Question 149

Solution

We know that two Na^{+} ions are replaced by each of the Sr^{2+} ions while SrCl_{2} is doped with NaCl. But in this case, only one lattice point is occupied by each of the Sr^{2+} ions and produce one cation vacancy.

Here 10 ^{– 3} mole of SrCl_{2} 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 10^{23}

=6.023 x 10^{-18} vacancies

Question 150

(i) Calculate the radius of the cation that just fits into the octahedral holes of this lattice of anions. (ii) Calculate the radius of the cation that just fits into the tetrahedral holes of this lattice of anions.

Solution

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.

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.

Question 151

Solution

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.

we have given that

z = 2

a = 300 pm

d = 5.2 gm cm

by using the given formula we obtaine mass of element i.e.

Question 152

Solution

Solution:

We have given

volume of the unit cell = (5 x 10^{-8}cm)^{3} = 1.25 x 10^{-22}cm^{3}

Density of FeO = 4g cm^{-3}

Density,

$\mathrm{\rho}=\frac{\mathrm{z}\times \mathrm{M}}{{\mathrm{a}}^{3}\times {\mathrm{N}}_{\mathrm{A}}}\phantom{\rule{0ex}{0ex}}4=\frac{\mathrm{z}\times 72}{(5\times {10}^{-8}{)}^{3}\times 6.02\times {10}^{23}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{z}=\frac{4\times 1.25\times {10}^{-22}\times 6.02\times {10}^{23}}{72}=4.18\cong 4$

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