In cubic `ZnS (II-VI)` compounds, if the radii of `Zn` and `S` atoms are `0.74 Å` and `1.70 Å`, the lattice parameter of cubic `ZnS` is

Options

A. `11.87 Å`
B. `5.634 Å`
C. `5.14 Å`
D. `2.97 Å`

Answer

B

Explanation

`(r_(o+))/(r_(ɵ)) = (i.e., (r_(Zn^(o+)))/(r_(S^(2-)))) = (1.74 Å)/(1.70 Å) = 0.44`
From radius ratio, it is expected that `Zn^(2+)` ion occupy `OVs`, however, the value of `0.44` is only slightly larger than `r_("void")//r_(ɵ) = 0.414` for `OV`. There is alos some covalent character in the `Zn^(2+) -S^(2-)` interation, which tends to shortent the interatom distance.
`:. (r_(Zn^(2+)) + r_(S^(2-))) = (sqrt(3))/(4)a`
`(0.74 + 1.70)Å = (sqrt(3))/(4)a rArr a = 5.634 Å`