Question

please guys...I have an exam tomorrow.

Answer 1

hello !

A forms FCC thus effective no. of atoms of A are 1/8×8+1/2×6=4

B is situated at alternate TD voids .
means effective no. of B atoms are 4
means 1 on each diagonal(alternate)sice there are 8 TD voids

now if we remove all atoms on 1 diagonal then then no of atoms decrease is one
hence no. of effective atoms of B is 3

thus Required formula of the crystal is A4B3.

hence you will like the answer.