A compound formed by elements `X` and `Y` has a cubic structure in which `X` atoms are at the corner of the cube and `Y` atoms are at the face centres. One atoms `X` is misssing from the corner.
a. Calculate (i). `Z_(eff')` (ii). total number of the atoms in the cube, and (iii). formula of the compound.
b. If alll the atoms are removed from one of the faces of the cube containing atoms at corners, as in (a) above, calculate (i) `Z_(eff')` (ii) total number of atoms in a cube, and (iii) formula of the compound.

Explanation

a. i. Number of effective `X` atoms
`= 7` corners ` xx (1)/(8)` per corner atom share
`= (7)/(8)` atoms/unit cell
Number of effective `Y` atoms `= 6` faces `xx (1)/(2)` per face atoms share
`= 3` atoms/units cell
`Z_(eff(X + Y)) = (7)/(8) + (3)/(1) = (31)/(8)`
ii. Total atoms in a cube `= 7 ("corners") + 6` faces
`= 13` atoms units cube
iii. Formula of the compound:
`Z_(eff(X)) = (7)/(8), Z_(eff(Y)) = 3`
Thus, formula of the compound is `X_((7)/(8)) Y_(3)`.
Simplifying: `X_(7) Y_(24)`.
b. i. Atoms removed from one face of the cube, containing all atoms at corners `= 4` corners `+ 1` face-centred atom
`= 5` atoms/units cell
Corner atom left `= 7 - 4 = 3` atoms at corner
Face-centre atom left `= 6 - 1 = 5` atoms at face centre.
`Z_(eff(X)) = 3` corners ` xx (1)/(8)` per corner atom share
`= (3)/(8)` atoms/unit cell
`= Z_(eff(Y)) = 5` face-centred atom `xx (1)/(2)` per face atom share
`=(5)/(2)` atoms/unit cell
ii. Total atoms in a cube `= 3("corner") + 5` (faces)
`= 8` atoms/unit cube
iii. Formula of compound:
`Z_(eff(X)) = (3)/(8), Z_(eff(Y)) = (5)/(2)`.
Formula: `X_((3)/(8)) Y_(5)/(2)`
Simplifying `= X_(3) Y_(20)`