Solid State

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 $=\frac{3}{7}*100\mathrm{\%}=43\mathrm{\%}$ $\frac{}{}$

$E_{S{n}^{2+}/Sn}^0=-0.140V=E_2^0$

$E_{S{n}^{4+}/Sn}^0=+0.010V=E_3^0$

Here, $\Delta G_3^0=\Delta G_1^0+\Delta G_2^0$

$-4F{E}_3^0=-2F{E}_1^0-2F{E}_2^0$

$E_3^0=\frac{E_1^0+E_2^0}{2}$

$+0.010=\frac{E_1^0+\left(-0.140\right)}{2}$

$E_1^0=E_{S{n}^{4+}/S{n}^{2+}}^0=+0.16$

= 16 * 10^-2 V

H₂ (g) + Cu²⁺ (aq)   $\stackrel{}{\to }$ 2H⁺ (aq) + Cu (s)

  $\therefore E_{cell}=E_{cell}^0-\frac{2.303RT}{nf}logQ$              

              0.31 = 0.34 -   $\frac{0.06}{2}log\frac{\left[H^{+}\right]}{C{u}^{2+}}$

$\left[C{u}^{2+}\right]=10^{-7}\left(?\left[H^{+}\right]=10^{-3}\right)$        

              x = 7

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100}*t=2.303lo{g}_{10}\frac{A_0}{A_t}$ ………….(i)

and $\frac{0.693}{50}*t=2.303lo{g}_{10}\frac{B_0}{B_t}$ ………….(ii)

from equation (i) and (ii)

$0.693t\left[\frac{1}{50}-\frac{1}{100}\right]=\left[log\frac{B_0}{B_t}-log\frac{A_0}{A_t}\right]*2.303$

Here; A0 = B0 and $A_t=4*B_t$

Therefore $0.693t\left[\frac{1}{100}\right]=2.303\left[log\left(\frac{B_0}{B_t}*\frac{A_t}{A_0}\right)\right]$

$\therefore t=\frac{2.303*0.3010*2*100}{693}=200s$