The Solid State

We know that two Na⁺ ions are replaced by each of the Sr²⁺ ions while SrCl₂ is doped with NaCl. But in this case, only one lattice point is occupied by each of the Sr²⁺ ions and produce one cation vacancy.

Here 10 ^{– 3} mole of SrCl₂ is doped with 100 moles of NaCl Thus, cation vacancies produced by NaCl = 10 ^{– 3} mol Since, 100 moles of NaCl produces cation vacancies after doping = 10 ^–3 mol

Therefore, 1 mole of NaCl will produce cation vacancies after doping
= 
therefore, total cationic vacancies
=10^-5 x Avogadro's number
=10^-5  x 6.023 x 10²³
=6.023 x 10^-18 vacancies
 

given:
Radius = 181 pm
thus,
Radius of octahedral void = 0.414 r = 0.414 x 181 pm = 74.934 pm
Cation having radius 74.934 pm will just fit into octahedral voids.

Radius of tetrahedral void
= 0.225 r = 0.225 x 181 pm = 40.725 pm
Cation having radius 40.725 pm will just fit into tetrahedral void.

solution:
we have given that
z = 2 
a = 300 pm
d = 5.2 gm cm^–3
by using the given formula we obtaine mass of element i.e.           

Solution:
We  have given
volume of the unit cell = (5 x 10^-8cm)³ = 1.25 x 10^-22cm³

Density of FeO = 4g cm^-3

Density,

$$\begin{gathered} \mathrm{\rho } = \frac{\mathrm{z}\times \mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}} \\ 4= \frac{\mathrm{z}\times 72}{(5\times {10}^{-8}{)}^3\times 6.02\times {10}^{23}} \\ \mathrm{z}= \frac{4\times 1.25\times {10}^{-22} \times 6.02\times {10}^{23}}{72} = 4.18\cong 4 \end{gathered}$$

Each unit cell has four units of FeO. So it has four Fe²⁺ and four O^2– ions.