could some one please help to calculate the fraction of octahedral holes occupied by aluminium in Al2O3
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http://en.wikipedia.org/wiki/Aluminium_oxide
Of course! To calculate the fraction of octahedral holes occupied by aluminum in Al2O3 (aluminum oxide), we need to consider the crystal structure of this compound.
Aluminum oxide has a corundum crystal structure, in which aluminum atoms occupy octahedral holes in a close-packed oxygen lattice. In this structure, there are two aluminum atoms for every three oxygen atoms.
The formula of aluminum oxide is Al2O3, indicating that there are two aluminum atoms per formula unit of the compound. Therefore, we can assume that there will be twice as many octahedral holes compared to the number of aluminum atoms.
To calculate the fraction of octahedral holes occupied by aluminum, we need to determine the ratio of the number of occupied octahedral holes to the total number of octahedral holes.
Let's denote the total number of octahedral holes by N and the number of occupied octahedral holes by n. Since there are two aluminum atoms in Al2O3 and one aluminum atom occupies one octahedral hole, the number of occupied octahedral holes by aluminum is also 2.
Therefore, the fraction of octahedral holes occupied by aluminum is:
n/N = 2/N
To find the value of N, we need to consider the packing arrangement of the oxygen lattice in Al2O3. In corundum structure, every oxygen atom is surrounded by six octahedral holes. Therefore, the total number of octahedral holes (N) is six times the number of oxygen atoms.
Since the formula of aluminum oxide is Al2O3, it implies that there are three oxygen atoms per formula unit. Thus, the total number of octahedral holes (N) is:
N = 6 * 3 = 18
Now we can substitute this value into the equation to find the fraction:
n/N = 2/18
Simplifying this ratio gives us:
n/N = 1/9
So, the fraction of octahedral holes occupied by aluminum in Al2O3 is 1/9.
Therefore, 1/9 of the octahedral holes in the corundum structure of Al2O3 are occupied by aluminum atoms.