Solid State

In Frenkel defect, cations are inserted in neighbouring site, therefore statement-I is correct, because vacancy developed by missing cations and interstitial sites are equal. Statement II is false because F-centre developed by metal excess defect due to anion vacancies.

$B{i}_2{S}_3?{\underset{2S}{2Bi}}^{+3}+{\underset{3S}{3S}}^{-2}$

$Ksp={\left(2S\right)}^2{\left(3S\right)}^3=4*27*{\left(S\right)}^5$

= 108 (S)5

${\left(S\right)}^5=\frac{1.08*10^{-73}}{108}=10^{-75}$ $\therefore S=10^{-15}M$

The interaction energy of London forces ; $E\propto \frac{1}{r^6}orE\propto r^{-6}$ . Hence x = -6.

$B{i}_2{S}_3?{\underset{2S}{2Bi}}^{+3}+{\underset{3S}{3S}}^{-2}$

$Ksp={\left(2S\right)}^2{\left(3S\right)}^3=4*27*{\left(S\right)}^5$

= 108 (S)5

${\left(S\right)}^5=\frac{1.08*10^{-73}}{108}=10^{-75}$ $\therefore S=10^{-15}M$

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x*\left(+2\right)+\left(1-x\right)*\left(+3\right)=$ O.S. of Fe atom

$or2x+3-3x=\frac{+200}{93}$

$\therefore \mathrm{\%}ofF{e}^{2+}=85\mathrm{\%}and\mathrm{\%}ofF{e}^{3+}=15\mathrm{\%}$

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x*\left(+2\right)+\left(1?x\right)*\left(+3\right)=$ O.S. of Fe atom

$or\text{?}\text{?}2x+3?3x=\frac{+200}{93}$

$?\mathrm{\%}\text{?}\text{?}of\text{?}\text{?}F{e}^{2+}=85\mathrm{\%}\text{?}\text{?}and\text{?}\text{?}\mathrm{\%}\text{?}\text{?}of\text{?}\text{?}F{e}^{3+}=15\mathrm{\%}$

$d=\frac{ZM}{N_A{a}^3}=\frac{4*58.5}{6.02*10^{23}*a^3}$

$43.1=\frac{4*58.5}{6.02*10^{23}*a^3}$

$\therefore a=2.08*10^{-8}cm$

$\therefore r_{N{a}^{+}}+r_{C{l}^{-}}=1.04*10^{-10}$ m

B.C.C structure

a = 300 pm = 300 * 10^-12 m

d = 6g/cm³

z = 2

$d=\frac{Z*M}{a^3}$            

$6=\frac{2*A}{{\left(300*10^{-10}\right)}^3}=\frac{2*A}{27*10^{-24}}$            

$\therefore AtomsofM$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

Unit cell = Hexagonal close packing (hcp)

No. of X in one unit cell = 6

No. of tetrahedral voids in one unit cell = 12

No. of Y in one unit cell = $\frac{2}{3}*12=8$ $\frac{}{}$

Formula => X₆ Y₈ => X₃ Y₄

% of X in unit cell $=\frac{3}{7}*100\mathrm{\%}=43\mathrm{\%}$ $\frac{}{}$