Question

Csbr has cubic structure with edge length 4.3 å. the shortest interionic distance between cs+ and br is

Answer 1

As they are in simple close packing where
${Cs}^{ + }$
is at body center of the unit cell and
${Br}^{-}$
occupy lattice sites which is corners of unit cell
So 2(r(Cs+) +r(Br-)) =
$\sqrt{3} a$
here a = 4.3*10^(-10) m
and we know that

(Cs+) +r(Br)
$\frac{r({Cs}^{ + })}{r({Br}^{ - } )} = 0.94$
so we can easily find the shortest interionic distance which is
r(Cs+) +r(Br-)




I HOPE THIS WILL HELP YOU

Answer 2

shortest interionic distance between Cs+ and Br- is 3.72 A°

CsBr has body - centered cubic lattice structure in which Cs+ ions are located at corner of lattice and Br- ions are located at the centre of cubic lattice.

so, diagonal of cubic lattice = √3 × edge length of cubic lattice

⇒2r_(Cs+) + 2r_(Br-) = √3 × 4.3 A°

⇒r_(Cr+) + r_(Br-) = √3 × 4.3/2 = 1.732 × 4.3/2 A°

= 0.866 × 4.3 A°

= 3.7238 A° ≈ 3.72 A°

hence shortest interionic distance between Cs+ and Br- is 3.72 A°

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