Place the following in order of increasing magnitude of lattice energy: K2O Rb2S
ID: 860622 • Letter: P
Question
Place the following in order of increasing magnitude of lattice energy:
K2O Rb2S Li2O
least lattice energy < --- --- ---> most lattice energy
I am really struggling with how to calculate lattice energy. So if you could, please explain how you got the answer and how that method could be generalized to other similar questions. I realize it has something to do with the charge product between the elements in the molecule, and secondary to that is the atomic radius... it's just not clicking for me to do on the fly.
Thank you in advance!!
Explanation / Answer
Lattice energy is given by the product of the charges on the ions, divided by the sum of their ionic radii.
Let's compare K2O and Li2O first.
K has a larger radius than Li because it's lower down in the periodic table. It also has the same charge (+1) as Li. Since the anion (O) is the same in both compounds, the lattice energy of Li2O is larger than K2O. This is because dividing the product of their charge by a larger radius gives a smaller value of lattice energy.
Rubidium is below K in the periodic table and thus its ionic radius is really huge. Plus, S is below O in the table and it has a larger ionic radius. Thus, even though Rb and S have the same charges as Li and O (+1, -2) the lattice energy of Rb2S will be much much lower than Li2O and K2O due to the huger ionic radius. After all, dividing a constant number (+1 times -2) by a larger value of radius will give a smaller number ie value of lattice energy.
Thus in ascending order, you should have Rb2S, K2O, Li2O ...........answer