An ionic crystal AB crystallises in bcc lattice. If the cationic radius is 1.7 Å and anionic radius is 1.8 Å,then find out the density of crystal AB (formula mass of AB is 168.5 g/mol; )
Answers
The density of the crystal is ρ = 4.238 g/cc
Explanation:
Given data:
- Radius of cation = 1.7 Å
- Radius of anion = 1.8 Å
- Formula mass of AB = 168.5 g/mol
Solution:
0.732 < r+ / r- < 1
- r+ = cation (cubical void)
- Anion = simple cubic
r^- + 2(r^)+ + r^- = √ 3 a
r^+ + r^- = √ 3 a / 2
1.7 + 1.8 = √ 3a / 2
√ 3a / 2 = 3.5
a = 7 / 3 Å = 7 / 3 x 10^-10 m
a = 7 /3 x 10^-8 cm
(a)^3 = (√ 7 /3 )^3 x 10^-24 cc
(a)^3 = 66 x 10^-24 cc
ρ = z x mω / a^3 x NA
ρ = 1 x 168.5 / 66 x 10^-24 x 6.022 x 10^-23
ρ = 4.238 g/cc
Thus the density of the crystal is ρ = 4.238 g/cc
Answer :-
The density of the crystal is ρ = 4.238 g/cc
Explanation:
Given data:
Radius of cation = 1.7 Å
Radius of anion = 1.8 Å
Formula mass of AB = 168.5 g/mol
Solution:
0.732 < r+ / r- < 1
r+ = cation (cubical void)
Anion = simple cubic
r^- + 2(r^)+ + r^- = √ 3 a
r^+ + r^- = √ 3 a / 2
1.7 + 1.8 = √ 3a / 2
√ 3a / 2 = 3.5
a = 7 / 3 Å = 7 / 3 x 10^-10 m
a = 7 /3 x 10^-8 cm
(a)^3 = (√ 7 /3 )^3 x 10^-24 cc
(a)^3 = 66 x 10^-24 cc
ρ = z x mω / a^3 x NA
ρ = 1 x 168.5 / 66 x 10^-24 x 6.022 x 10^-23
ρ = 4.238 g/cc
Thus the density of the crystal is ρ = 4.238 g/cc