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Solid State — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistrySolid State3 Marks
Question
Determine the density of Cesium chloride which crystallizes in $\text{BCC}$ type structure with the edge length $412. 1 \ pm.$ The atomic masses of $Cs$ and $C1$ are $133$ and $35.5$ respectively. Predict the co$-$ordination number of $Cs^+$ ion if $r_{cs+} = 1.69 A$ and $= r_{CI}-= 1.81 A.$
✓

Answer

Given: Edge length $(a) = 412.1 \ pm = 4.12 x 10^{-8} \ cm$
Molar mass $= 133 + 35.5 = 168.5 \ g \ mol^{-1}$
To find: Density $(d)$
Formulae: Mass of one molec $=$ Molar mass/Avogadro number
Volume of unit cell $= a^3$
Density $=$ Massof unit cell/Volumeof unit cell
Calculation: In the bcc type unit cell of $CsCl,$ there is one $Cs^+$ ion at the body centre
position and $8 Cl^-$ ions are at the $8$ corners
$\therefore$ Number of $Cs^+$ in unit cell $= 1$
Number of $Cl^-$ in unit cell $= 1/8 \times 8 = 1$
Hence, the unit cell contains one $CsCl$ molecule
Mass of one $CsCl$ molecule $=$ Molar mass/Avogadro number $= 168.5 \ g \ mol^{-1}/6.023 \times 10^{23} mol^{-1}$
$= 2.798 \times 10^{-22} g$
$\therefore$ Mass of unit cell $= 1 \times 2.798 \times 10^{-22}g = 2.798 \times 10^{-22}g$
Volume of unit cell $= a^3 = (4.12 \times 10^{–8}cm)^3= 6.993 \times 10^{-23}cm^3$
$\therefore$ Density $=$ Massof unit cell/Volumeof unit cell $= 2.798 10^{-22}/6.993 \times 10^{-23} \ cm^3 = 4.0 g \ cm^{–3}$
The radius ratio $(r^+/r^–)$ defines the coordination number of the cation
$r_{C s^{+}}=1.69 \mathring A , r_{C l^{-}}=1.81 \mathring A $
$\therefore \frac{r_{C s^{+}}}{r_{C l^{-}}}=\frac{1.69}{1.89}=0.9337$
Since, radius ratio is greater than $0.732,$ the coordination number of cation $(Cs^+)$ is $8.$


 

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