Question

Match the following:

$\begin{array}{cc}Compounds& Coordination(A:B)\\ \left(1\right)ZnS& \left(i\right)8:8\\ \left(2\right)CsCl& \left(ii\right)6:6\\ \left(3\right)NaCl& \left(iii\right)4:4\\ \left(4\right)N{a}_{2}O& \left(iv\right)8:4\\ \left(5\right)Ca{F}_2& \left(v\right)4:8\end{array}$

• A. 1 – (iii), 2 – (i), 3 – (ii), 4 – (v), 5 – (iv)
• B. 1 – (ii), 2 – (i), 3 – (iii), 4 – (v), 5 – (iv)
• C. 1 – (iii), 2 – (i), 3 – (ii), 4 – (iv), 5 – (v)
• D. 1 – (ii), 2 – (iii), 3 – (i), 4 – (v), 5 – (iv)

Answer 1

The correct option is A

1 – (iii), 2 – (i), 3 – (ii), 4 – (v), 5 – (iv)

Let’s try to recall all the given structures.

NaCl $\to$ It has a face-centered cubic (fcc) arrangement (also called cubic close packing, i.e., ccp) in which $C{l}^{⊝}$ ions occupy the corners and face centers of the cube which $N{a}^{⊝}$ ions are present at the body centre and edge centers.

NaCl structure

Each $N{a}^{\oplus }$ ion is surrounded by six $C{l}^{⊝}$ ions and each $C{l}^{⊝}$ ion is surrounded by six $N{a}^{\oplus }$ ions. In other words, the $N{a}^{\oplus }$ ions as well as $C{l}^{⊝}$ ions have coordination number of 6, i.e., this structure has 6 : 6 coordination (i.e., $C{l}^{⊝}$ ions being larger in size are present in the close packing while $N{a}^{\oplus }$ ions are present in all the octahedral voids.

CsCl $\to$

It has body-centered cubic (bcc) arrangement.

Each $C{s}^{\oplus }$ ion is surrounded by eight $C{l}^{⊝}$ ions and each $C{l}^{⊝}$ ion is surrounded by eight $C{s}^{\oplus }$ ions, i.e., this structure has 8: 8 coordination.

A unit cell of CsCl consists of only one unit of CsCl, i.e., one $C{s}^{\oplus }$ ion and one $C{l}^{⊝}$ ion (because each ion present at the corner is shared by eight unit cells) [The ratio $\frac{r_{Cs\oplus }}{r_{Cl\oplus }}$ = 0.93 which is greater than expected for exact fitting (0.732)].

ZnS → 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.

$Ca{F}_2$ $\to$ 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.

$N{a}_{2}O$ $\to$ It is an antiflourite type structure so the cations and anions’ positions will be reversed as they are in fluorite type ($Ca{F}_2$) type structure.

Coordination number of $A_{2}B$ -type structure:

Each $O^{2-}$ ions is surrounded by eight $N{a}^{\oplus }$ ions.

Each $N{a}^{\oplus }$ ion is surrounded by four $O^{2-}$.

Thus, the structure has 4 : 8 coordination.