Question

A solid crystal is composed of $X,Y$ and $Z$ atoms. $Z$ atoms (radius $200\text{pm}$) form cubic close packed structure, whereas $X$ atoms (radius $50\text{pm}$) and $Y$ atoms (radius $100\text{pm}$) occupy the respective voids. Assume all the voids to be occupied. If all the atoms along one body diagonal are removed, then the formula of the compound becomes:

• A. $X_8{Y}_4{Z}_5$
• B. $X_{4}Y{Z}_2$
• C. $X_{2}YZ$
• D. $X_6{Y}_3{Z}_4$

Answer 1

The correct option is A $X_8{Y}_4{Z}_5$

In the original compound, $Z$ atoms are arranged in CCP structure.
So, in the compound, octahedral and tetrahedral voids are present.
$\frac{r_X}{r_Z}=\frac{50}{200}=0.25$
As the radius ratio falls in the range of $0.225-0.414$, $X$ atoms are present in the tetrahedral voids.

$\frac{r_Y}{r_Z}=\frac{100}{200}=0.5$
As the radius ratio falls in the range of $0.414-0.732$, $Y$ atoms are present in the octahedral voids.

Assuming all the voids to be occupied, the formula of the original compound $=X_8{Y}_4{Z}_4$.
On removing atoms along one body diagonal,
we lose $2Z$ atoms from $2$ corners.
Now, the effective number of $Z$ atoms $=3+6\times \frac{1}{8}=\frac{30}{8}$
We lose $2X$ atoms from $2$ tetrahedral voids and $1Y$ atom from the body centre.

$\therefore$ The new formula becomes $X_6{Y}_3{Z}_{\frac{30}{8}}$
Hence, the simplest formula is $X_8{Y}_4{Z}_5$.