Problem 156
Question
Rubidium chloride has the sodium chloride structure at normal pressures but assumes the cesium chloride structure at high pressures. (See Exercise \(71 . )\) What ratio of densities is expected for these two forms? Does this change in structure make sense on the basis of simple models? The ionic radius is 148 \(\mathrm{pm}\) for \(\mathrm{Rb}^{+}\) and 181 \(\mathrm{pm}\) for \(\mathrm{Cl}^{-} .\)
Step-by-Step Solution
Verified Answer
The ratio of densities between Rubidium Chloride's CsCl and NaCl structures is approximately 1.6. This change in structure makes sense based on simple models, as the CsCl structure has a higher density than the NaCl structure. Higher pressures usually result in denser arrangements of atoms or ions in solid materials, which explains the transition from NaCl to CsCl structure under high pressure.
1Step 1: Determine cell parameters for NaCl and CsCl structures
We have the following information:
Ionic radius of Rb+ = 148 pm
Ionic radius of Cl- = 181 pm
NaCl structure:
In the NaCl structure, the cell edge length 'a' can be determined using the sum of the ionic radii of the two ions (since they touch each other along the edge). Therefore,
a (NaCl) = 2 × (Rb+ radius + Cl- radius) = 2 × (148 pm + 181 pm) = 658 pm
CsCl structure:
In the CsCl structure, the cell edge length 'a' can be determined using the diagonal of the unit cell, as the ions touch each other along the body diagonal.
The body diagonal is equal to the sum of the ionic radii of the two ions:
Body diagonal (CsCl) = Rb+ radius + Cl- radius = (148 pm + 181 pm) = 329 pm
Since the body diagonal is related to the edge length 'a' by:
Body diagonal = a × √3
We can find the edge length 'a' for the CsCl structure:
a (CsCl) = Body diagonal / √3 = 329 pm / √3 ≈ 190 pm
2Step 2: Calculate the unit cell volumes for NaCl and CsCl structures
Now that we have the edge lengths for both structures, we can calculate the volume of each unit cell.
Volume (NaCl) = a³ (NaCl) = (658 pm)³ = 2.84 × 10¹² pm³
Volume (CsCl) = a³ (CsCl) = (190 pm)³ = 6.86 × 10⁹ pm³
3Step 3: Calculate the densities of the two structures
The formula mass of Rubidium Chloride (RbCl) is:
Rb = 85.47 g/mol
Cl = 35.45 g/mol
RbCl = 120.92 g/mol
Using Avogadro's number, we can calculate the number of RbCl formula units per unit cell:
NaCl structure: 1 Rb+ and 1 Cl- per unit cell, so 1 RbCl formula unit per unit cell
CsCl structure: 1 Rb+ and 1 Cl- per unit cell, so 1 RbCl formula unit per unit cell
To calculate the densities, we use the following formula:
Density = (Formula mass × Number of formula units × Avogadro's number) / (Unit cell volume × 10^(-24)), as 1 cm³ = 10²¹ pm³
Density (NaCl) = (120.92 g/mol × 1 × 6.022 × 10²³ units/mol) / (2.84 × 10¹² pm³ × 10^(-24)) ≈ 2.58 g/cm³
Density (CsCl) = (120.92 g/mol × 1 × 6.022 × 10²³ units/mol) / (6.86 × 10⁹ pm³ × 10^(-24)) ≈ 4.14 g/cm³
4Step 4: Calculate the ratio of densities
Now that we have the densities for both crystal structures, we can calculate the ratio of densities.
Ratio of densities = Density (CsCl) / Density (NaCl) = 4.14 g/cm³ / 2.58 g/cm³ ≈ 1.6
5Step 5: Discuss whether the change in structure makes sense based on simple models
The change from the NaCl structure to the CsCl structure with increasing pressure makes sense because the CsCl structure has a higher density than the NaCl structure, meaning the ions are packed more closely in the CsCl structure. This makes sense, as higher pressures usually result in a denser arrangement of atoms or ions in solid materials.
Key Concepts
Density CalculationUnit CellIonic RadiusPhase Transition
Density Calculation
Understanding the density calculation in crystal structures is essential for predicting how a substance behaves under different conditions. Density is much like a substance's weight in relation to the space it occupies.
