In the ionic compounds LiF, NaCl, KBr, and RbI, the measured
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Question

In the ionic compounds LiF, NaCl, KBr, and RbI, the measured cation–anion distances are 201 pm (Li–F), 282 pm (Na–Cl), 330 pm (K–Br), and 367 pm (Rb–I), respectively. (b) Calculate the difference between the experimentally measured ion–ion distances and the ones predicted from Figure 7.8.

Accepted Answer

Hello everyone. So in this video we want to find the difference between the predicted and experimentally measured ion ion distances. So in the problem we're given the theoretical or the experimental values. And we're gonna use this table right here to go ahead and calculate for the theoretical values. So I'm gonna go ahead and scroll down to give us a little bit more space. So first one we're gonna do is lithium iodide. That's L. I. H. Or L. Ii. So let's see. lithium is going to be 0.9. And then we're gonna add that with our iodine. That's 2.6. Putting that into my calculator. I get a value of 2.96. Now for a difference in between our theoretical and experimental values take the ups of the value of a difference. So 2.96 -2.94 gives us a difference of 0.02 ppm. Again that's for our lithium iodide. Now moving on we'll do sodium bromide. That's going to be an A. B. R. Again. With the exact same process. We used the table given with its values. So 1.16 from the sodium And then 1.82 we add those two numbers up we get 2.98. Now finding the difference again is the absolute value between the difference of our experimental and theoretical value. So we have 2.98 -2.90. Giving us a difference of 0.08 ppm. So that's going to be our difference for our sodium bromide. Then we can go on and do potassium chloride which is K C. L. So we have 1.52 plus 1.67. Giving us a sum of 3.19. So again this part right here that's going to be calculating for our theoretical value. Now, taking the difference, we have 3.19 minus 3.20. Giving us a difference of 0.1 P p. M. Alright, the last small kill we'll go ahead and do is going to be rubidium fluoride. That's going to be our B. F. So we have values 1.66 plus 1.19. Giving us a sum of 2.85. And of course to gain the difference, then we have 2. And 2.88. Once we put that into calculator we get the value of 0. ppm. That's going to be the difference for rubidium fluoride. So we found our four differences here using the given experimental values and our tables. Thank you all so much for watching

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