Question

In a compound, atoms of element $Y$ form $ccp$ lattice and those of elements $X$ occupy $\dfrac{2}{3}{\text{rd}}$ of tetrahedral voids. The formula of the compound will be:
A.) ${X_4}{Y_3}$
B.) ${X_2}{Y_3}$
C.) ${X_2}Y$
D.) ${X_3}{Y_4}$

Answer 1

The total number of atoms for the $ccp$ lattice is four and the total number of tetrahedral voids can be given as twice the total number of atoms. The formula of the compound can be given by finding the ratio of the number of atoms in both the elements of the compound.

Complete step by step answer:
In this question, cubic close packing ($ccp$) lattice is that lattice in which the specific pattern of layers can be written as $ABCABC$ (If we consider the first layer as ‘$A$’, second layer as ‘$B$’ and third layer as ‘$C$’ ). This type of crystal pattern is also called face-centered cubic ($fcc$) In it, atoms are present at all the corners of the cube and at all the faces of the cube. Therefore, the total number of atoms in cubic closed packing ($ccp$) or face-centered cubic ($fcc$) can be given as:

Total atoms in $ccp = 8{\text{ corners}} \times \left( {\dfrac{1}{8}{\text{ per corner atom}}} \right) + 6{\text{ faces}} \times \left( {\dfrac{1}{2}{\text{ per face - centered atom}}} \right)$
$= 1 + 3 = 4$
Also, tetrahedral voids are the space left over by joining the centre of the sphere of the second layer which lie above the triangular voids of the first layer. The number of tetrahedral voids can be given as twice the total number of atoms.
Tetrahedral voids$= 2 \times$ total number of atoms
In the question, as $Y$ element forms $ccp$ lattice, therefore
Number of atoms of $Y$ element$=$total number of atoms in $ccp = 4$
Also, Number of tetrahedral voids $= 2 \times {\text{ Total number of atoms = 2}} \times {\text{4 = 8}}$
As given in question, that element $X$ occupies $\dfrac{2}{3}rd$ of tetrahedral voids thus
Number of atoms of $X$ element $= \dfrac{2}{3} \times {\text{tetrahedral voids}}$
${\text{ = }}\dfrac{2}{3} \times 8 \\\ = \dfrac{{16}}{3} \\\$
Now, The ratio of atoms of $X$ to the atoms of $Y$ $= \dfrac{{16}}{3}:4$
$= 4:3$
Therefore, the formula of compound can be given as : ${X_4}{Y_3}$

So, the correct answer is Option A .

Note:
Remember that cubic close packing ($ccp$) is an alternative name given to the face centered cubic ($fcc$) structure. The total number of atoms is the same in both. The layers of plane sheets of atoms are arranged in a slightly different manner in them. When we place the atoms in octahedral voids then packing is of $ABCABC$type hence it is called as $ccp$ and $fcc$ stands for the unit cell.