The Solid State

Chemistry I

Chemistry

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In the cubic crystal of CsCl (d = 3.97 g cm^{–3}), the eight corners are occupied by CI^{–}with a Cs^{+} at the centre and vice versa. Calculate the distance between the neighbouring Cs^{+} and CI^{–} ions. What is the radius ratio of the two ions ? (At. mass of Cs = 132.91 and CI = 35.45).

In a unit cell there are one Cs and 1 x 8/8 =1 chlorine (Cl^{-}) such that one CsCl molecule

therefore

As we have given

density = 3.97 g cm^{-3}

Mass of CsCl = 168.36g

Number of unit cell(Z) = 1

$\mathrm{apply}\mathrm{formula}\phantom{\rule{0ex}{0ex}}\mathrm{d}=\frac{\mathrm{Z}\mathrm{xM}}{{\mathrm{a}}^{3}\mathrm{x}{\mathrm{N}}_{\mathrm{A}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3.97=\frac{1\mathrm{x}168.36}{{\mathrm{a}}^{3}\mathrm{x}6.023\mathrm{x}{10}^{23}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{a}=4.13\mathrm{x}{10}^{-8}\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{a}=4.13{\mathrm{A}}^{0}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

for a cube of side length 4.13A^{0} diagonal

=$\sqrt{3\mathrm{a}}i.e.,\phantom{\rule{0ex}{0ex}}\sqrt{3x4.13}=7.15{A}^{0}\phantom{\rule{0ex}{0ex}}$

as it is a BCC with Cs^{+} at centre radius r^{+} and Cl^{-} at corner radius r^{-} so,

2r^{+} +2r^{-} =7.15 or r^{+} +r^{-} =3.57A^{0}

such that distance between neighbouring Cs^{+} and Cl^{-} =3.57A^{0}

now assume two Cl^{-} ion touch with each other so length of unit cell = 2r^{-} =4.13

r^{-} =2.06A^{0}

r^{+} =3.57 -2.06 =1.51

r+/r-=1.51/2.06 =0.73

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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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