If \(0.5 \mathrm{~mol}\) of a metal forms hexagonal close packed structure, the total number of voids, and…
If \(0.5 \mathrm{~mol}\) of a metal forms hexagonal close packed structure, the total number of voids, and tetrahedral voids respectively in mol are
\(1.5,1.0\)
\(1.0,0.5\)
\(1.0,1.5\)
\(0.5,1.0\)
Solution
Number of moles of metal in close packed structure \(=0.5 \mathrm{~mol}\)
No. of atoms in 1 mole of mole \(=N_A\)
\(\therefore\) Number of atoms in \(0.5 \mathrm{~mol}\) of metal
\(=0.5 \times 6.022 \times 10^{23}=3.011 \times 10^{23}\)
Also, number of tetrahedral voids \(=2 \times\) number of metal atoms in close packed structure
\(\therefore\) Number of tetrahadral voids \(=2 \times 3.011 \times 10^{23}\)
\(=6.022 \times 10^{23} \approx 1 \mathrm{~mol}\)
Number of octahedral void \(=\) number of metal atoms in close packed structure
So, that number of octahedral voids \(=3.011 \times 10^{23}\)
\(=0.5 \mathrm{~mol}\)
Total number of voids
\(\begin{aligned}
& =\text { tetrahedral void }+ \text { octahedral void } \\
& =6.022 \times 10^{23}+3.011 \times 10^{23} \\
& =9.05 \times 10^{23} \approx 1.5 \mathrm{~mol}
\end{aligned}\)
Therefore, total number of voids and tetrahedral voids in hexagonal close packed structure are \(1.5 \mathrm{~mol}\) and \(1.0 \mathrm{~mol}\) respectively. Hence, option (a) is the correct answer.