## Your thinking is almost correct, but there is a small mistake in the calculation. Let me guide you through the correct process step by step:

To find the distance between the chloride ions (Cl-) along a cell edge, you need to subtract the sum of their radii from the length of the cell edge.

Here's how you can calculate it:

1. Given data:

- Lattice type: face-centered cubic (FCC)

- Cell edge length: 0.513 nm

- Ionic radius of Cl- ion: 0.181 nm

2. Determine the space between the chloride ions along the cell edge by subtracting the sum of their radii:

Space = Cell edge length - (2 * ionic radius)

= 0.513 nm - (2 * 0.181 nm)

= 0.513 nm - 0.362 nm

= 0.151 nm

Therefore, the distance between the chloride ions along a cell edge is 0.151 nm.

Now that we have determined the distance between the chloride ions, we can proceed to answer part B of your question.

B) To determine if an Na+ ion (with a radius of 0.095 nm) or a K+ ion (with a radius of 0.133 nm) would fit into this space, we need to compare their radii with the distance between the chloride ions.

- For an Na+ ion (radius = 0.095 nm):

Since the distance between chloride ions along a cell edge is 0.151 nm, an Na+ ion would not be able to fit within this distance. It is too large.

- For a K+ ion (radius = 0.133 nm):

Similarly, the K+ ion is larger than the distance between chloride ions (0.151 nm), so it would not fit within this space either.

In conclusion, neither the Na+ ion nor the K+ ion would fit into the space between the chloride ions in the LiCl lattice structure with a cell edge length of 0.513 nm.

I hope this explanation helps! If you have any further questions, feel free to ask.