CsCl Structure: Calculate Cell Parameter & Identify Compound X

Oct 23, 2025 by SLV Team 63 views

Hey guys! Today, we're diving into a cool chemistry problem involving crystal structures, specifically the Cesium Chloride (CsCl) structure. We'll be calculating the cell parameter of a mysterious compound, let's call it X, and then put on our detective hats to figure out what X actually is. This problem combines some fundamental concepts of solid-state chemistry, so buckle up and let's get started!

1. Understanding the CsCl Structure

Before we jump into calculations, let's quickly recap the Cesium Chloride (CsCl) structure. It's a classic example of a cubic crystal system, but it's not a simple cubic structure. Instead, it's a body-centered cubic arrangement. This means that we have one type of ion (let's say the Cesium ion, Cs+) at the corners of the cube, and the other type of ion (the Chloride ion, Cl-) sits smack-dab in the center of the cube.

Think of it like this: imagine a cube made up of Cl- ions at each corner. Then, a single Cs+ ion sits in the very middle, surrounded by all those Cl- ions. This arrangement is highly symmetrical and leads to some interesting properties.

Key characteristics of the CsCl structure:

Coordination Number: Each ion (both Cs+ and Cl-) is surrounded by 8 ions of the opposite charge. This is called a coordination number of 8. The high coordination number contributes to the stability of the structure.
Unit Cell Composition: A single unit cell in the CsCl structure contains one Cs+ ion and one Cl- ion. Even though there are 8 Cl- ions at the corners, each corner ion is shared by 8 unit cells, so we only count 1/8 of each corner ion. (8 corners * 1/8 = 1 Cl- ion). The Cs+ ion in the center belongs entirely to that unit cell.
Relationship between ionic radii and cell parameter: This is where things get interesting for our problem! The cell parameter, often denoted as 'a', is the length of one side of the cube. In the CsCl structure, the ions are in contact along the body diagonal of the cube. This means that the body diagonal is equal to the sum of the radii of the two ions (Cs+ and Cl-) multiplied by 2. We'll use this relationship to calculate 'a' later.

2. Calculating the Cell Parameter (a)

This is the first part of our challenge. We're given the ionic radii of the cation (1.59 Å) and the chloride ion (1.74 Å), and we know that these ions are in contact along the body diagonal of the CsCl unit cell. Let's break down the calculation step-by-step.

Step 1: Relate the body diagonal to the cell parameter

Remember the Pythagorean theorem? We'll need it here! In a cube, the body diagonal (d) is related to the cell parameter (a) by the following equation:

$d = a

√3$

This comes from applying the Pythagorean theorem twice: first to find the face diagonal, and then again using the face diagonal and the height to find the body diagonal.

Step 2: Express the body diagonal in terms of ionic radii

As we discussed earlier, the ions are in contact along the body diagonal. This means the body diagonal is equal to twice the sum of the ionic radii:

$d = 2 * (r_+ + r_-)$

Where:

r+ is the radius of the cation (1.59 Å)
r- is the radius of the anion (1.74 Å)

Step 3: Combine the equations and solve for 'a'

Now we have two expressions for the body diagonal (d). Let's set them equal to each other:

√3 = 2 * (r_+ + r_-)$

Now we can solve for 'a':

$a = (2 * (r_+ + r_-)) /

√3$

Plug in the given values:

$a = (2 * (1.59 Å + 1.74 Å)) /

√3$

$a = (2 * 3.33 Å) /

√3$

$a ≈ 3.85 Å$

So, we've calculated the cell parameter 'a' to be approximately 3.85 Å. Awesome!

Key Takeaways for Calculating Cell Parameter:

Visualize the Structure: Understanding the geometry of the crystal structure (CsCl in this case) is crucial. How are the ions arranged? Which ions are in contact?
Use the Pythagorean Theorem: The relationship between the body diagonal, face diagonal, and cell parameter is derived from the Pythagorean theorem.
Relate Radii to Cell Parameter: The key is to find a direction in the unit cell where ions are in contact. This allows you to relate the ionic radii to the cell parameter.

