Problem 118 - Chemistry The Molecular Science

Question

Barium sulfide(s) has the \(\mathrm{NaCl}\) structure and a density of \(4.25 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the interionic distance and compare this value with the sum of the ionic radii \(\left(\mathrm{Ba}^{2+}=\right.\) \(\left.149 \mathrm{pm} ; \mathrm{S}^{2-}=170 \mathrm{pm}\right)\)

Step-by-Step Solution

Answer

Interionic distance calculated is 198 pm, compared to 319 pm from ionic radii sum.

Step 1: Understanding Given Data

First, let's summarize the given information. Barium sulfide (BaS) has a density of \(4.25 \mathrm{~g/cm^3}\) and follows the \(\mathrm{NaCl}\) structure. The ionic radii are given as \(\text{Ba}^{2+} = 149 \mathrm{~pm}\) and \(\text{S}^{2-} = 170 \mathrm{~pm}\). The task is to calculate the interionic distance and compare it with the sum of the ionic radii.

Step 2: Calculating Molar Mass

Calculate the molar mass of BaS. The molar mass of Ba is approximately \(137.33 \mathrm{~g/mol}\) and that of S is \(32.07 \mathrm{~g/mol}\). Therefore, the molar mass of BaS is \(137.33 + 32.07 = 169.40 \mathrm{~g/mol}\).

Step 3: Volume Calculation of Unit Cell

In a \(\mathrm{NaCl}\) structure, there are 4 formula units per unit cell. Using the given density and molar mass, compute the volume of the unit cell: \[ \text{Volume} = \frac{169.40 \text{ g/mol}}{4.25 \text{ g/cm}^3} \times \frac{1}{6.022 \times 10^{23} \text{ mol}^{-1}} = 6.64 \times 10^{-23} \text{ cm}^3 \]

Step 4: Calculating Edge Length of Unit Cell

The volume of the cubic unit cell is \(a^3\), where \(a\) is the edge length. Solving for \(a\), we have: \[ a = \sqrt[3]{6.64 \times 10^{-23} \text{ cm}^3} = 3.96 \times 10^{-8} \text{ cm} \]

Step 5: Converting Edge Length to Ionic Distance

Convert the edge length from \(\text{cm}\) to \(\text{pm}\): \[ a = 3.96 \times 10^{-8} \text{ cm} \times 10^{10} \text{ pm/cm} = 396 \text{ pm} \] In a \(\mathrm{NaCl}\) structure, the ionic distance is half the edge length, thus: \[ \text{Interionic distance} = \frac{396 \text{ pm}}{2} = 198 \text{ pm} \]

Step 6: Comparison with Sum of Ionic Radii

Sum the given ionic radii of \(\text{Ba}^{2+}\) and \(\text{S}^{2-}\): \[ 149 \text{ pm} + 170 \text{ pm} = 319 \text{ pm} \] Compare this value with the interionic distance of \(198 \text{ pm}\).

Key Concepts

Interionic Distance

Interionic distance refers to the physical space between two adjacent ions in a crystal structure. For barium sulfide (\(\text{BaS}\) in this context), this distance is crucial to understanding its crystal structure and stability. Knowing the density and ionic radii helps in computing the interionic distance.

• First, we calculate the volume of a unit cell using the molar mass and density.
• This volume helps derive the cubic edge length \(a\) of the unit cell, as \(a\) is the cube root of the volume.
• In a cubic crystal structure like NaCl, the interionic distance is half the edge length, because ions alternate along the cube's lines.

In this example, we calculate the interionic distance to be \(198\,\mathrm{pm}\). This simple calculation helps in understanding the arrangement of ions, providing insights into the compound's properties.

NaCl Crystal Structure

The NaCl crystal structure, also known as a face-centered cubic (FCC) lattice, is a common arrangement for ionic compounds like sodium chloride and barium sulfide. In this structure:

• Ions are arranged in a repeating pattern, where each sodium ion is surrounded by six chloride ions and vice versa, forming an octahedral geometry.
• This means the unit cell contains four formula units of the compound, involving a symmetrical and dense packing of ions.

The NaCl lattice is significant for its stability and optimal space utilization, balancing the forces of attraction and repulsion among ions effectively. By mimicking this structure, compounds exhibit high density, stability, and a clear correlation between structure and interionic distance.
For BaS, adopting the NaCl structure explains its ionic interactions and density consistency.

Barium Sulfide

Barium sulfide is a compound that showcases unique properties, influenced significantly by its crystal structure and ionic composition. Here are some key points about barium sulfide:

• It adopts the \(\text{NaCl}\) crystal structure, which imparts high density \((4.25 \,\text{g/cm}^{3})\) and mechanical strength.
• The ionic radii of \(\text{Ba}^{2+}\) and \(\text{S}^{2-}\) are crucial for predicting the compound's behavior when calculating interionic distances.
• Despite the predicted sum of ionic radii \(319\,\text{pm}\), the measured interionic distance is \(198\,\text{pm}\). This discrepancy demonstrates how crystal structure affects measured distances compared to theoretical sums.

Understanding these properties is essential for explanations in materials science and chemistry, as they underscore the interplay between structure and function in ionic compounds like barium sulfide.