Question

A solid crystal is composed of X, Y and Z atoms. Y atom is occupying 50% of octahedral void, where as X atoms are occupying the 100 % tetrahedral void with Z atoms in ccp arrangement, then the simplest formula of the compound in the given crystal is
A. ${X_8}{Y_2}{Z_4}$
B. ${X_5}{Y_{10}}{Z_8}$
C. ${X_4}Y{Z_2}$
D. ${X_{16}}{Y_4}{Z_8}$

Answer 1

The relation between the tetrahedral void and octahedral void is that the tetrahedral void is two times of octahedral void. The simplest formula of the compound in the given crystal is determined by calculating the ratio of X:Y:Z.

Complete step by step answer:
Given,
Y atoms occupy 50 % of the octahedral void.
X atoms occupy 100 % of the tetrahedral void.
Z atom is present in CCP arrangement.
The relation between tetrahedral void and octahedral void is shown below.
T = 2O
Where.
T is tetrahedral
O is octahedral
The tetrahedral void present is 8 so the octahedral void present is 4.
$50\% \;of\;4 \Rightarrow \dfrac{4}{{100}} \times 50$
$50\% \;of\;4 \Rightarrow 2$
$100\% \;of\;8 = \dfrac{8}{{100}} \times 100$
$100\% \;of\;8 = 8$
Y atoms present in an octahedral void is 2.
X atoms present in the tetrahedral void is 8.
In cubic close packing, arrangement of atoms is in corners and in centre.
The atoms present in the corners of a cubic close packing structure is 8 and the contribution of one atom is $\dfrac {1}{8}$. The atoms present at the centre is 6 and the contribution is $\dfrac{1}{2}$.
Thus, the calculation for Z atoms present in cubic close packing is done as shown below.
$Z = 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2}$
$\Rightarrow Z = 4$
Thus, the ratio of X:Y:Z is 8:2:4.
Therefore, the simplest formula of the compound in the given crystal is ${X_8}{Y_2}{Z_4}$.

So, the correct answer is Option A.

Note: The cubic close packing and face centered cubic cell are the different names given form the same lattice structure. The arrangement in the face centered unit cell is ABCABCA as three layers present.