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In cubic ZnS (II-VI) compounds, if the r...

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

A

11.87 Å

B

5.635 Å

C

5.14 Å

D

2.97 Å

Text Solution

Verified by Experts

`(r_(Zn^(2+)))/(r_(S)^(2-))=0.44`
With a radius ratio of 0.44, one might expect the Zinc (II) ions to occupy octahedral voids : however, the value of 0.44 is only slightly larger than `(r_("hole"))/(r)=0.414` for an octahedral hole. There is also some covalent character in the `Zn^(2+)-S^(2-)` interaction, which tends to shorten the inter atomic distance. Experimentally, one find that the Zinc (II) ions occupy tetrahedral holes.
`(sqrt(3))/(4)=r_(Zn^(2+))+r_(S^(2-))implies a =5.635 Å`
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