Question

a solid AB has ceasium chloride structure type the edge lenght of unit cell. is 404pm calculate tye distance of closet approach between cs and cl ion​

Answer 1

Answer:

In BCC lattice structure,

3

a=2(r

+

+r

−

)

∴(r

+

+r

−

)=

2

3

×4.04

=3.5

A

˚

Answer 2

Answer:

hm koini ki ptaa tkrr jie..Baki please gusa na kri Mai vese v pkki fdk to ni a Mai chahla pind a utho krdi a belong