Ionic Solids

Radius Ratio

Radius ratios: Generally ionic solids are more compact because voids are also occupied by ions (generally smaller ions). Pattern of arrangements and type of voids both depend upon relative size (ionic size) of cations and anions e.g when r⁺ = r^- the most probable and favourable arrangement will be BCC.

With the help of relative ionic radii (also called radius ratio) we can predict the most probable arrangement of ions,

From the expression it is obvious that larger the radius ratio larger the size of cation and would require to be surrounded by more number of anions i.e. higher co-ordination number (C.N.).
 

Radius Ratio	Voids	Coordination no. and no. of voids per unit cell	Location of cation	Example
0.155 - 0.225	Triangular void	3	At 2D plane in void	B203
0.225 - 0.414	Tetrahedral voids	4 *No. of Tetrahedral void = 2N	In Tetrahedral void of FCC unit cell	ZnS
0.414 - 0.732	Octahedral void	6 *No. of Octahedral void = N	In Octahedral void of FCC	NaCl
0.732 - 0.999	Cubic void	8	In cubic void of simple cube	Cscl

*N = Number of atoms per unit cell.
Note: When r⁺ > r^- location of cation and anion are exchanged. Cation constitute lattice and anion goes to voids.

Derivation of 

(i) For tetrahedral voids (T-voids)
Cation is located at centre of the cube
For limiting case,
r^- + r⁺ = 
and 2r^- = a√2

(ii) For octahedral voids (0-voids)
Applying Pythagoras Theorem on ΔABC, we get,

  = 2 r^- + 2r⁺

 = 0.414

(iii) For cubic void

2r^- = a

and 2r^- + 2r⁺ = √3 a
1 +  = √3

  = 0.732

1. Crystal structure of AB type solids

A. NaCl structure (rock salt structure)
where 0.414 ≤  < 0.732
e.g. NaCl, LiCl, AgBr, KBr, Rbl

B. CsCl structure
where, 0.732 ≤  < 1
e.g. CsBr, Csl, TlCl, Tli, TlCN etc.

C. ZnS (Zinc blende) structure
where, 0.225 ≤  < 0.414
Example: CuCl, CuBr, Cul, Agl, BeS

Features of Ionic Solids (AB type)
 

S.No.	Property	Rock salt structure	Zinc blende structure	CsCl structure
1.	Radius ratio
2.	Position of anion	Cl- forms CCP in FCC unit cell	S2- forms CCP in FCC	Cle forms PUC
3.	Position of cations	Na+ occupy all octahedral cell	Zn2+ occupy alternate Tetrahedral holes	Cs+ occupy the cubic void
4.	Coordination No. of A+ and B- (co. system)	6 : 6	4 : 4	8 : 8
5.	No of A+ and B- per unit cell	4 CIe, 4N+	4S2- 4 Zn2+	One Cle, one Cs+
6.	Formula of unit cell	Na4Cl4	Zn4S4	CsCl
7.	No. of (Z) formula unit per cell	Four	Four	One

2. AB₂ and A₂B type

A. Fluorite structure (CaF₂)
(i) Ca²⁺ ions are in CCP
(ii) F^e occupies all the tetrahedral voids
(iii) C. N. of F^e = 4; C. N. of Ca²⁺ = 8
(iv) No. of Ca²⁺ and F^e in unit cell = 4 and 8.
(v) i.e. No. of CaF₂ molecule in one unit cell (Z) = 4
(vi) Co. system = 8 : 4
(vii) Other example BaF₂, BaCl₂, CaCl₂

B. Antifluorite structure (Na₂O)
(i) 0^2- ions are in CCP
(ii) Na⁺ occupies all the tetrahedral voids
(iii) Co. No. of Na⁺ = 4, Co. No. ofO^2- = 8
(iv) Co system = 4 : 8
(v) No of Na⁺ and O^2- in one unit cell = 8 and 4 respectively i .e. 4 molecules of Na₂O per unit cell

3. Spine Structure (AB₂O₄), A²⁺, B³⁺, O^2-

(i) O^2- forms FCC

(ii) A²⁺ ion occupy 1/8 of the Tetrahedral voids (8 Tetrahedral voids) = One A²⁺

(iii) B³⁺ ion occupy 1/2 of the Octahedral voids (4 Octrahedral voids) = Two B³⁺

Typical Example: MgAl₂O₄, ZnAl₂O₄, Mgln₂O₄

4. Inverse spinel structure (Fe3O4):

(i) O^2- ions are in CCP.

(ii) Fe³⁺ ion occupy of the Tetrahedral voids (8 Tetrahedral voids) one Fe³⁺

(iii) Fe²⁺ ion occupy 1/4 of the Octahedral voids (4 Octahedral voids) = one Fe²⁺

(iv) Fe³⁺ ion occupy 1/4 of the Octahedral voids (4 Octahedral voids) = one Fe³⁺

Thus, formula ratio of Fe²⁺ : Fe³⁺ : O^2- = 1 : 2 : 4
Other Examples : Mn₃O₄, Pb₃O₄

Effect of Temperature and Pressure on Crystal Structure

1. On rising temperature of CsCI structure C. N. decrease from 8 : 8 to 6 : 6

2. On subjecting the NaCl structure to high pressure it would increase C. N. from 6 : 6 to 8 : 8