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

Q: Why is the contribution of a corner atom 1/8?

A corner atom is shared equally by 8 adjacent unit cells meeting at that corner. Therefore, only 1/8th of its volume belongs to any single unit cell.

Q: Why are there 4 octahedral voids in an FCC lattice?

Octahedral voids are located at the 12 edge centers (each shared by 4 unit cells, contributing 1/4) and 1 at the body center (unshared, contributing 1). Total = 12*(1/4) + 1 = 4.

Q: Is CCP the same as FCC?

Yes, geometrically, a cubic close-packed (CCP) arrangement results in a face-centered cubic (FCC) unit cell. The terms are often used interchangeably when discussing unit cell calculations.

Q: What if the question mentioned tetrahedral voids instead?

In an FCC lattice, there are 8 tetrahedral voids (located near each corner). If cations occupied all of them, the effective number of cations would be 8, leading to a formula of $A_8B_4$ or $A_2B$.

Q: Why do we simplify the ratio to find the empirical formula?

An empirical formula represents the simplest whole-number ratio of atoms in a compound. So, a ratio of 4:4 must be simplified to 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 in $A_x B$. The value of $x$ is ........ (Integer answer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are diving into the microscopic world of a crystal lattice. The problem presents us with a compound where anions (let's call them B) form a **cubic close-packed (ccp)** arrangement.

It is crucial to remember that a ccp lattice is geometrically identical to a **face-centered cubic (fcc)** unit cell. In this structure, the B atoms are positioned at all eight corners of the cube and at the centers of all six faces.

### The Master Equation for Anions

To find the empirical formula, we first need to determine the effective number of B atoms per unit cell. We can't just count the atoms we see; we must account for how they are shared with neighboring unit cells.

Each corner atom is shared by eight adjacent cubes, so its contribution is $\frac{1}{8}$. Each face-centered atom is shared by two cubes, contributing $\frac{1}{2}$.

Let's set up the math:

$$Z_B = \left(8 	imes \frac{1}{8}ight) + \left(6 	imes \frac{1}{2}ight)$$

$$Z_B = 1 + 3 = 4$$

So, there are exactly **4 effective B atoms** in our unit cell.

### Locating the Cations

Now, let's shift our focus to the cations, A. The problem states they occupy **all the octahedral voids**. But where exactly are these voids hiding in an fcc lattice?

Octahedral voids are located at the center of every single edge of the cube, plus one perfectly hidden right in the body center.

### Final Calculation for Cations

Just like the atoms, these voids are shared. A void on an edge is shared by four unit cells, so it contributes $\frac{1}{4}$. The void in the body center is entirely within our unit cell, contributing a full $1$.

Let's calculate the effective number of A atoms:

$$Z_A = \left(12 	imes \frac{1}{4}ight) + \left(1 	imes 1ight)$$

$$Z_A = 3 + 1 = 4$$

We have exactly **4 effective A atoms**.

### The Empirical Formula

We now have the effective numbers for both ions: 4 A atoms and 4 B atoms. The ratio of A to B is $4:4$.

An empirical formula must represent the simplest whole-number ratio. Simplifying $4:4$ gives us $1:1$.

Therefore, the empirical formula is **$AB$**.

The question tells us the formula is $A_xB$. By comparing our result, it is beautifully clear that **$x = 1$**.

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 in $A_{x} B$ . The value of $x$ is ........ (Integer answer)

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Visualized Solution

Visualizing the FCC Lattice

Effective Number of Anions

Calculating $Z_{B}$

Locating Octahedral Voids

Effective Number of Cations

Calculating $Z_{A}$

Empirical Formula

What if Tetrahedral Voids?

The Sigma Insight: Solid State

Analyzing the Setup

The Master Equation for Anions

Locating the Cations

Final Calculation for Cations