Total no. of voids in `0.5` mole of a compound forming hexagonal closed packed structure are :

Options

A. `6.022 xx 10^(23)`
B. `3.011 xx 10^(23)`
C. `9.033 xx 10^(23)`
D. `4.516 xx 10^(23)`

Answer

C

Explanation

No. of effective particles in hcp lattice `1= (6 + 6) xx 1//6 + 3 xx 1 + 2 xx 1//2 = 6`
No. of particles in 1/2 mole `=1/2 xx 6.022 xx 10^(23)`
No. of unit cells in `1/2` mole `=1/(2 xx 6) xx 6.022 xx 10^(23)`
The no. of tetrahedral voids is equal to double the no. of particles in a unit cell. Thus the no. of tetrahedral voids.
`T = 2 xx 6 xx 1/(2 xx 6) xx 6.022 xx 10^(23) = 6.022 xx 10^(23)`
The no. of octahedral voids is equal to the no. of particles in a unit cell. Thus the no. of octahedral voids.
`O = 6 xx 1/(2 xx 6) xx 6.022 x 10^(23) rArr O = 3.011 xx 10^(23)`
Total no. of voids = T + O
`= 6.022 xx 10^(23) + 3.011 xx 10^(23) = 9.033 xx 10^(23)`