In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedra

Q: Why do we assume the number of Y atoms is N? Can I use a real number?

Yes, absolutely! $N$ is just a general variable. You can assume it to be $100$, $4$ (for a single fcc unit cell), or any number. If you assume $100$ $Y$ atoms, you get $200$ tetrahedral voids, and $X$ would be $\frac{2}{3} \times 200 = \frac{400}{3}$. The ratio of $X:Y$ would be $\frac{400}{3} : 100$, which still simplifies perfectly to $4:3$.

Q: How do I remember the number of tetrahedral and octahedral voids?

For $N$ atoms in a close-packed lattice, there are $N$ Octahedral Voids (OV) and $2N$ Tetrahedral Voids (TV). A quick memory trick: Think of 'Tetra' as 'Two' times the atoms!

Q: What does 'ccp' mean? Is it different from 'fcc'?

ccp stands for cubic close-packed. Geometrically, a ccp lattice is identical to a face-centered cubic (fcc) unit cell. They are often used interchangeably in solid-state problems.

Q: Why did we multiply the ratio by 3 at the end?

Chemical formulas must have whole numbers as subscripts to represent discrete atoms. Our initial ratio was $\frac{4}{3} : 1$. Multiplying both sides by $3$ clears the denominator, giving us $4 : 3$, which are clean whole numbers.

Q: What if the fraction was greater than 1, like 4/3 of voids occupied?

That is physically impossible! The maximum occupancy is $1$ (or $100\%$). You cannot have more atoms stuffed into voids than the total number of voids that actually exist in the geometry.

Question

In a compound, atoms of element $Y$ form ccp lattice and those of element $X$ occupy 2/3rd of tetrahedral voids. The formula of the compound will be

Options:

(A) $X_4Y_3$

(B) $X_2Y_3$

(C) $X_2Y$

(D) $X_3Y_4$

Sigma Percentile · Accepted Answer

### Visualizing the Lattice

Imagine you are shrinking down to the atomic level and looking at a beautiful, highly ordered crystal. In this compound, the atoms of element $Y$ form the foundational structure, specifically a **cubic close-packed (ccp)** lattice.

To make our mathematical journey easier, let's assign a variable to the total number of these foundational atoms. Let the number of atoms of element $Y$ in the lattice be $N$.

$$N_Y = N$$

### The Secret of Voids

Now, no matter how tightly you pack spheres together, there will always be empty spaces left between them. In crystallography, we call these empty spaces **voids**. There is a beautiful, unbreakable geometric rule for close-packed structures: for every single atom in the lattice, there are exactly **two tetrahedral voids**.

Since we established that there are $N$ atoms of element $Y$, the total number of tetrahedral voids available in this crystal must be double that amount.

$$N_{TV} = 2N$$

### Placing the X Atoms

The problem gives us a crucial piece of information: the atoms of element $X$ do not fill up every single tetrahedral void. They only occupy $\frac{2}{3}$ of them.

To find the effective number of $X$ atoms, we simply multiply the total number of tetrahedral voids by this occupancy fraction.

$$N_X = \frac{2}{3} 	imes 2N$$

$$N_X = \frac{4N}{3}$$

### Deriving the Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of its constituent atoms. We now have the effective number of both $X$ and $Y$ atoms, so let's set up their ratio.

$$	ext{Ratio of } X : Y = \frac{4N}{3} : N$$

First, we can cancel out the $N$ from both sides, as it represents the same base quantity. This leaves us with a fractional ratio.

$$	ext{Ratio of } X : Y = \frac{4}{3} : 1$$

Chemical formulas cannot have fractions as subscripts. To convert this into a whole-number ratio, we multiply both sides by $3$.

$$	ext{Ratio of } X : Y = 4 : 3$$

With this clean, whole-number ratio, we can confidently write the final empirical formula of the compound.

**Formula = $X_4Y_3$**

Animated Solution for Chemistry - States of Matter: In a compound, atoms of element $Y$ form ccp lattice and those of element $X$ occupy 2/3rd of tetrahedral voids. The formula of the compound will be

Select Answer:

Visualized Solution

$N_{Y} = N$

$N_{T V} = 2 N$

$N_{X} = \frac{2}{3} \times 2 N$

$N_{X} = \frac{4 N}{3}$

$X : Y = \frac{4 N}{3} : N$

$X_{4} Y_{3}$

$What if Octahedral Voids?$

The Sigma Insight: Solid State

Visualizing the Lattice

The Secret of Voids

Placing the X Atoms

Deriving the Empirical Formula

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