Question

If the anions (b) form hexagonal close packing and cations (a) occupy only \(2 / 3\) octahedral holes in it, then the general formula of the compound is (a) \(\mathrm{AB}_{2}\) (b) \(\mathrm{A}_{2} \mathrm{~B}_{3}\) (c) \(\mathrm{A}_{3} \overrightarrow{\mathrm{B}}_{3}\) (d) \(\mathrm{A}_{2} \mathrm{~B}_{2}\)

Short answer

The general formula of the compound is \(\mathrm{A}_{2} \mathrm{B}_{3}\) (option b).

Step-by-step solution

Understand Hexagonal Close Packing (HCP)

In hexagonal close packing, each unit cell contains 6 atoms of anions (B). This structure has octahedral and tetrahedral holes where cations can be accommodated.

Determine Number of Octahedral Holes

The number of octahedral holes in hexagonal close packing is equal to the number of anions present in the structure. Therefore, in our case, there are also 6 octahedral holes available.

Calculate Number of Cations

According to the problem, cations (A) occupy \(\frac{2}{3}\) of the available octahedral holes. Since there are 6 octahedral holes, the number of filled holes is \(\frac{2}{3} \times 6 = 4\). Thus, there are 4 cations.

Formulate the Chemical Composition

Based on the previous steps, we have 4 cations (A) and 6 anions (B). Therefore, the formula of the compound becomes \(\mathrm{A}_{4} \mathrm{B}_{6}\).

Simplify the Formula

Simplify \(\mathrm{A}_{4} \mathrm{B}_{6}\) by dividing both subscripts by their greatest common divisor, which is 2. This gives the chemical formula \(\mathrm{A}_{2} \mathrm{B}_{3}\).

Key concepts

Octahedral Holes

In the context of hexagonal close packing (HCP), octahedral holes play an essential role in determining how cations are distributed among anions. After constructing a structure where anions form a tightly-packed hexagonal arrangement, certain spaces called holes are left vacant. These holes or voids come in two types: octahedral and tetrahedral. Knowing how to calculate and use octahedral holes can simplify predicting the chemical formula of compounds.
- Octahedral holes in hexagonal close packing are equal in number to the anions.
- If there are 6 anions in the HCP unit, there are also 6 octahedral holes. Under typical scenarios, cations are placed in these octahedral holes. If only \(\frac{2}{3}\) of these octahedral holes are occupied, we need to multiply \(\frac{2}{3}\) by the number of available holes to find out how many are filled, which was calculated as 4 in the original step-by-step solution.

Chemical Formula Simplification

The simplification of chemical formulas is vital to make them more readable and easier to work with. In chemistry, this means reducing a complex formula to its simplest form while maintaining the same ratio of atoms.
To achieve simplification, one must identify the greatest common divisor (GCD) of the subscripts in the chemical formula and use it to reduce the numbers.
Here's how you do it: - Start with a detailed formula, such as \(A_4B_6\).
- Identify the GCD of the subscripts (in this case, 4 and 6) which is 2.
- Divide each subscript by this GCD. So, \(A_4B_6\) simplifies to \(A_2B_3\). This step makes the formula neat without changing its inherent properties.

Cation and Anion Ratio

Understanding the ratio of cations to anions is crucial in determining the properties and behavior of chemical compounds. The ratio not only defines the empirical formula but also affects the crystal structure and stability of the compound.
In this particular example, identifying the cation (A) to anion (B) ratio involves: - Counting the number of cations and anions involved in the structure.
- In the solution provided, there are 4 cations and 6 anions, leading to a \(4:6\) ratio.
- Simplifying this ratio leads to \(2:3\), which aligns with the simplified chemical formula \(A_2B_3\). The balance between cations and anions and their resulting ratios offer insights into the stoichiometry and balance of the compound.