InorganicVII

Chapters / Chapter 1 · Solid state / Topic 7 of 14

Voids and the radius ratio rule

Chapter 1 · Solid state and crystal structures

Tetrahedral void: 4 spheresr(void) = 0.225 R, 2 per sphereOctahedral void: 6 spheresr(void) = 0.414 R, 1 per sphere
A tetrahedral void is bounded by four spheres; an octahedral void by six.

The empty spaces left between packed spheres are called interstitial voids. Two types matter.

VoidSurrounded byPer sphereLargest sphere that fits
TetrahedralFour spheres at the corners of a tetrahedron20.225 R
OctahedralSix spheres at the corners of an octahedron10.414 R

R is the radius of the packed sphere. An fcc unit cell has 4 atoms, so it has 8 tetrahedral voids and 4 octahedral voids.

CsCl: body diagonal = 2r+ + 2r-sqrt(3) a = 2r+ + 2r-a = 2r- (anions touch)r+/r- = sqrt(3) - 1 = 0.732NaCl (octahedral): 0.414ZnS (tetrahedral): 0.225
CsCl geometry: the cation touches anions along the body diagonal.

Radius ratio rule

The radius ratio r+/r− predicts which hole a cation occupies, and therefore its coordination number. The limits are the ratios at which the cation just touches the surrounding anions.

r+/r−CoordinationGeometryExample
0.225 to 0.4144TetrahedralZnS
0.414 to 0.7326OctahedralNaCl
0.732 to 1.0008CubicCsCl

Worked limits, as written in the notes

CsCl, 8:8. Side of cube a = 2r−. Body diagonal √3 a = 2r+ + 2r−. Hence √3 = (r+ + r−)/r−, so r+/r− = √3 − 1 = 1.732 − 1 = 0.732.

NaCl, octahedral, 6:6. Face diagonal √2 a = 2r+ + 2r−, with a = 2r−. Hence r+/r− = √2 − 1 = 0.414.

ZnS, tetrahedral, 4:4. Half the body diagonal gives r+ + r− = √3 a/2, with the face diagonal √2 a = 2r−. The limiting ratio is 0.225.