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Nickel has an fec unit cell with an edge length of \(350.7 \mathrm{pm}\). Calculate the radius of a nickel atom.

Short Answer

Expert verified
Answer: The atomic radius of a Nickel atom in the given fcc unit cell is approximately 123.9 pm.

Step by step solution

01

Understanding the fcc unit cell

An fcc unit cell contains atoms at the corners of the cube and at the center of each face. There are 8 corner atoms and 6 face atoms, but only a fraction of each atom is within the unit cell. Each corner atom belongs to 8 adjacent unit cells, and thus, contributes 1/8th atom per unit cell. On the other hand, each of the 6 face atoms contributes 1/2 atom per unit cell. So the fcc unit cell contains the equivalent of 4 atoms per unit cell.
02

Find the relationship between edge length and atomic radius

In order to find the atomic radius, we can analyze the relationship between atoms adjacent along a face diagonal. A face diagonal can be represented as a line that connects two opposite face atoms passing through the center of the edge atom. The length of the face diagonal consists of 4 atomic radii, i.e., two for each face atom and one for the center atom. The face diagonal can be obtained by using Pythagorean Theorem as the diagonal connects two opposite corner atoms in a unit square being part of the base as well as the diagonal of the unit cell. So in terms of the edge length (a), the face diagonal is given by: Face diagonal = \(\sqrt{2}a\) Now, the relationship is: Face diagonal = 4 * Atomic Radius
03

Calculate the atomic radius of Nickel

Given that the edge length (a) is 350.7 pm, we can now find the radius of a Nickel atom using the relationship derived in Step 2. 4 * Atomic Radius = \(\sqrt{2}a\) Atomic Radius= \(\dfrac{\sqrt{2}a}{4}\) Substitute the value of a (350.7 pm) into the equation: Atomic Radius = \(\dfrac{\sqrt{2}(350.7 \mathrm{pm})}{4}\) Now, calculate the value: Atomic Radius ≈ \(\dfrac{350.7\mathrm{pm} \times 1.414}{4} = \dfrac{495.5 \mathrm{pm}}{4} ≈ 123.9 \mathrm{pm}\) Therefore, the radius of a Nickel atom is approximately 123.9 pm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Face-Centered Cubic Unit Cell
The face-centered cubic (fcc) unit cell is a type of crystal structure often seen in metals like nickel. In an fcc unit cell, atoms are positioned at each of the eight corners and at the center of each of the six faces of the cube. This arrangement leads to a highly efficient packing structure. Because these are whole atoms shared with adjacent unit cells, the contribution of these corner and face atoms must be considered:
  • Each corner atom is shared by eight neighboring unit cells, hence it only contributes 1/8th of its volume to the given unit cell.
  • Each face-centered atom is shared with one neighboring cell, thus contributing 1/2 to the volume of the unit cell.
This results in a total equivalent of four full atoms per face-centered cubic unit cell. Understanding this configuration is vital as it forms the basis for calculating characteristics like atomic radius and density when considering materials with an fcc arrangement.
Atomic Structure
The atomic structure involves understanding how atoms, the basic building blocks of matter, are arranged in space within a material. In crystalline solids, like metals, the atoms are arranged in a orderly repeating pattern termed a lattice. Each atom acts as a point in this lattice, creating a specific structure like the face-centered cubic (fcc) structure mentioned earlier. The atoms within the lattice interact with each other through chemical bonds, ultimately determining the properties of the material such as density and rigidity.
When calculating properties related to atomic structure, like the atomic radius in an fcc unit cell, one utilizes geometric relationships between atoms. For instance, the atoms are packed similarly to how spheres are arranged, and thus the relationships derived from geometry (for instance, using the Pythagorean theorem for diagonals) help in deducing the atomic dimensions. This geometric insight is crucial when converting the theoretical models into actual physical dimensions.
Unit Cell Geometry
Geometry within a unit cell is all about understanding and applying mathematical relationships to the structure of the cell. A unit cell is the simplest repeating unit in a crystalline structure, and its geometry effectively depicts the arrangement and distance between atoms.
In an fcc unit cell, the key geometric feature is the face diagonal, which links opposite corner atoms in a face. Utilizing the Pythagorean Theorem, one finds that if the edge length of the cube is denoted as \(a\), the face diagonal is \(\sqrt{2}a\). Since the diagonal spans across face-centered atoms and central atoms, it neatly equates to four times the atomic radius, as there are two half-atoms on each end plus a full atom passing through the center.
By understanding the link between face diagonal and atomic radius through unit cell geometry, it's possible to deduce metrics such as atomic radius (important in identifying potential reactivity and bonding characteristics) from given structures, crucial for both academic and practical applications in materials science.

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Most popular questions from this chapter

Europium, one of the lanthanide elements used in television screens, crystallizes in a crystal lattice built on bcc unit cells, with a unit cell edge of \(240.6 \mathrm{pm}\). Calculate the radius of a europium atom.

The center of Earth is composed of a solid iron core within a molten iron outer core. When molten iron cools, it crystallizes in different ways depending on pressure-in a bcc unit cell at low pressure and in a hexagonal unit cell at high pressures like those at Earth's center. a. Calculate the density of bec iron given that the radius of an iron atom is \(126 \mathrm{pm}\) b. Calculate the density of hexagonal iron given a unit cell volume of \(5.414 \times 10^{-23} \mathrm{cm}^{3}\) c. Seismic studies suggest that the density of Earth's solid core is only about \(90 \%\) of that of hexagonal Fe. Laboratory studies have shown that up to \(4 \%\) by mass of Si can be substituted for Fe without changing the hcp crystal structure built on hexagonal unit cells. Calculate the density of such a crystal.

Ice under Pressure Kurt Vonnegut's novel Cat's Cradle describes an imaginary, high-pressure form of ice called "ice-nine." With the assumption that ice-nine has a cubic closest-packed arrangement of oxygen atoms with hydrogen atoms in the appropriate holes, what type of hole will accommodate the \(\mathrm{H}\) atoms?

A unit cell consists of a cube that has an ion of element X at each corner, an ion of element \(Y\) at the center of the cube, and an ion of element \(Z\) at the center of each face. What is the formula of the compound?

If the unit cell of a substitutional alloy of copper and tin has the same unit cell edge as the unit cell of copper, will the alloy have a greater density than copper? Explain your answer.

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