Atomic radius for a cubic closest-packed metal.

Question

Question

what is the minimum atomic radius required for a cubic closest-packed metal to accommodate boron atoms (radius 88 pm) in its octahedral holes?

Answer 1

Answer #1

size of octahedral hole a =0.414r
r= redius of metal
a=88pm =radius of boron

min. radius of metal r=212.56pm

answered by: malisa

Answer 2

Answer #2

There is a simple relation between octahedral void and radius of the atom that is,

The radius of atom = 0.732 X radius of octahedral void

So, Radius of octahedral void required = 88 / 0.732 = 120.22 ppm

answered by: Amypp

Answer 3

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