The empirical formula for a compound with a cubic close packed arrangement of anions and with cation

Q: Why is the effective number of atoms in a ccp lattice equal to 4?

A ccp lattice is geometrically identical to a face-centered cubic (fcc) lattice. In an fcc unit cell, there are 8 corner atoms (each shared by 8 cells, contributing $8 \times \frac{1}{8} = 1$) and 6 face-centered atoms (each shared by 2 cells, contributing $6 \times \frac{1}{2} = 3$). Total $= 1 + 3 = 4$.

Q: Where exactly are the octahedral voids located in a ccp unit cell?

In a ccp (or fcc) unit cell, octahedral voids are located at the center of each of the 12 edges and exactly one at the body center of the cube.

Q: How do we calculate the effective number of octahedral voids?

There are 12 edge-center voids, each shared by 4 unit cells ($12 \times \frac{1}{4} = 3$), and 1 body-center void which is entirely within the cell ($1 \times 1 = 1$). Total $= 3 + 1 = 4$.

Q: What if the question said tetrahedral voids instead of octahedral?

For a lattice with $N$ atoms, there are $2N$ tetrahedral voids. If cations occupied all tetrahedral voids, there would be 8 cations for every 4 anions, making the formula $A_2B$.

Q: Why do we simplify the ratio $A_4B_4$ to $AB$?

An empirical formula represents the simplest whole-number ratio of atoms in a compound. Therefore, $4:4$ is simplified to its lowest terms, which is $1:1$, giving the formula $AB$.

Question

The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites is $A_xB$. The value of $x$ is ........ (Integer answer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are shrinking down to the atomic level and walking through a crystal lattice. The problem tells us that the anions are forming the main framework of this crystal, specifically in a **cubic close-packed (ccp)** arrangement.

In solid-state chemistry, a ccp structure is geometrically identical to a face-centered cubic (fcc) unit cell. To find the empirical formula, our first mission is to determine exactly how many anions are present in one unit cell.

In an fcc unit cell, atoms are located at the 8 corners and the 6 face centers. 
- The corner atoms are shared by 8 adjacent unit cells, so their contribution is $8 	imes \frac{1}{8} = 1$.
- The face-centered atoms are shared by 2 adjacent unit cells, contributing $6 	imes \frac{1}{2} = 3$.

Adding these together, the effective number of anions (let's call them $B$) in the unit cell is $1 + 3 = 4$.

### The Master Equation for Voids
Now, where do the cations go? The problem states they occupy **all the octahedral sites**.

Here is a golden rule of solid-state chemistry that you must always remember: If a close-packed lattice is formed by $N$ atoms, it will generate exactly $N$ octahedral voids and $2N$ tetrahedral voids.

Since our lattice is formed by 4 anions ($N = 4$), there must be exactly 4 octahedral voids in the unit cell. Geometrically, these voids are located at the 12 edge centers (each contributing $\frac{1}{4}$) and 1 at the body center.

### Final Calculation
The problem specifies that the cations (let's call them $A$) occupy *all* of these octahedral sites. Therefore, the effective number of cations in the unit cell is also 4.

We now have the exact count of both ions in our unit cell:
- Number of $A$ cations $= 4$
- Number of $B$ anions $= 4$

The ratio of $A$ to $B$ is $4:4$. Since an empirical formula represents the simplest whole-number ratio of the elements, we simplify $4:4$ to $1:1$.

This gives us the empirical formula **$AB$**.

The question asks us to compare this with the given formula $A_xB$. By direct comparison, it is crystal clear that the value of $x$ is **$1$**.

This is a classic, high-yield concept for JEE. Always pay close attention to the type of void (octahedral vs. tetrahedral) and the fraction of those voids that are actually occupied!

Animated Solution for Chemistry - States of Matter: The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites is $A_{x} B$ . The value of $x$ is ........ (Integer answer)

Enter Numerical Value:

Visualized Solution

$B \to cc p$

$Z_{B} = 4$

$A \to O V$

$N_{O V} = Z_{B}$

$Z_{A} = 4$

$A_{x} B \Rightarrow x = 1$

$A \to T V ?$

The Sigma Insight: Solid State

Analyzing the Setup Imagine you are shrinking down to the atomic level and walking through a crystal lattice

The Master Equation for Voids

Final Calculation The problem specifies that the cations (let's call them $A$ ) occupy all of these octahedral sites

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