Question

Question No. 7
Percentage of tetrahedral voids occupied by Zn2+ ions per unit cell of ZnS is
O 50
100
O 75
o 25

Answer 1

Answer:

100

because it is a tetrahedral voids

Answer 2

Answer:

The tetrahedral void's percentage occupied by the $Zn^{2+}$ ions per unit cell of ZnS observed is $50\%$.

Therefore, option a) $50\%$ is correct.

Explanation:

Given data,

The $Zn^{2+}$ ions occupied tetrahedral voids.

The tetrahedral void's percentage occupied by the $Zn^{2+}$ions per unit cell of ZnS =?

As we know,

• ZnS is known as the zinc blende having $Zn^{2+}$ and $S^{2-}$ ions.

And,

• In ZnS, the stoichiometric ratio of zinc to sulphur is $1:1$.

Also,

• ZnS has cubic close packing ( ccp ) arrangement.

As mentioned above, the stoichiometric ratio of zinc to sulphur is $1:1$.

Therefore, in tetrahedral voids, the  $Zn^{2+}$ ions occupy half voids, i.e., alternate tetrahedral voids.

Thus, the percentage is half, i.e., $50\%$.

Hence, the tetrahedral void's percentage occupied by the $Zn^{2+}$ ions per unit cell of ZnS observed is $50\%$.