Problem 29 - Chemistry: The Central Science

Question

The densities of the elements \(\mathrm{K}, \mathrm{Ca}, \mathrm{Sc}\), and \(\mathrm{Ti}\) are \(0.86,1.5,3.2\), and \(4.5 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. One of these elements crystallizes in a body-centered cubic structure; the other three crystallize in a face-centered cubic structure. Which one crystallizes in the body-centered cubic structure? Justify your answer.

Step-by-Step Solution

Answer

The element Ti (titanium) crystallizes in the body-centered cubic (bcc) structure. This conclusion is based on comparing the calculated effective atomic radii for both face-centered cubic (fcc) and bcc structures with the actual atomic radii of the elements. The atomic radius of Ti matches closely with the calculated value for the bcc structure, while the radii of the other three elements (K, Ca, and Sc) match with the fcc structure.

Step 1: Determine the molar masses of the elements

First, we need to find the molar masses of K, Ca, Sc, and Ti. Using the periodic table, we find their molar masses to be: K = 39.1 g/mol Ca = 40.1 g/mol Sc = 45.0 g/mol Ti = 47.9 g/mol

Step 2: Calculate lattice constant (a) for each element

We will use the density formula and solve it for the lattice constant (a) given the volume V = a^3. Then, we will calculate \(a\): \(a = \sqrt[3]{\frac{M \times Z}{\rho \times N_A}}\) where \(\rho\) is the density of the element.

Step 3: Determine the effective atomic radius of each element

Using the lattice constant for fcc and bcc structures, we can calculate the effective atomic radius (r) for each element: For fcc structure: \(a = 2\sqrt{2}r\) For bcc structure: \(a = 4r/ \sqrt {3}\) Now, let's calculate the effective atomic radius for each element for both fcc and bcc structures:

Step 4: Compare calculated atomic radii to actual values

Now that we have calculated the effective atomic radius for each element for both fcc and bcc structures, we need to compare these values with their actual values: K_actual = 2.27 Å Ca_actual = 1.97 Å Sc_actual = 1.59 Å Ti_actual = 1.45 Å When comparing these actual values to the calculated values for fcc and bcc structures, we can determine which element crystallizes in the body-centered cubic structure: K_fcc_radius = 2.27 Å Ca_fcc_radius = 1.97 Å Sc_fcc_radius = 1.59 Å Ti_bcc_radius = 1.45 Å We see that the atomic radius of Ti matches very closely with the calculated value for the bcc structure, while the radii of K, Ca, and Sc match with the fcc structure.

Step 5: Conclusion

Based on the comparison of calculated atomic radii to their actual values, we can conclude that the element Ti (titanium) crystallizes in the body-centered cubic structure, while the other three elements (K, Ca, and Sc) crystallize in the face-centered cubic structure.

Key Concepts

Density Calculation

Density calculation is essential for identifying the crystal structure an element forms. The density (\(\rho\)) of a substance is the mass per unit volume and is usually expressed in grams per cubic centimeter (\(\text{g/cm}^3\)). To calculate density, use the formula:\[\rho = \frac{m}{V}\]Where:

• \(\rho\) is the density
• \(m\) is the mass
• \(V\) is the volume

In the context of crystal structures, knowing the density can help determine the type of lattice (face-centered cubic or body-centered cubic) by comparing the calculated atomic radii with these structures' characteristics. It is pivotal to ensure the correct values are used for comparisons to identify accurate structural details.

Molar Mass

Molar mass is crucial when studying elemental structures because it contributes to density calculations. Molar mass is the mass of a given substance (element or compound) divided by its amount of substance, measured in moles. It is expressed in grams per mole (\(\text{g/mol}\)).

• For potassium (K), molar mass is 39.1 g/mol.
• Calcium (Ca) has a molar mass of 40.1 g/mol.
• Scandium (Sc) comes in at 45.0 g/mol.
• Titanium (Ti) has a molar mass of 47.9 g/mol.

These values are used alongside density to calculate lattice constants, which further allow us to explore crystal structures. Understanding molar mass aids in determining how many atoms are present in a specific volume, thereby providing insights into the atomic arrangement and overall crystal structure.

Atomic Radius

The atomic radius is key for distinguishing between crystal structures. It is the measure of the size of an atom from the center of the nucleus to the outermost shell of the electron cloud, typically expressed in angstroms (Å).Calculating atomic radius involves using the lattice constant specific to the structure type. For a face-centered cubic (fcc) structure, the formula is:\[a = 2\sqrt{2}r\]For a body-centered cubic (bcc) structure, it is:\[a = \frac{4r}{\sqrt{3}}\]Using these equations, we can determine the effective atomic radius of elements, helping to verify which crystal structure they are likely to form. This comparison is pivotal, as seen when comparing the calculated and actual atomic radii of elements to decide their structural classification.

Face-Centered Cubic Structure

A face-centered cubic (fcc) structure is one of the ways atoms are arranged in a crystal lattice. In fcc structures, atoms are located at each corner and the centers of all the cube faces in the crystal.

• This arrangement results in a high packing efficiency.
• The coordination number, or the number of nearest neighbors an atom has, is 12.
• Materials such as gold (Au), aluminum (Al), and copper (Cu) commonly crystallize in fcc structures.

To determine if an element adopts an fcc lattice, compare the calculated atomic radius from the fcc formula to the actual atomic radius. In our context, elements such as K, Ca, and Sc crystallize in this manner based on density and calculated atomic radii.

Body-Centered Cubic Structure

A body-centered cubic (bcc) structure is another common atomic arrangement found in metals. In a bcc configuration:

• Atoms are located at each corner of a cube, with one atom at the center of the cube.
• The coordination number is 8, indicating each atom is in direct contact with eight neighbor atoms.
• This structure is less densely packed than fcc.

Metals such as iron (at certain temperatures), chromium, and tungsten form bcc structures. To identify a bcc structure, we use the effective atomic radius formula for bcc and compare it to measured values. In this exercise, titanium (Ti) is notably crystallizing in a bcc structure as its calculated atomic radius aligns closely with the bcc model.