Problem 49 - Chemistry and Chemical Reactivity

Question

Silver crystallizes in a face-centered cubic unit cell. Each side of the unit cell has a length of 409 pm. What is the radius of a silver atom? (Hint: Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is, each face atom is touching the four corner atoms.)

Step-by-Step Solution

Answer

The radius of a silver atom in an FCC unit cell is approximately 144.5 pm.

Step 1: Understanding the Face-Centered Cubic (FCC) Structure

In an FCC unit cell, atoms are located at each of the corners and at the center of each face of the cube. The atoms touch along the face diagonal rather than along the edge.

Step 2: Relate Diagonal Length to Atomic Radius

In the FCC unit cell, the face diagonal has four atomic radii. The diagonal can be expressed as \( 4r \), where \( r \) is the atomic radius. The relationship is derived from the geometry of the FCC structure.

Step 3: Calculate the Face Diagonal

The face diagonal of a cube is related to its side length (a). For a cube with side length 409 pm, the face diagonal can be calculated using the Pythagorean theorem: \( \sqrt{a^2 + a^2} = a\sqrt{2} \).

Step 4: Calculate the Atomic Radius

Equate the expression for the face diagonal to \( 4r \): \[ a\sqrt{2} = 4r \]. Substituting the side length, \( 409\text{ pm} \), we have: \[ 409\sqrt{2} = 4r \]. Solve for \( r \): \( r = \frac{409\sqrt{2}}{4} \approx 144.5 \text{ pm} \).

Step 5: Solution Verification

Verify the calculation by reviewing each step and ensuring the relationship between face diagonal and radius is correctly applied. The calculation confirms that the radius of a silver atom is approximately 144.5 pm.

Key Concepts

Atomic Radius

The atomic radius is an essential concept when examining crystals such as face-centered cubic (FCC) structures. It refers to the distance from the nucleus of an atom to the outer boundary of its surrounding electron cloud. In the context of metallic crystals like silver (Ag), knowing the atomic radius helps us understand how atoms pack in the unit cell.
The atomic radius can be calculated using geometric or mathematical relationships derived from the unit cell’s dimensions. In the FCC structure, atoms touch each other along the face diagonal, and this relationship directly influences how we determine the atomic radius given by the equation: \[\text{Face Diagonal} = 4r\]where \(r\) is the atomic radius.

Unit Cell Geometry

The unit cell is the basic repeating structural unit that creates the overall crystal lattice. For face-centered cubic (FCC) unit cells, the arrangement involves atoms at each of the eight corners of the cube and one atom at the center of each of the cube's six faces.
The geometry of the FCC unit cell implies that the atoms do not touch along the edges but rather along the face diagonals. This means calculating features like the atomic radius requires understanding this geometry. The edge length of the cubic unit cells, such as the one given for silver with an edge of 409 pm, plays a crucial role in further calculations.

Silver Atom

Silver (Ag) is a metal with fascinating crystallography, particularly because it crystallizes in a face-centered cubic lattice. This metallic element is significant because of its high electrical conductivity and malleable properties, making it useful in various technological applications.
In its crystalline form, each silver atom plays a part in forming the stable structure of the FCC lattice. When examining silver in this context, understanding the atomic radius and how the atoms are arranged within the unit cell elevates our grasp of its physical properties. Thus, this exercise of calculating atomic radius directly relates to predicting how silver behaves under different conditions.

Diagonal Calculation

In the examination of face-centered cubic (FCC) structures, diagonal calculations become paramount. The face diagonal is crucial to determining the atomic radius because atoms touch along the diagonal, not the cube edges.
Using the Pythagorean theorem, the face diagonal \(d_f\) for a cubic unit cell with side length \(a\) is given by:\[d_f = a\sqrt{2}\]
With the face diagonal containing four atomic radii, the relationship becomes \[a\sqrt{2} = 4r\]Solving this equation allows one to find the atomic radius \(r\), connecting geometrical calculations and physical atomic dimensions. For example, substituting the edge size 409 pm produces the silver atomic radius, approximately 144.5 pm.