The ionic radius of the Ca2+ ion is 0.106nm. The ionic radius of the O2- ion is 0.132nm.
a.Calculate the coulomb force of repulsion between nearest neighbor O2- ions in CaO (which has the NaCl-type structure).
b. Calculate the colombia force of repulsion between nearest neighbor Ca2+ ions in CaO (which has the NaCl-type structure).
c. Calculate the coulomb force of attraction between Ca2+ ions and O2- ions in CaO (which has the NaCl-type structure).
Z represents the valences of the two ions
a is the interatomic spacing
q is the magnitude of charge of the electron
The Attempt at a Solution
I believe I can do c:
Fc= - k(+2 x 1.602x10-19c)(-2 x 1.602x10-19c)/(0.238x10^-9m)2 = 1.63x10-8 N
I have a couple questions:
1. What is the significance of the problem telling us "which has the NaCl-type structure" in each part? Is this just telling us that CaO is bonded ionically?
2. I know that repulsive force(Fr)+attractive force(Fc)=0, therefore Fr=-Fc
So does this mean that I can calculate the attractive force between two nearest neighbor O2- ions, then flip the sign and call it the repulsive force?
3. If this is true then are we saying there is an attractive force between two negatively charged things (O2-'s? I realize within CaO the Ca2+ and O2- have an attractive and repulsive force which equals zero (that is why they are held together within the molecule).
I guess I want to know if the same process can be used for part C on part A and B, and if so why?