An ionic compound is made up …

Number of atoms of A per unit cell = contribution from corners + contribution from alternate face centres (these are 2) =  1/8 X 8 + 1/2 X 2 = 2

Number of atoms of B per unit cell = contribution from body centre + contribution from edge centres

= 1 + 1\4 X 12 = 4

> Formula of compound = A₂B₄

or exactly it is  AB₂

_{Since charge on B is -1 therefore Charge on A will be +2}

A occupy all the corners and alternate face centres.

Therefore, No. of atoms of A = {tex}(8\times {1\over 8} )+ (3\times {1\over 2}) = 1 + {3\over 2}{/tex}

= {tex}5\over 2{/tex}

Now, B occupy all the body centre and the edge centre.

So, No. of atoms of B : {tex}1+({1\over 4}\times 12) = 1+3=4{/tex}

Therefore, formula of compound will be

{tex}A_{5\over 2} B_4 = A_5B_8{/tex}

Charge on B = -1

Let charge on A = x

Then

5x +8(-1) = 0

=> 5x -8=0

=> 5x = 8

=> x = {tex}8\over 5{/tex}