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**Hint:**In fcc packing atoms occupy at the corners and face centres of the cubic unit cell. If in a structure the number of closed spheres is “n”, then the number of octahedral voids is also “n”. Therefore the number of octahedral voids in a lattice is equal to the number of closed packed particles.

**Complete step by step answer:**

Now let us write down the details given in the question.

It is given that the fcc packing lattice is formed by the atoms of X. fcc is face-centred cubic close packing where the atoms are placed in corners and face centres of the cubic unit cell.

Therefore the X atoms occupy corners and face centres in the cubic unit cell of the crystalline solid.

So first we have to calculate the number of X atoms in fcc packing. As mentioned earlier the atoms at corners and face centres contribute to fcc packing. Considering this we can calculate the number of X atoms in fcc packing as follows,

The number of atoms of X in fcc packing $ = $ contribution by the number of atoms of X at corners $ + $ contribution by the number of atoms at face-centred.

To calculate the number of atoms of X at corners, consider the cubic unit cell.

A simple cubic has eight atoms at the corner.

In a unit cell, an atom at the corner is shared by eight other unit cells.

Therefore the contribution of each atom present at the corner is $\dfrac{1}{8}$.

Therefore the number of atoms present in the corners of the unit cell $ = \dfrac{1}{8} \times 8 = 1$.

Thus contribution by atoms present at the corners $ = 1$

Now let’s find out the contribution of atoms on the faces.

There are six faces in a simple cubic lattice. Each face has one atom. Therefore altogether there are six atoms on the faces.

An atom on each face is shared with two unit cells.

Therefore the contribution of each atom on the face is $\dfrac{1}{2}$.

Thus contribution by atoms on the faces $ = \dfrac{1}{2} \times 6 = 3$

Therefore the total number of atoms of X in fcc packing = $8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 1 + 3 = 4$.

The number of octahedral voids is equal to the number of atoms of X present in it.

Since Y atoms occupy the octahedral voids, their number would be equal to the number of X atoms present.

Therefore the number of Y atoms at octahedral voids $ = 4$.

Thus we have calculated the number of X atoms that forms fcc packing and the number of Y atoms occupying octahedral voids as four.

Thus atoms of X and Y are present in $1:1$ ratio and the formula should be XY.

But in the question, it is said that atoms along one body diagonal are removed.

We know that along with one of the body diagonal there are two X atoms and one Y atom.

So we have to find out the number of X atoms and the number of Y atoms after the removal of atoms along one body diagonal.

We have found out that the number of X atoms $ = 4$

Two X atoms along one of the body diagonal are removed.

Therefore the number of atoms of X after removal $ = 4 - 2 \times \dfrac{1}{8} = \dfrac{{15}}{4}$.

The number of atoms of Y after removal $ = 4 - 1 = 3$.

Therefore the ratio $X:Y = \dfrac{{15}}{4}:3 = 5:4$.

Therefore the formula is ${X_5}{Y_4}$.

**Thus the answer is an option (C) ${X_5}{Y_4}$.\[\]**

**Note:**

X atoms are present at corners and face centres in fcc packing. Contribution by each atom at the corner of the unit cell is $\dfrac{1}{8}$. Along with one of the body diagonal, there are two X atoms present and the contribution of the two X atoms present at the corners along the body diagonal of a unit cell $ = 2 \times \dfrac{1}{8}$. So as we remove X atoms along one body diagonal, this contribution of the two X atoms along the body diagonal has to be subtracted from the total number of X atoms in fcc packing. Therefore the number of X atoms in fcc packing $ - $ contribution of the two X atoms along one body diagonal. That is the number of X atoms after removal $ = 4 - 2 \times \dfrac{1}{8} = \dfrac{{15}}{4}$. When atoms along one body diagonal are removed, one of the Y atoms is removed. Therefore the number of Y atoms after removal is the number of Y atoms in octahedral voids $ - $ one Y atom present along the body diagonal $ = 4 - 1 = 3$

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