An ionic compound is made up of A carions and B anions.if A cations are present at the alternate corners and B anions is present on the body of the diagonal...what is compound's fromula
Dear student,
In a unit cell,
according to the question, No. of A cations present at alternate corners =4 (since no. of corners in a unit cell is 8)
Each corner contribution = 1\8
Total no. of cations = 1\8 4 = 0.5
No. of B anions present at body diagonal = 1
Here the anion contribution = 1
The ratio of no. of A cations to the no. of B anions = 0.5 : 1 = 1 : 2
Therefore compound formula is AB2.
Regards
In a unit cell,
according to the question, No. of A cations present at alternate corners =4 (since no. of corners in a unit cell is 8)
Each corner contribution = 1\8
Total no. of cations = 1\8 4 = 0.5
No. of B anions present at body diagonal = 1
Here the anion contribution = 1
The ratio of no. of A cations to the no. of B anions = 0.5 : 1 = 1 : 2
Therefore compound formula is AB2.
Regards