Is the lattice energy of Al2O3 greater than that of AlF3?

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The discussion centers on comparing the lattice energy of Al2O3 and AlF3, emphasizing the principles of Coulomb attraction and the significance of ionic charges and atomic radii. It is established that lattice energy increases with larger charges and smaller atomic radii. Al2O3 features aluminum and oxygen ions, while AlF3 consists of aluminum and fluoride ions. The presence of multiple charges in Al2O3 (two Al and three O) raises questions about its total ionic character and how this affects lattice energy. Additionally, the packing of ions in the crystal structure is highlighted as a crucial factor influencing lattice energy. Overall, the comparison hinges on the ionic nature of the compounds and the arrangement of ions within their lattices.
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will the lattice energy of Al2O3 be greater than the lattice energy of AlF3. If so, why? If not, why?
 
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Think about Coulomb attraction and charges.
 
The basic rules of lattice energy that I've been taught are that the lattice energy is greater if the charges of the atoms are larger and if the atomic radii are smaller. I haven't done this with two or more of the same atoms (like how there are two Al and 3 O) yet so I don't know how much of a difference that makes. Hope I was able to help in a way.
 
MysticDude said:
The basic rules of lattice energy that I've been taught are that the lattice energy is greater if the charges of the atoms are larger and if the atomic radii are smaller. I haven't done this with two or more of the same atoms (like how there are two Al and 3 O) yet so I don't know how much of a difference that makes. Hope I was able to help in a way.
What you say, assuming Al2O3 is totally ionic. Is it? Furtermore, in the ionic case, you also have to consider the kind of packing of ions.
 

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