Question

The cubic phase of hafnium oxide (HfO2) has an FCC lattice with the basis Hf at 0 0 0, and O at ¼ ¼ ¼ and ¾ ¾ ¾.
Useful information: Lattice parameter of HfO2 is 5.08Å; Atomic weights: Hf = 175.5 g/mole; O = 16 g/mole; A = 6.022x1023 atoms/mole.

a. How many formula units of HfO2 are there per unit cell?
b. Assuming the O is close packed, calculate the density of cubic HfO2 in g/cm3.

Answer 1

HfO2 is in Flourite structure. Hf²⁺resent in FCC and O^2- iin tetrahedral voids (position ¼ ¼ ¼ and ¾ ¾ ¾).

Total no of Hf²⁺ is 4 (8 in cornsres, 6 in face center so 8*1/8 + 6*1/2 = 4), and total no of O^2- is 8 (equal to Td voids).

The formula is Hf₄O₈ or  4 HfO₂ .

So the Z = 4

density = Z*M/N*a³

Z = 4; M = 175.5 + 2(16) = 207.5 g/mole

N = 6.023*10²³; a = 5.08 A = 5.08* 10^-12 cm

a³ = (5.08* 10^-12 cm)³ = 131.09 * 10^-36

so

d = 4*207.5 / (6.023*10²³)(131.09 * 10^-36)

d= 830/ 789* 10^-13 = 1.05* 10^-13

density = 1.05* 10^-13 g/cm³ or

density = 1.05* 10^-16 kg/m³