The structure of the compound CuI is best described as a cubic closest packed array of iodide ions with the copper ions in tetrahedral holes. What percent of the tetrahedral holes are occupied in this solid

12.5 %

Explanation:

In CCP, the effective number of anion is 4

That is there are 4 I- present in 1 unit cell

Number of tetrahedral void = 2*effective number of anion

= 2*4

= 8

In ZrI4, for every 4 anion, there are only 1 Zr atom.

So, one tetrahedral void is occupied per unit cell out of 8

% tetrahedral void occupied = 1*100/8

= 12.5 %

12.5 %

sorry idk the answer well to you i have a explaination

explanation: a formula which shows the arrangement of atoms in the molecule of a compound

ill try to . no promises tho.

explanation: