Question

Atoms of the element 'A' form HCP and atoms of element ' \(\mathrm{C}\) ' occupy only two-third of octahedral voids in it, then the general formula of the compound is (a) \(\mathrm{CA}\) (b) \(\mathrm{CA}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{~A}_{3}\) (d) \(\mathrm{C}_{3} \mathrm{~A}_{2}\)

Short answer

The general formula of the compound is \(\mathrm{C}_{2} \mathrm{~A}_{3}\).

Step-by-step solution

- Calculate the number of atoms in HCP

In a hexagonal close-packed (HCP) structure, there are 6 atoms at corners shared by 6 unit cells, 2 atoms at top and bottom faces completely inside the unit cell, and 12 atoms on the edges shared by 4 unit cells, giving \(\frac{6}{6} + 2 + \frac{12}{4} = 6\) atoms per unit cell for element A.

- Determine the number of octahedral voids

In an HCP structure, the number of octahedral voids is equal to the number of atoms in the unit cell, which is 6 for element A.

- Calculate the number of C atoms in octahedral voids

With 6 octahedral voids and only two-thirds being occupied by element C, the number of C atoms in the unit cell is \(\frac{2}{3} \times 6 = 4\).

- Determine the ratio of C to A atoms in the compound

The ratio of C atoms to A atoms in the compound is 4:6, which can be simplified by dividing by their greatest common divisor, 2, to get the ratio 2:3.

- Write the general formula

With the simplified ratio of 2:3, the general formula for the compound formed by element C occupying two-thirds of the octahedral voids in an HCP of element A is \(\mathrm{C}_{2} \mathrm{~A}_{3}\).

Key concepts

HCP (hexagonal close-packed)

Understanding the hexagonal close-packed (HCP) structure is essential in the field of crystallography and materials science. In an HCP structure, the atoms align in a way that allows them to occupy the maximum space possible, similar to stacking oranges as tightly as possible on a market stand. This leads to high packing efficiency.

An HCP crystal consists of two layers, where the top layer atoms fit into the depressions of the bottom layer, creating a repeating ABAB pattern. If you were to look straight down onto one of the layers, you would see that each atom is surrounded by six others, forming a hexagon, which is why we call it 'hexagonal' close packing.

There's a bit of math involved when we count the atoms within a single unit cell of an HCP structure. We take into account that some atoms at the corners or edges are shared between multiple unit cells, leading to fractions of atoms when calculated for just one unit cell. As a result, we find that there are effectively 6 full atoms per unit cell in an HCP structure.

Octahedral Voids

As part of understanding crystal structures, knowing about octahedral voids is just as important as understanding the atomic arrangement. These voids are spaces between atoms that can potentially be occupied by other atoms or ions, depending on the compound's chemical makeup. The shape of these spaces resembles an octahedron – a polyhedron with eight faces.

In an HCP structure, each unit cell has a set number of octahedral voids which in a perfectly packed structure, corresponds to the number of atoms in the unit cell itself. It's as though every atom wants to be neighborly and provides one room for another periodic guest. In simpler terms, if we have 6 atoms per unit cell in an HCP, we also have 6 octahedral voids ready to be occupied based on the stoichiometry of the specific compound being formed.

Stoichiometry

Stoichiometry is a section of chemistry that feels like a mix between a detective's puzzle and a baker's recipe. Here, we're not baking a cake, but we're definitely mixing things in specific ratios to get a new compound. Stoichiometry allows us to understand the proportions of each element that take part in a chemical reaction or form a crystalline structure.

In the context of our HCP structure, stoichiometry comes into play when we begin to fill the octahedral voids with atoms of a different element. We don't just randomly sprinkle them in; there's a precise ratio that must be followed to maintain the integrity of the crystal. This ratio is determined by the number of octahedral voids and the portion of these voids that are filled. In the given problem, this is where the two-thirds comes into play, highlighting the precision needed to solve such crystal structure chemistry problems.

Unit Cell

In the vast crystal lattices that extend in three dimensions, we identify a repetitive building block called the 'unit cell'. Think of it as a Lego piece from which the entire structure is built by stacking and repeating it in space. Each unit cell in an HCP structure is like a three-dimensional puzzle, with atoms at specific locations that, when repeated, form the entire crystalline substance.

Diving deeper, examining the unit cell helps us to uncover the number of atoms present, positions of voids, and how another element might fit into the crystal when forming a compound. With this understanding, we're able to calculate specific formulas for compounds based on how many unit cells are occupied and how, akin to predicting how many bricks are needed to build a wall.