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Byju's Answer

Standard XII

Chemistry

Density

CsCl has cubi...

Question

$C s C l$ has cubic structure of ions in which $C s^{+}$ ions is present in the body-centre of the cube. If density is $3.99 g c m^{- 3}$ .
(a) Calculate the length of the edge of a unit cell.
(b) What is the distance between $C s^{+}$ and $C l^{-}$ ions?
(c) What is the radius of $C s^{+}$ ion if the radius of $C l^{-}$ ion is $180 p m$ ?

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Solution

We know that,
(a) $Z = \frac{a^{3} \times ρ \times N}{M}$
$Z = 1$
$M = 168.5$
$1 = \frac{a^{3} \times 3.99 \times 6.023 \times 10^{23}}{168.5}$
$a = 4.123 \times 10^{- 8} c m = 412 p m$

(b) Distance between $C s^{+}$ and $C l^{-}$ ions
$= \frac{a \sqrt{3}}{2} = \frac{412 \times \sqrt{3}}{2} = 356.8 p m$

(c) $r_{C s^{+}} + r_{C l^{-}} = 356.8$
$r_{C s^{+}} + 180 = 356.8$
$r_{C s^{+}} = 176.8 p m$

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CsCl has cubic structure of ions in which Cs+ ions is present in the body-centre of the cube. If density is 3.99 g cm−3.(a) Calculate the length of the edge of a unit cell.(b) What is the distance between Cs+ and Cl− ions?(c) What is the radius of Cs+ ion if the radius of Cl− ion is 180 pm?

We know that,(a) Z=a3×ρ×NMZ=1M=168.51=a3×3.99×6.023×1023168.5a=4.123×10−8cm=412 pm(b) Distance between Cs+ and Cl− ions=a√32=412×√32=356.8 pm(c) rCs++rCl−=356.8rCs++180=356.8rCs+=176.8 pm