Question

Determine if each statement is true or false: (a) Substitutional alloys are solid solutions, but interstitial alloys are heterogenous alloys. (b) Substitutional alloys have "solute" atoms that replace "solvent" atoms in a lattice, but interstitial alloys have "solute" atoms that are in between the "solvent" atoms in a lattice. (c) The atomic radii of the atoms in a substitutional alloy are similar to each other, but in an interstitial alloy, the interstitial atoms are a lot smaller than the host lattice atoms.\(12.32\) Sodium metal (atomic weight \(22.99 \mathrm{~g} / \mathrm{mol}\) ) adopts a bodycentered cubic structure with a density of \(0.97 \mathrm{~g} / \mathrm{cm}^{3}\). (a) Use this information and Avogadro's number \(\left(N_{A}=6.022 \times 10^{23} / \mathrm{mol}\right)\) to estimate the atomic radius of sodium. (b) If sodium didn't react so vigorously, it could float on water. Use the answer from part (a) to estimate the density of Na if its structure were that of a cubic close-packed metal. Would it still float on water?

Short answer

In summary, we found the atomic radius of sodium by utilizing its given information (atomic weight, density, and crystal structure). We also estimated the hypothetical density of sodium if its structure were cubic close-packed (CCP) and compared it to the density of water to determine if it could still float. The atomic radius of sodium (r) and the hypothetical density can be obtained by following the step-by-step solution and calculations.

Step-by-step solution

Find the Volume of One Unit Cell

First, let's use the given density and atomic weight of Na to find the volume and length of one unit cell. The formula to find the volume of one unit cell is: Volume = \(mass \div density\) Density of Na = 0.97 g/cm³ Atomic weight of Na = 22.99 g/mol Avogadro's number \(N_A = 6.022 \times 10^{23} / mol\) Now, we can find the mass of one Na atom: \(Mass = \frac{Atomic\_Weight}{N_A} = \frac{22.99}{6.022 \times 10^{23}}\) And, the volume of one unit cell is: \(Volume = \frac{Mass}{Density} = \frac{22.99}{0.97 (6.022 \times 10^{23})}\)

Find the Length of One Unit Cell

Sodium has a body-centered cubic (BCC) structure, which means we have 2 atoms per unit cell. With this information, we can find the length (a) of one unit cell using the formula: \(a = \sqrt[3]{\frac{Volume}{2}}\) By substituting the Volume we found in step 1, we can calculate the value of 'a'.

Calculate the Atomic Radius of Sodium

To find the atomic radius of sodium (r), we can use the relationship for a BCC structure: \(a = \frac{4r}{\sqrt{3}}\) By substituting the value of 'a' found in step 2, we can find the atomic radius of sodium.

Estimate the Hypothetical Density of Sodium

If sodium's structure was cubic close-packed (CCP), there would be 4 atoms per unit cell instead of 2, as in BCC. The formula to find density with the new structure is: \(Density = \frac{mass \times 4}{volume}\) By substituting the values for mass and volume from step 1, we can find the hypothetical density of sodium.

Determine if Sodium Would Still Float on Water

To determine if sodium would still float on water, compare its hypothetical density to the density of water (1 g/cm³). If the hypothetical density of sodium is less than the density of water, it will float. Otherwise, it will sink.

Key concepts

Substitutional Alloys

Substitutional alloys are a type of alloy where the atoms of one element replace some of the atoms of another in a crystal lattice. They typically form between elements with similar chemical properties and atomic sizes, such as copper and zinc in brass. Substitutional alloying is predominant when the solute has an atomic radius within about 15% of that of the solvent. This creates a solid solution that maintains the overall structure but introduces different physical and chemical properties, like increased strength or corrosion resistance.

For accurate comprehension, remember that unlike interstitial alloys, substitutional alloys are homogeneous at the atomic level, meaning there's a regular pattern to how the foreign atoms replace the host atoms.

Interstitial Alloys

In contrast, interstitial alloys are formed when smaller atoms fill the spaces or 'interstices' between larger atoms in a lattice. Common examples include carbon atoms in the interstices of iron in steel. Due to the size difference, the smaller atoms do not replace the larger lattice atoms but rather squeeze in between them. This often leads to distorted crystal structures, which can greatly increase the hardness and strength of the material.

The smaller interstitial atoms occupy the voids in the lattice without disturbing the larger host atoms. This difference is important when considering mechanical and thermal properties, as interstitial alloying typically prevents slippage in the lattice, making these alloys harder, and sometimes more brittle.

Body-Centered Cubic Structure

Metals such as sodium often crystallize in specific geometric arrangements called crystal lattices. The body-centered cubic (BCC) structure is one form where atoms are located at the eight corners of a cube and a single atom is at its center. This arrangement accounts for a lower packing density compared to other structures like face-centered cubic (FCC), which results in different physical properties.

For BCC, the coordination number — which is the number of nearest neighbors to an atom — is eight. Understanding the BCC structure is key when estimating atomic radii and calculating density because it directly affects these properties. Furthermore, the BCC arrangement influences the mechanical properties of a metal, such as ductility and tensile strength.

Avogadro's Number

Avogadro's number, roughly equal to \(6.022 \times 10^{23}\), is a fundamental constant and represents the number of atoms, ions, or molecules in one mole of a substance. It serves as a bridge between the atomic scale and the macroscopic scale, allowing chemists to count particles by weighing them.

When dealing with atomic radii estimation or density calculations for a crystalline solid, Avogadro's number is crucial. It helps convert mass measurements into a count of atoms, facilitating volume or density calculations in terms of individual atoms or molecules within a lattice structure.

Atomic Radius Estimation

Estimating the atomic radius of an element within a crystal lattice involves understanding the geometry of the lattice and applying geometric and algebraic relationships. As seen in the sodium exercise, we can calculate the edge length of the BCC unit cell and then use the relationship between the edge length and the atomic radius. This process often requires knowledge of the material's density and crystal geometry, reflecting how closely atoms are packed.

For BCC lattice structures, where there are two atoms per unit cell, the atomic radius is inferred from the cube's dimensions that the atoms occupy. Factors such as coordination number and the number of atoms per unit cell play pivotal roles in these calculations. Moreover, once the radius is known, it allows for other properties to be inferred, like the metallic bond strength.

Density Calculations

Density calculations for a crystal lattice require the mass of atoms present in a unit cell and the volume of the cell. By incorporating Avogadro’s number, the mass of a single atom can be determined. You calculate the volume of a unit cell by taking the atomic radius and applying it to the geometric formula that defines the unit cell's shape.

With sodium's exercise, it becomes clear how altering the number of atoms in a lattice (such as going from a BCC to a CCP structure) changes the density. In practical terms, if the density of a metal is less than that of water, the metal will float; hence, the consideration of structure is fundamental not only to the intrinsic properties of the metal but also to its behavior in different environments.