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
                                         
 

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