Question

Which of the following are TRUE for $ZnS$ type and $Ca{F}_2$ type structure?

• A. $ZnS$ – 6:6 coordination; $Ca{F}_2$ – 8:4 coordination
• B. $ZnS$ – 4:4 coordination; $Ca{F}_2$ – 8:4 coordination
• C. $ZnS$ – $Z{n}^{2+}$ at half of tetrahedral voids; $Ca{F}_2$ – F^- ions are at all tetrahedral voids
• D. $ZnS$ – $Z{n}^{2+}$ at all of tetrahedral voids; $Ca{F}_2$ – F^- ions are at all tetrahedral voids

Answer 1

Answer: (B), (C)

$ZnS$ – $Z{n}^{2+}$ at half of tetrahedral voids; $Ca{F}_2$ – F^- ions are at all tetrahedral voids

Just recall the structure of $Ca{F}_2$ here. $Ca{F}_2$ is an example of fluorite type structure only.

The calcium ions, $C{a}^{2+}$, are located at the face-centered cubic lattice points and, therefore, have cubic closed-packed arrangement. The fluoride ions ($F^{⊝}$ ions) occupy all the 8 tetrahedral voids. Thus, there are 4 $C{a}^{2+}$ ions and 8 $F^{⊝}$ ions in the unit cell of calcium fluoride.

Since each $C{a}^{+2}$ surrounded by 8 $F^{⊝}$ ions and

each $F^{⊝}$ is surrounded by 4 $C{a}^{+2}$ ions.

Thus, the structure has 8 : 4 coordination.

Now think about structure of $ZnS$.

$ZnS$ $\to$ The arrangement as possessed by ZnS is called cubic close packing (ccp) in which $S^{2-}$ ions are present at the corners as well as at the centre of each face of the cube.

Each $Z{n}^{2+}$ ion is surrounded tetrahedrally by four $S^{2-}$ ions and each $S^{2-}$ ion is surrounded tetrahedrally by four $Z{n}^{2+}$ ions. Thus, this structure has 4 : 4 coordination.

Ans: (b) and (c)