to surround itself with as many ions of opposite sign as possible to reduce the potential energy. This tendency promotes the formation of close-packed structures.

(4)The ratio of the cation to that of the anion is called RADIUS RATIO.

(5) Eventually greater is the radius , the larger is the size of cation and hence greater is it's coordination number.

(6)The relationship between the radius and coordination number and structural arrangement are called radius ratio rule and are given as table below.

RADIUS RATIO AND COORDINATION NUMBER:

ILLUSTRATIVE EXAMPLE (1): The two ions A⁺ and B^- have radii 88 and 200 pm respectively. In the close packed crystal of compound AB, predict the coordination number of A⁺.

SOLUTION:

                        It lies in the range of 0.414 – 0.732

                        Hence, the coordination number of A⁺ = 6

ILLUSTRATIVE EXAMPLE(2): Br^- ion forms a close packed structure. If the radius of Br^- ions is 195 pm. Calculate the radius of the cation that just fits into the tetrahedral hole. Can a cation A⁺ having a radius of 82 pm be slipped into the octahedral hole of the crystal A⁺ Br^-?

SOLUTION: (1)  Radius of the cations just filling into the tetrahedral hole

                              = Radius of the tetrahedral hole = 0.225´195

                              = 43.875 pm

                        (2)  For cation A⁺ with radius = 82 pm

As it lies in the range 0.414 – 0.732, hence the cation A⁺ can be slipped into the octahedral hole of the crystal A⁺ Br^-.

ILLUSTRATIVE EXAMPLE(3):  Why is co-ordination number of 12 not found in ionic crystals?

SOLUTION:  Maximum radius ratio in ionic crystals lies in the range 0.732 – 1 which corresponds to a coordination number of 8. Hence coordination number greater than 8 is not possible in ionic crystals.

LOCATING TETRAHEDRAL AND OCTAHEDRAL VOIDS:

(1) The close packed structures have both octahedral and tetrahedral voids. In a ccp structure, there is 1 octahedral void in the centre of the body and 12 octahedral void on the edges. Each one of which is common to four other unit cells. Thus, in cubic close packed structure.

Octahedral voids in the centre of the cube =1

Effective number of octahedral voids located at the 12 edge of = 12*1/4=3

Total number of octahedral voids = 4

(2) In ccp structure, there are 8 tetrahedral voids. In close packed structure, there are eight spheres in the corners of the unit cell and each sphere is in contact with three groups giving rise to eight tetrahedral voids

TETRAHEDRAL VOIDS LOCATION:

(3) Circles labelled T represent the centers of the tetrahedral interstices in the ccp arrangement of anions.  The unit cell "owns" 8 tetrahedral sites.

OCTAHEDRAL VOIDS LOCATION:

Circles labelled O represent centers of the octahedral interstices in the ccp arrangement of anions (FCC unit cell).  The cell "owns" 4 octahedral sites.