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will the lattice energy of Al2O3 be greater than the lattice energy of AlF3. If so, why? If not, why?
The lattice energy of Al2O3 is indeed greater than that of AlF3 due to the higher charges of the ions involved and the smaller atomic radii of the aluminum and oxide ions compared to aluminum and fluoride ions. The discussion emphasizes that lattice energy increases with larger ionic charges and decreases with larger atomic radii. Additionally, the packing of ions in the crystal structure plays a significant role in determining lattice energy, particularly in ionic compounds.
PREREQUISITESChemistry students, materials scientists, and professionals involved in solid-state chemistry or ionic compound analysis will benefit from this discussion.
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.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.