Interstitial Voids

INTERSTITIAL VOIDS

In hcp as well as ccp only 74% of the available space is occupied by spheres. The remaining space is vacant and constitutes interstitial voids or interstices or holes. These are of two types

(a) Tetrahedral voids (b) Octahedral voids

Tetrahedral voids

In close packing arrangement, each sphere in the second layer rests on the hollow (triangular void) in three touching spheres in the first layer. The centres of theses four spheres are at the corners of a regular tetrahedral. The vacant space between these four touching spheres is called tetrahedral void. In a close packing, the number of tetrahedral void is double the number of spheres, so there are two tetrahedral voids for each sphere

Radius of the tetrahedral void relative to the radius of the sphere is 0.225

i.e. $\frac{r_{void}}{r_{sphere}}=0.225$

In a multi layered close packed structure , there is a tetrahedral hole above and below each atom hence there is twice as many tetrahedral holes as there are in close packed atoms

Octahedral voids

As already discussed the spheres in the second layer rest on the triangular voids in the first layer. However, one half of the triangular voids in the first layer are occupied by spheres in the second layer while the other half remains unoccupied. The triangular voids 'b' in the first layer are overlapped by the triangular voids in the second layer. The interstitial void, formed by combination of two triangular voids of the first and second layer is called octahedral void because this is enclosed between six spheres centres of which occupy corners of a regular octahedron In close packing, the number of octahedral voids is equal to the number of spheres. Thus, there is only one octahedral void associated with each sphere. Radius of the octahedral void in relation to the radius of the sphere is 0.414 i.e. $\frac{r_{Void}}{r_{sphere}}=0.414$

LOCATING TETRAHEDRAL AND OCTAHEDRAL VOIDS

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 = $12\times \frac{1}{4}=3$

$\therefore$ Total number of octahedral voids = 4

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

Illustration 1 : In a solid, oxide ions are arranged in ccp. Cations A occupy one – sixth of the tetrahedral voids and cations B occupy one third of the octahedral voids. What is the formula of the compound?

Solution: In ccp with each oxide there would be 2 tetrahedral voids and one octahedral voids 1/3^rd octahedral voids is occupied by B and 1/6^th tetrahedral void by A. Therefore the compound can be

$A_{2/6}{B}_{1/3}OorAB{O}_3$

SIZES OF TETRAHEDRAL AND OCTAHEDRAL VOIDS

(i) Derivation of the relationship between the radius (r) of the octahedral void and the radius (R) of the atoms in close packing.

A sphere into the octahedral void is shown in the diagram. A sphere above and a sphere below this small sphere have not been shown in the figure. ABC is a right angled triangle. The centre of void is A.

Applying Pythagoras theorem.

$B{C}^2=A{B}^2+A{C}^2$

${\left(2R\right)}^2={\left(R+r\right)}^2+{\left(R+r\right)}^2=2{\left(R+r\right)}^2$ $\Rightarrow \frac{4R^2}{2}={\left(R+r\right)}^2$

$\Rightarrow \left(\sqrt{2}\right)={\left(R+r\right)}^2$

$\Rightarrow \sqrt{2}R=R+r$

$\Rightarrow r=\sqrt{2}R-R=\left(1.414-1\right)R$

r = 0.414 R

(ii) Derivation of the relationship between radius (r) of the tetrahedral void and the radius (R) of the atoms in close packing: To simplify calculations, a tetrahedral void may be represented in a cube as shown in the figure. In which there spheres form the triangular base, the fourth lies at the top and the sphere occupies the tetrahedral void.

Let the length of the side of the cube = a

From right angled triangle ABC, face diagonal

$AB=\sqrt{A{C}^2+B{C}^2}=\sqrt{a^2+a^2}=\sqrt{2}a$

As spheres A and B are actually touching each other, face diagonal AB = 2R

$\therefore 2R=\sqrt{2}aorR=\frac{1}{\sqrt{2}}a...(i)$

Again from the right angled triangle ABD

$AD=\sqrt{A{B}^2+B{D}^2}=\sqrt{{\left(\sqrt{2}a\right)}^2+a^2}=\sqrt{3}a$

But as small sphere (void) touches other spheres, evidently body diagonal AD = 2(R + r).

$\therefore 2\left(R+r\right)=\sqrt{3}a$

$\Rightarrow R+r=\frac{\sqrt{3}}{2}a...\left(ii\right)$

Dividing equation (ii) by equation (i)

$\frac{R+r}{R}=\frac{\sqrt{3}/2\times a}{a/\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$

$\Rightarrow 1+\frac{r}{R}=\frac{\sqrt{3}}{\sqrt{2}}=1.225$ $\Rightarrow \frac{r}{R}=1.225-1=0.225$

$\Rightarrow$r = 0.225 R

RADIUS RATIO IN 1:1 OR AB TYPE STRUCTURE

Illustration 2. 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: $\frac{r^{+}}{r^{-}}=\frac{88}{200}=0.44$

It lies in the range of 0.414 – 0.732

Hence, the coordination number of A⁺ = 6

Illustration 3. 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: (i) Radius of the cations just filling into the tetrahedral hole

= Radius of the tetrahedral hole = 0.225´195

= 43.875 pm

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

Radius ratio $\frac{r^{+}}{r^{-}}=\frac{82}{195}=0.4205$

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^-.