- #1

cookiemnstr510510

- 162

- 14

## Homework Statement

The ionic radius of the Ca

^{2+}ion is 0.106nm. The ionic radius of the O

^{2-}ion is 0.132nm.

a.Calculate the coulomb force of repulsion between nearest neighbor O

^{2-}ions in CaO (which has the NaCl-type structure).

b. Calculate the colombia force of repulsion between nearest neighbor Ca

^{2+}ions in CaO (which has the NaCl-type structure).

c. Calculate the coulomb force of attraction between Ca

^{2+}ions and O

^{2-}ions in CaO (which has the NaCl-type structure).

## Homework Equations

F

_{c}=-k(Z

_{1}q

_{1})(Z

_{2}q

_{2})/a

^{2}

k=constant=9x10

^{9}Nm

^{2}/C

^{2}

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:

a=0.106nm+0.132nm=0.238nm

F

_{c}= - k(+2 x 1.602x10

^{-19}c)(-2 x 1.602x10

^{-19}c)/(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(F

_{r})+attractive force(F

_{c})=0, therefore F

_{r}=-F

_{c}

So does this mean that I can calculate the attractive force between two nearest neighbor O

^{2-}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 (O

^{2-}'s? I realize within CaO the Ca

^{2+}and O

^{2-}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?

Thanks!