Structures of Ionic Solids

The ionic solids are a type of crystalline solid, formed by ions, which are held together by strong electrostatic forces known as ionic bonds. The general composition of these solids includes a repetitive pattern of cations (positively charged ions) and anions (negatively charged ions). The geometry for most ionic solids is connected with several factors, such as the dimensions and charges of the involved ions and how these ions can be packed into their most stable arrangement.

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This Story also Contains

1. NaCl (Rock Salt)
2. CsCl
3. ZnS (Zinc-Blende)
4. CaF2 (Fluorite Structure)
5. Na2O (Anti-Fluorite Structure)
6. Some Solved Examples
7. Summary

NaCl (Rock Salt)

• In it CI^- ions have CCP structure that is, FCC and Na⁺ ions occupy octahedral holes (voids) here.
• Both Na⁺ and CI^- have coordination number six. The number of formula units per unit cell is four. For example, halides of alkali metals (except CsX) like LiX, NaX, and KX, other halides like AgCl, AgBr, NH₄Cl oxides like TiO, FeO, NiO.
• r_c + r_a = a/2 and r_Na+/r_cl- must be 0.414 however, it is 0.525.

CsCl

• Here Cl^- ions are at the comers of a cube and Cs⁺ ions are in the cubic void (centered position) i.e. BCC like structure.
• Here coordination number of both Cs⁺ and Cl^- is 8 and here the number of formula units per unit cell is 1. r_c + r_a = √3a/2.
• Here r_Cs⁺/r_Cl⁺ should be 0.732 but it is 0.93, e.g. CsX, TiBr, NH₄Cl, NH₄Br
  
  

ZnS (Zinc-Blende)

• Here S^2- occupies CCP or FCC arrangement while Zn²⁺ ions occupy alternate tetrahedral voids.
• Only half of the total voids are occupied.
• Here coordination number of both Zn²⁺ and S^2- is four and here the number of formula units per unit cell is four. For example, ZnS, CuBr, Cul, Agl, etc.
  r_c + r_a = √3a/4
  

CaF₂ (Fluorite Structure)

• Here Ca²⁺ occupies CCP and F^- ions occupy all the tetrahedral voids.
• Here coordination number of Ca²⁺ is 8 while for F^- it is four and the effective number of Ca²⁺ and F^- ions in a unit cell are 4 and 8 respectively, respectively the number of formula units per unit cell is four.
  r_c + r_a = √3a/4
  

Na₂O (Anti-Fluorite Structure)

• Here negative ions (O^2-) occupy CCP while cations Na⁺ occupy all the tetrahedral voids.
• Here coordination number of Na⁺ is four while for O^2- it is eight, here the number of formula units per unit cell is four, and the effective number of Li⁺ and O^2- ions in a unit cell are 8 and 4 respectively.
  r_c + r_a = √3a/4
  

Normal Spinel Structure (XY₂O₄)

• Spinel is MgAl₂O₄, and spinel, in general, has the formula AB₂O₄. Here A is a divalent cation (Mg²⁺, Ca²⁺) and B is a trivalent cation Al³⁺.
• Here Mg²⁺ occupy 1/8th of tetrahedral voids while oxide ions 1/ 8 of Al³⁺ occupy 1/2 of the octahedral voids.
• As magnetic materials, these are used in telephones, and memory loops of computers.

Inverse Spinel Structure (Fe₃O₄, Magnetite)

• In Fe₃O₄, Fe³⁺ and Fe²⁺ are present in 2:1 ratio.
• Here oxides ions are in CCP while Fe²⁺ ions occupy 1 octahedral void while Fe³⁺ occupies 1 octahedral and 1 tetrahedral void.
• Here O^2- ions form the FCC unit cell.
• Here the formula ratio of
  Fe²⁺ : Fe³⁺ : O^2- = 1 : 2 : 4
• MgFe₂O₄, Pb₃O₄, and Mn₃O₄ have also this type of structure.

