Question

In a solid XY having the rock salt structure, Y atoms form the FCC arrangement and X atoms occupy all the octahedral voids. If the X atom at body centred octahedral void is missing, then the formula of the compound will be
A. XY
B. ${{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{4}}}$
C. ${{\text{X}}_{\text{4}}}{{\text{Y}}_{\text{3}}}$
D. ${{\text{X}}_{\text{3}}}{\text{Y}}$

Answer 1

Hint: To answer this question we should know the structure of the FCC unit cell and the number of voids present in the FCC unit cell. First, we will determine the number of Y atoms that are forming the FCC lattice. Then we will determine the number of X atoms according to the voids. Then we will take the ratio of X and Y atoms to determine the formula of the compound.


Complete solution:
When a lattice forms one type of atoms form the main lattice and the second type of atoms goes into the voids. The ratio of both the atoms gives the general formula of the compound.
It is given that in the solid XY, Y atoms form the FCC arrangement. In FCC eight atoms are present at corners and contribute $1/8$ in a lattice. Six atoms are present at each face-centre and contribute $1/2$ in a lattice.
So, eight Y will be at the corner and the six Y will be at each face-centre.
So, the total number of atoms Y in a FCC unit cell is,
Total Y atoms \[ = \,8\, \times \,\dfrac{1}{8}\, + \,6\, \times \,\dfrac{1}{2}\]
Total Y atoms \[ = \,1\, + \,3\]
Total Y atoms \[ = \,4\]
So, a FCC unit of compound XY has four Y atoms.
A FCC unit cell has four atoms, so the number of octahedral voids will be four. A void occupied by only one atom, so we can say that four voids will be occupied by four X atoms.
Now, it is given that the X atom at body centred octahedral void is missing. So, if out of four X atoms, one is missing then one FCC unit cell of XY compound has only three X atoms.
So, the ratio of X and Y atoms are,
X : Y
$3:4$
So, the formula of the compound will be ${{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{4}}}$.

Therefore, option (B) is correct.


Note: We should remember the total number of atoms and their position in a unit cell. The total number of atoms in the FCC unit cell is four, in BCC is two and Simple cubic unit cell is one. The atoms that are forming lattice will always be in fixed numbers if they are forming a regular unit cell. The number of other atoms can be determined by the help of voids. Any unit cell has a number of octahedral voids and $2$ n number of tetrahedral voids. Where, n is the number of atoms in a unit cell.