[student8607]

Order the lattice energies of AlBr3, MgBr2, LiBr and CaO from least to greatest.
I know from class that as the charges increase the U increases. And, when the distance increases the U decreases.

But with AlBr3 and MgBr2, the charges are not the same for each atom, so is that fair for comparing the compounds? Also, they don't all have 1 atom in common so how can you compare radii?

I'm guessing it would go AlBr3 --> MgBr2 --> CaO --> LiBr?

[Borek]

I have forwarded your question somewhere else (mailing list populated mainly by college and university teachers) and I got a real (experimental) answer:

LE (kj/mol)
AlBr3   = 5361
CaO     = 3414
MgBr2   = 2440
LiBr    = 807

But there was (more or less) general agreement that the question is poorly thought and doesn't make much sense, as it requires you to compare things that are not easily comparable.

[Astrokel]

But there was (more or less) general agreement that the question is poorly thought and doesn't make much sense, as it requires you to compare things that are not easily comparable.

True enough. Lattice energy is proportional to charge density(charge/distance between ions), right? Can i ask which is the distance between ions in the case of AlBr₃ and MgBr₂, is it Mg2+ to Br- OR Br- to Mg2+ to Br-? Thanks

[Borek]

Can i ask which is the distance between ions in the case of AlBr₃ and MgBr₂, is it Mg2+ to Br- OR Br- to Mg2+ to Br-?

No idea, Perhaps AWK will know or will be able to easily locate this information, crystallography is his area.

[student8607]

No problem. I appreciate you guys helping me out when you can.

If it's a poor question, not your fault.

I understand the concept, just didn't get this whacky question.
Thanks