# Second question:

In the LiCl structure shown in 9.18 [LiCl's lattice, where the "Cl- ions are in contact with each other, forming a face centered cubic lattice."], the chloride ions form a face-centered cubic unit cell 0.513 nm on an edge. The ionic radius of Cl- is 0.181 nm.
A) Along a cell edge, how much space is in between the Cl- ions?
B) Would an Na+ ion (r=0.095 nm) fit into this space? A K+ ion (r=0.133 nm)?

I guess I just need help with letter A, since B will come as a result of that. I don't know how to do it though. Any help would be appreciated

According to the pictures I can find, we have one atom at the left corner with its radius and one atom at the right corner with its radius and a space between. The radius + radius + space = a, the cell edge. Therefore, if we take (cell edge - 2*radius) = 0.513 nm -(2*1.81 nm) = the distance of the space. Check my thinking.

8 months ago

## 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.

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