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

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

Atomic radii of Ti, Zn and Hf vary.

Atomic radii of Ti, Zn and Hf vary.

All Zn(+II) compounds are white because:

All Zn(+II) compounds are white because:

In the zinc blende structure, zinc ions occupy alternate tetrahedral voids and S^(2-) ions exist as ccp the radii of Zn^(2+) and S^(2-) ions are 0.83Å and 1.74 Å respectively.The edge length of the ZnS unit cell is :

In the zinc blende structure, zinc ions occupy alternate tetrahedral voids and S^(2-) ions exist as ccp the radii of Zn^(2+) and S^(2-) ions are 0.83Å and 1.74 Å respectively.The edge length of the ZnS unit cell is :

The crystal structure of Zinc blende (ZnS) consists of cubic close-packed (ccp) array of S^(2-) ions. If the radii of Zn^(2+)andS^(2-) ions are 0.74 and 1.84Å respectively, then which type of voids (tetrahedral or octahedral) formed in ccp array of S^(2-) ions are occupied by Zn^(2+) ions? What fraction of total number of this void remains unoccupied?