asked 168k views
2 votes
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

asked
User Eliwedel
by
7.7k points

1 Answer

6 votes

Answer:

12.5 %

Step-by-step 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 %

Answer: 12.5 %

answered
User Moolsbytheway
by
8.3k points
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