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
for a cube of side length 4.13A0 diagonal
=
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- =3.57A0
such that distance between neighbouring Cs+ and Cl- =3.57A0
now assume two Cl- ion touch with each other so length of unit cell = 2r- =4.13
r- =2.06A0
r+ =3.57 -2.06 =1.51
r+/r-=1.51/2.06 =0.73
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1/3rd of tetrahedral voids?
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An element with density 11.2 g cm-3 forms a f.c.c. lattice with edge length of 4 x10-8
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