If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of

Q: Why are there exactly 4 oxygen atoms in a ccp unit cell?

A ccp lattice is geometrically identical to a face-centered cubic (fcc) lattice. It has atoms at the 8 corners (each contributing 1/8) and 6 face centers (each contributing 1/2). So, $8 \times (1/8) + 6 \times (1/2) = 1 + 3 = 4$ atoms.

Q: How do we know the number of voids without drawing them?

It's a fundamental geometric property of close-packed structures. If there are $N$ atoms in the lattice, there are exactly $N$ octahedral voids and $2N$ tetrahedral voids. Since $N=4$ here, we get 4 octahedral and 8 tetrahedral voids.

Q: Why do we multiply the number of ions by their charges?

Because we are balancing the total electrical charge, not just the number of atoms. One $\text{Al}^{3+}$ ion contributes $+3$ to the total charge, so $4m$ of them will contribute $4m \times 3 = 12m$.

Q: Can the fractions $m$ and $n$ be greater than 1?

No. The fractions $m$ and $n$ represent the portion of available voids that are occupied. You cannot occupy more voids than what actually exists in the lattice, so $m$ and $n$ must be between 0 and 1.

Q: What if multiple options satisfied the charge balance equation?

In standard JEE problems of this type, usually only one option will mathematically satisfy the electrical neutrality constraint. If multiple did, you would need additional information about the crystal's density or molar mass to find the unique solution.

Question

If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with $m$ fraction of octahedral holes occupied by aluminium ions and $n$ fraction of tetrahedral holes occupied by magnesium ions $m$ and $n$ respectively, are -

Options:

(A) $\frac{1}{2}, \frac{1}{8}$

(B) $1, \frac{1}{4}$

(C) $\frac{1}{2}, \frac{1}{2}$

(D) $\frac{1}{4}, \frac{1}{8}$

Sigma Percentile · Accepted Answer

### The Architecture of the Crystal

Imagine you are an architect tasked with building a microscopic crystal. The foundation of this structure is laid by the oxygen ions ($	ext{O}^{2-}$). The problem tells us that these oxygen ions form a **cubic close-packed (ccp)** array.

In the world of solid-state chemistry, a ccp lattice is identical to a face-centered cubic (fcc) lattice. If we count the effective number of atoms in one such unit cell, we find there are exactly $4$ oxygen ions.

### Uncovering the Hidden Voids

Now, no matter how closely you pack spheres together, there will always be empty spaces left between them. These empty spaces are called **voids**.

For any close-packed structure with $Z$ effective atoms, nature dictates a strict geometric rule:
- The number of **octahedral voids** is exactly equal to $Z$.
- The number of **tetrahedral voids** is exactly double, $2Z$.

Since our unit cell has $4$ oxygen atoms, it must contain $4$ octahedral voids and $8$ tetrahedral voids.

### The Cation Invaders

The problem introduces two types of cations that want to occupy these voids: Aluminium ($	ext{Al}^{3+}$) and Magnesium ($	ext{Mg}^{2+}$).

However, they don't fill all the voids. The aluminium ions occupy an $m$ fraction of the octahedral voids. Therefore, the effective number of $	ext{Al}^{3+}$ ions is $4m$.

Similarly, the magnesium ions occupy an $n$ fraction of the tetrahedral voids. This means the effective number of $	ext{Mg}^{2+}$ ions is $8n$.

### The Master Key: Electrical Neutrality

Here is where the magic happens. Any stable ionic crystal must be **electrically neutral**. The total positive charge provided by the cations must perfectly balance the total negative charge from the anions.

Let's calculate the total negative charge first. We have $4$ oxygen ions, each carrying a charge of $-2$. 
Total negative charge magnitude = $4 	imes 2 = 8$.

Now for the positive charge. We have $4m$ aluminium ions (each $+3$) and $8n$ magnesium ions (each $+2$).
Total positive charge = $(4m 	imes 3) + (8n 	imes 2) = 12m + 16n$.

### The Final Equation

Equating the positive and negative charges, we get our master equation:
$$12m + 16n = 8$$

To make our lives easier, let's divide the entire equation by $4$:
$$3m + 4n = 2$$

This is a beautiful, simple linear equation. Since $m$ and $n$ represent fractions of voids, they must be positive numbers between $0$ and $1$. To find the exact values, we simply test the given options.

Let's check Option (A), where $m = \frac{1}{2}$ and $n = \frac{1}{8}$:
$$3\left(\frac{1}{2}ight) + 4\left(\frac{1}{8}ight) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$$

The equation balances perfectly! Thus, the correct fractions are $m = \frac{1}{2}$ and $n = \frac{1}{8}$.

Animated Solution for Chemistry - States of Matter: If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with $m$ fraction of octahedral holes occupied by aluminium ions and $n$ fraction of tetrahedral holes occupied by magnesium ions $m$ and $n$ respectively, are -

Select Answer:

Visualized Solution

$Crystal Lattice Setup$

$Number of Voids$

$Occupancy of Voids$

$Electrical Neutrality$

$Charge Equation$

$Solving for m and n$

$Verifying Options$

The Sigma Insight: Solid State

The Architecture of the Crystal

Uncovering the Hidden Voids

The Cation Invaders

The Master Key

The Final Equation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Crystal Lattice Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Voids

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Occupancy of Voids

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electrical Neutrality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Charge Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for m and n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Verifying Options