The Solid State

Chemistry I

Chemistry

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An element X with an atomic mass of 60 g/mol has density of 6.23 g/cm^{–3}. If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this element.

Solution:

we have given that

Density, d = 6.23 g/cm

M = 60g/mol

Volume = (a

$\mathrm{Z}=\frac{\mathrm{d}\times {\mathrm{a}}^{3}\times {\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}$

$\frac{{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}=\frac{6.023\times {10}^{23}{\mathrm{mol}}^{-1}}{60\mathrm{g}/\mathrm{mol}}\phantom{\rule{0ex}{0ex}}$

$\mathrm{Z}=\frac{6.023\mathrm{g}{\mathrm{cm}}^{-3}\times (400{)}^{3}\times {10}^{-30}{\mathrm{cm}}^{3}\times 6.023\times {10}^{23}{\mathrm{mol}}^{-1}}{60\mathrm{g}{\mathrm{mol}}^{-1}}\phantom{\rule{0ex}{0ex}}=\frac{6.023\times 64\times 6.023}{600}\phantom{\rule{0ex}{0ex}}=\frac{2401.49}{600}=4.$

Hence, the type of cubic unit cell is FCC

$\mathrm{Radius}=\frac{\mathrm{a}}{\mathrm{a}\sqrt{2}}=\frac{400}{2\sqrt{2}}=\frac{200}{\sqrt{2}}$

$=\frac{200}{1.414}=141.4\mathrm{pm}.$

So radius of element is 141.4 pm

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Refractive index of a solid which have the same value along all directions are isotropic in nature. It would not show cleavage property.

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