Here's how you calculate it:
Here's how you calculate it:
- First, find the formula mass, which is the total mass of all atoms in a compound. For example, in rubidium chloride (RbCl), Rb has a molar mass of 85.47 g/mol, and Cl has 35.45 g/mol, totaling 120.92 g/mol.
- Next, determine how many formula units fit into a unit cell; here, there's one per unit cell.
- Then, find each unit cell's volume, given in pm³.
- Using the density formula: Density = \( \frac{{\text{{Formula mass}} \times \text{{Number of formula units}} \times \text{{Avogadro's number}}}}{{\text{{Unit cell volume}} \times 10^{-24}}} \), where 1 cm³ equals \( 10^{21} \, \text{pm}^3 \).
Unit Cell
A unit cell is the smallest repeating pattern that shows the full symmetry of the structure of a substance.
By understanding unit cells, we illuminate how substances form their crystal lattices.
Especially in structures, like sodium chloride, ions adhere to each other along the edges, while in cesium chloride, they align along the body's diagonal.
Finding the edge length 'a' is paramount:
By understanding unit cells, we illuminate how substances form their crystal lattices.
Especially in structures, like sodium chloride, ions adhere to each other along the edges, while in cesium chloride, they align along the body's diagonal.
Finding the edge length 'a' is paramount:
- For NaCl, ions touch along the edge, so calculate as \( a = 2 \times (\text{Rb}^+ \text{radius} + \text{Cl}^- \text{radius}) \).
- For CsCl, ions meet along the body diagonal, meaning the diagonal equals the sum of both radii. Solve \( a = \frac{{\text{Body diagonal}}}{\sqrt{3}} \).
Ionic Radius
The ionic radius is an ion's size, expressed in picometers (pm), determined by an atom's electron shell.
It's crucial in estimating how ions will fit together within a crystal structure. For rubidium chloride, we have:
As ions join, they must share space, with both ionic radii contributing to distances like edge lengths and diagonals.
Under conditions like high pressure, the ability of ions to snugly fit in closer proximity hints why rubbing chloride adopts a different structure.
With pressures altering spatial dynamics, smaller radii pack more, explaining changes from NaCl to CsCl structure.
It's crucial in estimating how ions will fit together within a crystal structure. For rubidium chloride, we have:
- \( \text{Rb}^+ \) with 148 pm
- \( \text{Cl}^- \) with 181 pm
As ions join, they must share space, with both ionic radii contributing to distances like edge lengths and diagonals.
Under conditions like high pressure, the ability of ions to snugly fit in closer proximity hints why rubbing chloride adopts a different structure.
With pressures altering spatial dynamics, smaller radii pack more, explaining changes from NaCl to CsCl structure.
Phase Transition
A phase transition in crystal structures, like rubidium chloride's shift from NaCl to CsCl, demonstrates atomic or ionic rearrangement when external conditions, like pressure, change.
Under rising pressure, atoms or ions adjust into denser formations for stability. Rubidium chloride exemplifies this with its structure variations.
Explaining a few scenarios promotes understanding of phase transitions:
These insights from phase transitions offer a picture of how materials respond robustly and precisely within varied environments, frequently predicted through conceptual models.
Under rising pressure, atoms or ions adjust into denser formations for stability. Rubidium chloride exemplifies this with its structure variations.
Explaining a few scenarios promotes understanding of phase transitions:
- An increased pressure compresses ion structures, favoring tighter configurations.
- Tighter structures, as seen in the CsCl layout, use space more efficiently, resulting in higher densities.
These insights from phase transitions offer a picture of how materials respond robustly and precisely within varied environments, frequently predicted through conceptual models.
Other exercises in this chapter
Problem 153
Some water is placed in a sealed glass container connected to a vacuum pump (a device used to pump gases from a container), and the pump is turned on. The water
View solution Problem 155
For a simple cubic array, solve for the volume of an interior sphere (cubic hole) in terms of the radius of a sphere in the array.
View solution Problem 159
One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at \(600 .\) C followed by condensation of the mercury vapor. Give
View solution Problem 160
General Zod has sold Lex Luthor what Zod claims to be a new copper-colored form of kryptonite, the only substance that can harm Superman. Lex, not believing in
View solution