Effect of T and P on Crystal Structure

• On increasing the temperature of the CsCl structure coordination number decreases from 8:8 to 6:6.
  $\mathrm{CsCl} \xrightarrow{\Delta} \mathrm{NaCl}$ Structure
  8:8 6:6
• On subjecting the NaCl structure to high pressure it will increase coordination number from 6:6 to 8:8.$\mathrm{NaCl} \xrightarrow{\text { highly pressurized }} \mathrm{CsCl}$ Structure
• 6:6 8:8

Recommended topic video on(Structure of ionic solid)

Some Solved Examples

Example 1: The Ca²⁺ and F^- are located in CaF₂ crystal respectively at face-centred cubic lattice points and in

1) Tetrahedral voids

2)Half of tetrahedral voids

3)Octahedral voids

4)Half of the octahedral voids

Solution

As we have learned,

CaF₂ (Fluorite Structure)

• Here Ca²⁺ occupies CCP and F^- ions occupy all the tetrahedral voids.
• Here coordination number of Ca²⁺ is 8 while for F^- it is four and the effective number of Ca²⁺ and F^- ions in a unit cell are 4 and 8 respectively, here the number of formula units per unit cell is four.
  r_e + r_a = √3a/4
  

-

In $\mathrm{CaF}_2$ crystal $\mathrm{Ca}^{2+}$ occupy all corners and face centre of the cube whereas $F^{-}$ occupy all tetrahedral
voids.

Therefore, Option(1) is correct

Example 2: n antifluorite structure, the negative ions:

1) Are arranged in ccp
2)Occupy tetrahedral voids

3)Are arranged in hcp

4)Occupy octahedral voids

Solution

As we have learned,

Na₂O (Anti-Fluorite Structure)

• Here negative ions (O^2-) occupy CCP while cations Na⁺ occupy all the tetrahedral voids.
• Here co-ordination number of Na' is four while for O^2- it is eight, here number of formula units per unit cell is four and the effective number of Li⁺ and O^2- ions in a unit cell are 8 and 4 respectively.
  r_e + r_a = √3a/4
  

-

In antifluorite crystal i.e Na₂O, the anions are arranged in cubic close packing while the cations occupy all the tetrahedral voids.
Therefore, Option(1) is correct

Example 3: The melting point of RbBr is $682^{\0} \mathrm{C}$, while that of NaF is $988^{\0} \mathrm{C}$. The principal reason that the melting point of NaF is much higher than that of RbBr is that:

1) The internuclear distance $r_c+r_a$ is greater for RbBr than that for NaF

2)These two crystals are not isomorphous

3)The bond in RbBr has more covalent character than the bond in NaF

4)The molar mass of NaF is smaller than that of RbBr

Solution

The internuclear distance is lesser for NaF than RbBr due to which there is a strong electrostatic force of attraction in NaF. Thus, the melting point of NaF is greater than RbBr.
Hence, the answer is the option (1).

Example 4: Ionic radii of cation $\mathrm{A}^{+}$ and anion $\mathrm{B}^{-}$ are 102 and 181ppm respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for $\mathrm{B}^{-} \cdot \mathrm{A}^{+}$ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is ____________ pm. (Nearest Integer)

1) 566

2)666

3)457

4)654

Solution

This crystal has cubic close paking for $\mathrm{B}^{-}$,
Means $\mathrm{B}^{\Theta}$ is present all corner and face centres , and $\mathrm{A}^{+}$ is present in all octahedral voids.


a = edge length of the unit cell of the crystal AB
$\begin{aligned} & \mathrm{a}=2 \mathrm{r}^{-}+2 \mathrm{r}^{+} \\ & \mathrm{a}=2 \times 181+2 \times 102 \\ & \mathrm{a}=362+204 \\ & \mathrm{a}=566 \mathrm{pm}\end{aligned}$
Ans = 566

Example 5: KF has a NaCl structure. What is the distance (in pm) between K⁺ and F^- in KF. If density is 2.48 gm/cm³?

1) 268.8

2)537.5

3)155.3

4)5.375

Solution

Density formula :

$d=\frac{Z \times M}{a^3 \times N_A}$

For NaCl structure Z = 4

M for KF=58

$\begin{aligned} & 2.48=\frac{4 \times 58}{a^3 \times 6.02 \times 10^{23}} \\ & a^3=\frac{4 \times 58}{2.48 \times 6.02 \times 10^{23}}=150 \times 10^{-24} \\ & a=537.5 \mathrm{pm}\end{aligned}$

Rock salt structure or NaCl type structure -

$\mathrm{Cl}^{-}$ located at all corners and face center

$\mathrm{Na}^{+}$ located at all octahedral voids

Edge length $=2\left(r_{N a^{+}}+r_{C l^{-}}\right)$

Coordination number = 6:6
Thus, the distance between K⁺ and F^- is $\frac{a}{2}=268.8 \mathrm{pm}$

Hence, the answer is (268.8 pm).

Summary

Ionic solids are composed of positive and negative ions held together by ionic bonds. The geometrical structure that the ions take in these solids is determined by the charge and size of the ions, thereby giving different stable structures such as simple cubic, body-centered cubic, and face-centered cubic forms, among others. Some important structures include patterns of rock salt, NaCl, and that of cesium chloride, CsCl,