Question

Match the List-I with List-II and List-III:

List-I (solids)	List-II (unit cell)	List-III (coordination number)
(a)- Rock salt	(p)- Face centered cubic, anion in the tetrahedral void	(w)- Coordination 6
(b)- Fluorite	(q)- Face centered cubic, cation in the octahedral void	(x)- Cation (8), anion (4)
(c)- AgI, ZnS	(r)- Face centered cubic, cation in the alternate tetrahedral void	(y)- Cation (4), anion (8)
(d)- $N{{a}_{2}}O$	(s)- Face centered cubic, cation in the tetrahedral void	(z)- Cation (4), anion (4)

Answer 1

Hint: The number of oppositely charged ions surrounding each other is called coordination number. In rock salt, both the anions and cations are surrounded by 6 ions. In fluorite, cations are surrounded by 8 anions. In zinc blend (ZnS) both the cations and anions are surrounded by 2 ions. $N{{a}_{2}}O$ has an antifluorite type of crystal structure.

Complete step by step answer:
Let us study the structure one by one:
(a)- Rock salt: It has an fcc arrangement in which $C{{l}^{-}}$ ions occupy the corners and face centers of a cube while $N{{a}^{+}}$ ions are present at the edge centers and body center.
The coordination number of $N{{a}^{+}}$= 6 and of $C{{l}^{-}}$= 6.
(b)- Fluorite: It has ccp arrangement in which $C{{a}^{2+}}$ ions surrounded by 8 ${{F}^{-}}$ ions and each ${{F}^{-}}$ions by 4 $C{{a}^{2+}}$. ${{F}^{-}}$ions occupy all the tetrahedral voids.
The coordination number of $C{{a}^{2+}}$= 8 and of ${{F}^{-}}$= 4
(c)- AgI, ZnS: They have a ZnS type crystal structure. It has a ccp arrangement in which ${{S}^{2-}}$ ions form fcc and each $Z{{n}^{2+}}$ ion is surrounded tetrahedrally by four ${{S}^{2-}}$ ions and vice versa.
So the coordination number of both cations and anions are 4.
(d)- $N{{a}_{2}}O$: It has a crystal structure of Anti fluorite type. Here negative ions form the ccp arrangement so that each positive ion is surrounded by 4 negative ions and each negative ion is surrounded by 8 positive ions. The cations occupy half of the tetrahedral voids.
The coordination number of $N{{a}^{+}}$= 4 and of ${{O}^{2-}}$ = 8

So, from the above discussion, we can match the list as:

List-I (solids)	List-II (unit cell)	List-III (coordination number)
(a)- Rock salt	(q)- Face centered cubic, cation in the octahedral void	(w)- Coordination 6
(b)- Fluorite	(p)- Face centered cubic, anion in the tetrahedral void	(x)- Cation (8), anion (4)
(c)- AgI, ZnS	(r)- Face centered cubic, cation in the alternate tetrahedral void	(z)- Cation (4), anion (4)
(d)- $N{{a}_{2}}O$	(s)- Face centered cubic, cation in the tetrahedral void	(y)- Cation (4), anion (8)

Note: In the above discussion ccp is cubic close packing, and fcc is face-centered cubic. The structure of the diamond is similar to that of the zinc blende, in which all the cations and anions are replaced with the carbon atom.