Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is con

Q: Why did we assume anions are in the octahedral voids and not forming the FCC lattice?

In a standard NaCl structure, $Cl^-$ forms the FCC lattice and $Na^+$ occupies the octahedral voids. However, the first instruction says "Remove all the anions except the central one". An FCC lattice does not have a central atom; its center is an octahedral void! This phrasing forces us to conclude that for this specific solid MX, the anions must be in the octahedral voids (which includes the body center) and the cations form the FCC lattice.

Q: Why do edge-centered atoms only contribute 1/4 to the unit cell?

Imagine a single edge of a cube. That edge is shared by four adjacent cubes in a 3D lattice (two on the bottom layer, two on the top layer). Therefore, any atom sitting exactly on that edge is split equally among four unit cells, meaning only 1/4 of it belongs to our specific unit cell.

Q: In step (ii), when we replace face-centered cations with anions, why does the anion count increase by 3?

There are 6 faces on a cube. An atom at a face center is shared between 2 adjacent unit cells, so its contribution is 1/2. When we place anions on all 6 faces, the effective number of anions added is $6 \times \frac{1}{2} = 3$.

Q: Does the size of the atoms matter when we swap them?

In reality, swapping a small cation for a large anion would severely distort the lattice and likely cause it to collapse. However, the problem explicitly states to "Neglect the charge balance" and implies we should treat this purely as a geometric counting exercise. We are just tracking positions and effective numbers!

Q: What happens to the overall charge of the unit cell?

The final unit cell has 1 cation ($M^+$) and 3 anions ($X^-$), which means it is highly negatively charged. In the real world, such a crystal could not exist due to massive electrostatic repulsion. But again, the problem gives us a "cheat code" by saying "Neglect the charge balance", allowing us to focus entirely on the mathematical logic of the lattice points.

Question

Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.
(i) Remove all the anions (X) except the central one
(ii) Replace all the face centered cations (M) by anions (X)
(iii) Remove all the corner cations (M)
(iv) Replace the central anion (X) with cation (M)
The value of $\left(\frac{	ext{number of anions}}{	ext{number of cations}}ight)$ in Z is____.

Sigma Percentile · Accepted Answer

The problem presents us with a fascinating geometric puzzle disguised as a solid state chemistry question. We are given an ionic solid MX with an NaCl-type structure and asked to perform a series of surgical modifications to its unit cell to create a new structure, Z.

The key to solving this problem lies in meticulous bookkeeping of the effective number of atoms at each step. Let's break it down.

### Decoding the Initial Structure

The very first sentence tells us we have an NaCl structure. In a standard NaCl crystal, the chloride anions ($Cl^-$) form a Face-Centered Cubic (FCC) lattice, while the sodium cations ($Na^+$) occupy all the Octahedral Voids (O.V.).

However, there is a massive clue hidden in the first instruction: **"Remove all the anions (X) except the central one."**

An FCC lattice does not have an atom at its body center; the body center of an FCC lattice is an octahedral void! Therefore, for there to be a "central anion", the anions ($X^-$) must be the ones occupying the octahedral voids, and the cations ($M^+$) must be forming the FCC lattice.

Let's establish our initial baseline:
*   **Cations ($M^+$) in FCC:** 8 corners $	imes \frac{1}{8} = 1$ atom, and 6 face centers $	imes \frac{1}{2} = 3$ atoms. Total $M^+ = 4$.
*   **Anions ($X^-$) in O.V.:** 1 body center $	imes 1 = 1$ atom, and 12 edge centers $	imes \frac{1}{4} = 3$ atoms. Total $X^- = 4$.

### Step (i): Stripping the Edges

The first instruction is to remove all anions except the central one.

Our anions are located at the edge centers and the body center. By following this instruction, we strip away all 12 edge-centered anions.

*   Removed $X^-$: 12 edge centers $	imes \frac{1}{4} = 3$.
*   Remaining $X^-$: 1 (at the body center).
*   Remaining $M^+$: 4 (untouched).

### Step (ii): The Face Swap

Next, we must replace all the face-centered cations ($M^+$) with anions ($X^-$). This is a two-part operation.

First, we remove the cations from the six faces. 
*   Removed $M^+$: 6 face centers $	imes \frac{1}{2} = 3$.
*   Remaining $M^+$: $4 - 3 = 1$ (these are the corner cations).

Second, we place anions in those exact face-centered spots.
*   Added $X^-$: 6 face centers $	imes \frac{1}{2} = 3$.
*   Total $X^-$: 1 (from the center) + 3 (newly added) = 4.

### Step (iii): Clearing the Corners

The third instruction is straightforward: remove all the corner cations ($M^+$).

We have 8 cations sitting at the corners of our unit cell. 
*   Removed $M^+$: 8 corners $	imes \frac{1}{8} = 1$.
*   Remaining $M^+$: $1 - 1 = 0$.

At this point, the entire cation lattice has been temporarily wiped out! Our anion count remains safely at 4.

### Step (iv): The Final Centerpiece

Finally, we replace the central anion ($X^-$) with a cation ($M^+$).

We take that single anion sitting right in the middle of the body center and remove it.
*   Removed $X^-$: 1 body center $	imes 1 = 1$.
*   Remaining $X^-$: $4 - 1 = 3$ (these are the face-centered anions).

In its place, we drop in a brand new $M^+$ cation.
*   Added $M^+$: 1 body center $	imes 1 = 1$.
*   Total $M^+$: $0 + 1 = 1$.

### The Final Tally

We have successfully built the new structure Z. Let's look at the final tally of our effective atoms:
*   **Number of anions ($X^-$):** 3
*   **Number of cations ($M^+$):** 1

The question asks for the value of the ratio $\left(\frac{	ext{number of anions}}{	ext{number of cations}}ight)$.

$$ 	ext{Ratio} = \frac{3}{1} = 3 $$

Through careful, step-by-step spatial reasoning and basic fraction arithmetic, we arrive at a clean integer answer of **3**.

Enter Numerical Value:

Visualized Solution

$Initial NaCl Structure$

$Step (i): Remove Edge Anions$

$Step (ii): Replace Face Cations$

$Step (iii): Remove Corner Cations$

$Step (iv): Replace Central Anion$

$Final Ratio Calculation$

The Sigma Insight: Solid State

Decoding the Initial Structure

Step (i)

Step (ii)

Step (iii)

Step (iv)

The Final Tally

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