3. Determining the Identity of Compound X

Now for the fun part – detective work! We know compound X has a CsCl structure, a density of 7010 kg/m³, and we've calculated its cell parameter (a ≈ 3.85 Å). We also know the ionic radii of the cation and chloride ion. Let's use this information to identify X.

Step 1: Calculate the volume of the unit cell

The unit cell is a cube, so its volume (V) is simply the cell parameter cubed:

$V = a^3$

$V = (3.85 Å)^3$

But wait! We need to be careful with units. Density is given in kg/m³, so we need to convert the cell parameter from Ångströms to meters:

$1 Å = 10^{-10} m$

$a = 3.85 Å = 3.85 * 10^{-10} m$

Now we can calculate the volume:

$V = (3.85 * 10^{-10} m)^3$

$V ≈ 5.71 * 10^{-29} m^3$

Step 2: Calculate the mass of the unit cell

We know the density (ρ) and the volume (V), so we can calculate the mass (m) using the following formula:

$ρ = m / V$

$m = ρ * V$

$m = 7010 kg/m³ * 5.71 * 10^{-29} m^3$

$m ≈ 4.00 * 10^{-25} kg$

Step 3: Calculate the molar mass of compound X

Remember that each unit cell in the CsCl structure contains one formula unit (one cation and one chloride ion). So, the mass we just calculated is the mass of one formula unit. To find the molar mass (M), we need to multiply this mass by Avogadro's number (NA):

$M = m * N_A$

Where:

$N_A$ is Avogadro's number, approximately $6.022 * 10^{23} mol^{-1}$

$M = 4.00 * 10^{-25} kg * 6.022 * 10^{23} mol^{-1}$

Again, watch out for units! We need to convert the mass from kg to grams:

$M = 4.00 * 10^{-25} kg * (1000 g / 1 kg) * 6.022 * 10^{23} mol^{-1}$

$M ≈ 241 g/mol$

Step 4: Identify the cation

We know that compound X is a chloride, so it contains Cl- ions. The molar mass of Cl is approximately 35.5 g/mol. To find the molar mass of the cation, we subtract the molar mass of Cl from the molar mass of X:

M(cation) = M(X) - M(Cl)

M(cation) = 241 g/mol - 35.5 g/mol

M(cation) ≈ 205.5 g/mol

Now we need to look at the periodic table and see which element has a molar mass close to 205.5 g/mol. A quick search reveals that Bismuth (Bi) has a molar mass of approximately 209 g/mol. That's pretty close!

Step 5: Confirm the identity

Based on our calculations and the ionic radii provided, it's highly likely that compound X is Bismuth Chloride (BiCl). However, Bismuth Chloride doesn't typically crystallize in the CsCl structure under normal conditions. This result suggests there might be a slight discrepancy in the provided data or the compound might exist under specific high-pressure/high-temperature conditions that favor the CsCl structure. Nonetheless, based on the provided information and calculations, Bismuth Chloride is the most plausible answer.

Key Takeaways for Identifying the Compound:

Use Density and Cell Volume: These properties are crucial for calculating the mass of the unit cell and, subsequently, the molar mass of the compound.
Consider the Crystal Structure: The CsCl structure has one formula unit per unit cell, which simplifies the molar mass calculation.
Combine Information: Use all the available information (density, cell parameter, ionic radii, molar mass) to narrow down the possibilities and identify the compound.

Conclusion

So, there you have it! We successfully calculated the cell parameter of compound X (approximately 3.85 Å) and, through some detective work, identified it as most likely Bismuth Chloride (BiCl). This problem highlighted the importance of understanding crystal structures, unit cell geometry, and how to relate macroscopic properties like density to microscopic properties like ionic radii. Keep practicing these types of problems, guys, and you'll become solid-state chemistry whizzes in no time!