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Chapter 15: Problem 27

The number of hexagonal faces that are present in a truncated octahedron is ___ .

Short Answer

Expert verified
The truncated octahedron has 8 hexagonal faces.

Step by step solution

01

Understanding the Truncated Octahedron

A truncated octahedron is a polyhedron that can be obtained by truncating (cutting off) the corners of an octahedron. It consists of faces that are either hexagonal or square.
02

Components of a Truncated Octahedron

A truncated octahedron is made up of two different types of faces: hexagons and squares. We need to identify each type to determine how many of them there are.
03

Identifying the Hexagonal Faces

For the truncated octahedron, each vertex of the original octahedron is truncated, creating a hexagonal face. Since an octahedron has 6 vertices and each is truncated to form a hexagonal face in the truncated form, the truncated octahedron contains 8 hexagonal faces.

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Key Concepts

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

Understanding Polyhedrons
A polyhedron is a three-dimensional shape with flat polygonal faces, straight edges, and vertices. These fascinating structures can take on numerous forms and have a rich presence in both mathematics and nature.
Think of a polyhedron as a multi-sided die where each face is a polygon. Polyhedrons are often classified based on their face shapes, numbers of edges, and vertices.
  • Convex Polyhedrons: All faces point outward, like cubes.
  • Concave Polyhedrons: Have some inward-pointing faces.
  • Regular Polyhedrons: Faces and angles are identical, like tetrahedrons.
  • Truncated Polyhedrons: Made by cutting (truncating) parts off other polyhedrons.
Polyhedrons feature prominently in geometric studies and have practical applications in fields ranging from architecture to molecular biology.
The Role of Hexagonal Faces
Hexagonal faces are a crucial aspect of many polyhedrons, including the truncated octahedron. A hexagon is a six-sided polygon, and in three-dimensional shapes, these faces provide stability and aesthetic appeal.
In the context of a truncated octahedron, hexagonal faces emerge from the process of truncating an octahedron's vertices. Each vertex cut generates a hexagonal face, replacing the original triangular faces of the octahedron.
  • Hexagons allow more efficient space packing, which is why honeycombs use them.
  • They maintain equal angles and side lengths, providing uniformity.
  • Truncated octahedrons specifically have 8 hexagonal faces, making them unique among polyhedrons.
Understanding hexagonal faces helps in visualizing and constructing complex geometrical models which have implications in various scientific and engineering disciplines.
Exploring the Octahedron
An octahedron is a type of polyhedron that has eight faces. It's a fascinating shape that can be visualized as two pyramids base-to-base.
This shape naturally features equilateral triangle faces, and it's known for its symmetry properties.
  • An octahedron has 6 vertices.
  • Its symmetrical nature allows equal division and even shaping.
  • It can be truncated to form other polyhedrons, like the truncated octahedron.
The octahedron's ease of division and application in modeling molecular geometry make it a staple in both academic and practical applications. Learning how truncating these shapes alters their properties and structures is key to understanding more advanced geometric concepts.

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Most popular questions from this chapter

A compound of formula \(\mathrm{A}_{2} \mathrm{~B}_{3}\) has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms: (a) hcp lattice \(-\mathrm{A}, \frac{2}{3}\) Tetrahedral voids \(-\mathrm{B}\) (b) hcp lattice \(-\mathrm{A}, \frac{1}{3}\) Tetrahedral voids \(-\mathrm{B}\) (c) hcp lattice \(-\mathrm{B}, \frac{2}{3}\) Tetrahedral voids \(-\mathrm{A}\) (d) hcp lattice \(-\mathrm{B}, \frac{1}{3}\) Tetrahedral voids \(-\mathrm{A}\)

A metal crystallises into two cubic phases, face centered cubic (FCC) and body centred cubic (BCC), whose unit cell lengths are \(3.5\) and \(3.0\) \(\AA\), respectively. Calculate the ratio of densities of \(\mathrm{FCC}\) and \(\mathrm{BCC}\).

Read the following statement (Assertion) and explanation (Reason) and answer each question as per the options given below : (a) If both assertion and reason are correct, and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct, but reason is not the correct explanation of the assertion. (c) If assertion is correct but reason is incorrect. (d) If assertion is incorrect but reason is correct. Assertion : In any ionic solid \([M X]\) with Schottky defects, the number of positive and negative ions are same. Reason : Equal number of cation and anion vacancies are present.

An element with molar mass \(2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}\) forms a cubic unit cell with edge length \(405 \mathrm{pm}\). If its density is \(2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\), the radius of the element is approximately \(\times 10^{-12} \mathrm{~m}\) (to the nearest integer).

In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space- filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. The number of atoms in the HCP unit cell is (a) 4 (b) 6 (c) 12 (d) 17
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