In cubic ZnS if the radii of `Zn^(2+) and S^(2-)` atoms are 0.74 Å and 1.70 Å, the lattice parameter of 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_(ө)(i.e.,(r_(Zn^(o+)))/(r_(S^(2-))))=(0.74Å)/(1.70Å)=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_(ө)=0.414` for OV. There is also some covalent character in the `Zn^(2+)-S^(2-)`. interaction, which tends to shorten the interatomic distance. Experimentally, it was found that `Zn^(2+)` ions occupy TVs.
`therefore(r_(Zn^(2+))+r_(s^(2-)))=sqrt3/4a," "(0.74+1.70)A^(@)=sqrt3/4a rArra=5.634A^(@)`