A crystal is made of particles X and Y. X forms fcc packing and Y occupies all the octahedral voids. If all the particles along one body diagonal are removed then the formula of the crystal would be
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53
Hi there
The rank of FCC is Z=4
The no of tetrahedral void=2Z=8
The no of octahedral void=Z=4
As X forms FCC packing,X=4
and Y occupies all the octahedral voids,Y=4
Now all the atoms along 1 body diagonal are removed then,from 2 corners 1/8 X atoms are removed and from the Octahedral void at the centre 1 Y atom is removed.
Now,final X=4-2(1/8)=15/4
and, final Y=4-1=3
so the formula vis X(15/4)Y3 or
or X15Y12 Answer
Hope it helps.
The rank of FCC is Z=4
The no of tetrahedral void=2Z=8
The no of octahedral void=Z=4
As X forms FCC packing,X=4
and Y occupies all the octahedral voids,Y=4
Now all the atoms along 1 body diagonal are removed then,from 2 corners 1/8 X atoms are removed and from the Octahedral void at the centre 1 Y atom is removed.
Now,final X=4-2(1/8)=15/4
and, final Y=4-1=3
so the formula vis X(15/4)Y3 or
or X15Y12 Answer
Hope it helps.
Answered by
6
Answer:
Explanation:there is a very simple funda to solve these kinds of question that u have to simply have to know that tetrahedral and octahedral voids are located on body diagonals. The no. Of tetrahedral voids in one body diagonal is 2 and the no. Of octahedral void is 1 so u can use these thing and can solve ur Q easily . also as body diagonal also contain two corner atoms so dont forget them.
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