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In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atom B is missing from one of the face centered points, the formula of the ionic compound is:
(A) $A{{B}_{2}}$
(B) ${{A}_{5}}{{B}_{2}}$
(C) ${{A}_{2}}{{B}_{3}}$
(D) ${{A}_{2}}{{B}_{5}}$

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Last updated date: 25th Jun 2024
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Answer
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Hint: The face-centred cubic contains 4 atoms per unit cell. The FCC structure has a total 8 atoms at the 8 corners and 6 atoms at the 6 faces of the structure. Each atom at the corner contributes $\text{ }\dfrac{1}{8}\text{ }$ and atoms at the face of structure contributes $\text{ }\dfrac{1}{2}\text{ }$ towards the total number of atoms. To determine the formula of the compound determines the contribution of each type of atom towards the compound.

Complete Solution :
We have been provided with face-centered cubic lattice atoms A are at the corner points and atoms B at the face-centered points.
If one B atom is missing from one of the face-centered points, we need to find the formula of the ionic compound,
So, for that:
- Let's consider the general formula for the compound formed by the A and B in the face-centered cubic structure as $\text{ }{{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\text{ }$ , where x and y correspond to the number of atoms of A and B present per Face centered cubic structure.
- We know that in a face-centered cubic lattice, atoms A are at the corner points,
Therefore, the contribution of A would be: $8\times \dfrac{1}{8}=\text{ 1 }$,
If B is at the face-centered points, then the contribution of B towards the ionic compound is,
$6\times \dfrac{1}{2}=\dfrac{6}{2}\text{ }$,

Now, according to the question,
- Let's consider that one of the B atoms is missing from the unit cell, then the total contribution of the B atoms decreases from $\text{ }\dfrac{6}{2}\text{ }$to $\dfrac{5}{2}$ per lattice.
- Thus the formula for the compound would be written as,
$\text{ A : B }\Rightarrow \text{ 1 : }\dfrac{5}{2}\text{ }\Rightarrow \text{ 2 : 5 }$
So, the formula of the ionic compound would be: ${{A}_{2}}{{B}_{5}}$.
So, the correct answer is “Option D”.

Note: Note that, if the number of an atom is not missing from the face of face-centered cubic structure then the ionic formula for the compound would be equal to, $\text{ A : B }\Rightarrow \text{ 1 : }\dfrac{6}{2}\text{ }\Rightarrow \text{ 2 : 3 }$
- That is ${{A}_{2}}{{B}_{3}}\text{ }$. Usually the formula is determined by considering the cation and anion and their position in the structure. Some of the examples of FCC crystal structure are sodium chloride and zinc blend.