A solid has a structure in which X atoms are located at cubic corners of a unit cell, O atoms are at the edge centres and Y atoms at cube centre. Then the formula of a compound is $X_{a} Y_{b} O_{c}$ . If two atoms of are missing from any of two edge centres per unit cell, then the molecular formula is $X_{x} Y_{y} O_{z}$ . Then, find the value of $(x + y + z) - (a + b + c)$ .

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Solution

A. Number of $X$ atoms = $8 \times \frac{1}{8} = 1 / u n i t c e l l .$
Number of $Y$ atoms = 1 unit cell
Number of atoms $= 12 \times \frac{1}{4} = 3 / u n i t c e l l$
Formula is: $X Y O_{3} \Rightarrow X_{a} Y_{b} O_{c}$
B. Number of atoms missing from two edge centers
Per unit cell $= 2 \times \frac{1}{4} = \frac{1}{2} / u n i t c e l l$
Number of atoms left $= 3 - \frac{1}{2} = 2.5 / u n i t c e l l$
$∴$ The value of $(x + y + z) - (a + b + c) = (2 + 2 + 5) - (1 + 1 + 3) = 4$

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