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A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Option: 1
$9.03\times 10^{23},6.022\times 10^{23}$

Option: 2
$8.26\times 10^{23},6.022\times 10^{23}$

Option: 3
$9.03\times 10^{23},6.022\times 10^{22}$

Option: 4
$9.03\times 10^{23},5.34\times 10^{23}$

Answers (1)

best_answer

Given, the number of atoms in close packaging = 0.5 moles

$\therefore$ Number of close-packed particles $=0.5\times6.022\times10^{23}=3.011\times10^{23}$

Now, we know the ?number of tetrahedral voids

= 2 X number of atoms in close-packed structures.

Therefore, the number of tetrahedral voids $=2\times 0.311\times10^{23}$

$=6.022\times10^{23}$

Again, the number of octahedral voids $=$ number of atoms in close packaging

Thus, the number of octahedral voids $= 3.011\times10^{23}$

Now, the Total number of voids

= Tetrahedral voids + octahedral voids
$=6.022\times10^{23}+3.011\times10^{23}$ $=9.03\times10^{23}$

Therefore, Option(1) is correct

Posted by

Suraj Bhandari

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