A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction o

Q: Why did we assume B forms the lattice and not A?

In multiple-choice questions like this, you often have to test the possibilities. We started with B forming the lattice, which led us to a fraction of 1/3, perfectly matching an option. If we had assumed A formed the lattice, the fraction would have been 3/4, which wasn't an option. It's a process of elimination!

Q: How do we know there are exactly $2N$ tetrahedral voids?

This is a fundamental geometric property of close-packed structures (both HCP and CCP). Every sphere in the lattice is in contact with three spheres in the layer above and three in the layer below, creating two tetrahedral voids per sphere. Hence, $N$ spheres create $2N$ tetrahedral voids.

Q: What if the question mentioned octahedral voids instead?

If the atoms were occupying octahedral voids, the math would change slightly. The number of octahedral voids is exactly equal to the number of lattice atoms ($N$). So you would divide the number of void-occupying atoms by $N$ instead of $2N$.

Q: Does it matter if the lattice is HCP or CCP for this calculation?

No, it doesn't! Both Hexagonal Close Packing (HCP) and Cubic Close Packing (CCP/FCC) have the exact same void ratio: $N$ lattice atoms generate $2N$ tetrahedral voids and $N$ octahedral voids. The formula derivation works identically for both.

Q: Why did the $N$ cancel out at the end?

Because the chemical formula represents a ratio, not an absolute number of atoms. Whether you have 100 lattice atoms or 1 mole of lattice atoms, the fraction of voids occupied remains constant. The $N$ is just a placeholder that naturally cancels out when finding a ratio or fraction.

Question

A compound of formula $A_2B_3$ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms ?

Options:

(A) hcp lattice- A, $\frac{2}{3}$ tetrahedral voids-B

(B) hcp lattice-A, $\frac{1}{3}$ tetrahedral voids-B

(C) hcp lattice-B, $\frac{1}{3}$ tetrahedral voids-A

(D) hcp lattice-B, $\frac{2}{3}$ tetrahedral voids-A

Sigma Percentile · Accepted Answer

### Decoding the Crystal Formula

Welcome to the fascinating world of Solid State Chemistry! When we are given a chemical formula like **$A_2B_3$**, it is not just a random collection of letters and numbers. It is a precise architectural blueprint of a crystal lattice.

The formula immediately tells us the molar ratio of the atoms present in the crystal. For every $2$ atoms of element $A$, there are exactly $3$ atoms of element $B$. Mathematically, we can write this as:

$$A : B = 2 : 3$$

In close-packed structures like Hexagonal Close Packing (HCP) or Cubic Close Packing (CCP), one type of atom forms the main structural framework (the lattice), while the other type of atom sneaks into the empty spaces, known as **voids**, left between the lattice atoms.

### Setting Up the Lattice

To solve this, we need to make an educated assumption and test it. Let's assume that atom **$B$** forms the main HCP lattice.

If $B$ forms the lattice, let's denote the total number of $B$ atoms in the crystal as $N$.

$$N_B = N$$

Now, we must find out how many $A$ atoms are present. Using our established ratio of $2:3$, the number of $A$ atoms must be two-thirds the number of $B$ atoms.

$$N_A = \frac{2}{3}N$$

These $A$ atoms are the ones occupying the voids within the $B$ lattice.

### The Magic of Tetrahedral Voids

Here is a golden rule of solid-state geometry: For any close-packed lattice containing $N$ atoms, the number of **tetrahedral voids (TV)** generated is always exactly double the number of lattice atoms.

$$	ext{Total Tetrahedral Voids} = 2N$$

We know that the $A$ atoms are sitting in these tetrahedral voids. But do they occupy all of them? To find out, we calculate the fraction of tetrahedral voids occupied by $A$ atoms by dividing the number of $A$ atoms by the total number of available tetrahedral voids.

$$	ext{Fraction Occupied} = \frac{	ext{Number of } A 	ext{ atoms}}{	ext{Total TV}}$$

$$	ext{Fraction Occupied} = \frac{\frac{2}{3}N}{2N}$$

Notice how beautifully the $N$ cancels out from the numerator and the denominator. The $2$ also cancels out, leaving us with a clean, elegant fraction:

$$	ext{Fraction Occupied} = \frac{1}{3}$$

### The Final Verdict

Our calculation shows that if $B$ forms the HCP lattice, $A$ will occupy exactly **$\frac{1}{3}$** of the tetrahedral voids. Looking at our options, this matches perfectly with option (c)!

As a quick mental check, what if $A$ formed the lattice instead? If $A$ was $N$, then $B$ would be $\frac{3}{2}N$. The fraction of tetrahedral voids occupied by $B$ would be $\frac{\frac{3}{2}N}{2N} = \frac{3}{4}$. Since $\frac{3}{4}$ is nowhere to be found in the given options, we can be absolutely certain that our initial assumption was correct. Atom $B$ forms the lattice, and atom $A$ occupies one-third of the tetrahedral voids.

Animated Solution for Chemistry - States of Matter: A compound of formula $A_{2} B_{3}$ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms ?

Select Answer:

Visualized Solution

$A_{2} B_{3}$

$A : B = 2 : 3$

$N_{B} = N$

$N_{A} = \frac{2}{3} N$

$Total TV = 2 N$

$Fraction = \frac{1}{3}$

$Cross-check$

The Sigma Insight: Solid State

Decoding the Crystal Formula

Setting Up the Lattice

The Magic of Tetrahedral Voids

The Final Verdict

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