Atoms of element B form hcp lattice and those of the element A occupy 2/3 rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B?
Understanding Crystal Structures and Compound Formulas
This question asks us to determine the chemical formula of a compound based on how its constituent atoms are arranged in a crystal lattice. One element forms the main lattice structure, while the other occupies specific positions called voids within that structure.
Analyzing the Given Information
- Element B forms a hexagonal close-packed (hcp) lattice.
- Element A occupies 2/3rd of the tetrahedral voids in this lattice.
Step-by-Step Solution
Step 1: Determine the Number of Atoms in the hcp Lattice
In a hexagonal close-packed (hcp) structure, the effective number of atoms per unit cell is 6. These atoms form the lattice framework. Since element B forms the hcp lattice, the number of B atoms per unit cell is 6.
Number of B atoms = 6
Step 2: Determine the Total Number of Voids in the hcp Lattice
In any close-packed structure (including hcp and ccp), there are two types of voids: tetrahedral voids and octahedral voids.
- The number of tetrahedral voids is twice the number of atoms in the lattice.
- The number of octahedral voids is equal to the number of atoms in the lattice.
Number of atoms in lattice (N) = 6 (for element B)
Total number of tetrahedral voids = 2 × N = 2 × 6 = 12
Total number of octahedral voids = N = 6
In this problem, element A occupies tetrahedral voids.
Step 3: Calculate the Number of Atoms of Element A
Element A occupies 2/3rd of the tetrahedral voids. The total number of tetrahedral voids is 12.
Number of A atoms = (Fraction of voids occupied) × (Total number of tetrahedral voids)
Number of A atoms = $\frac{2}{3} \times 12$
Number of A atoms = $8$
Step 4: Determine the Ratio of Atoms A to Atoms B
We have found that there are 8 atoms of element A and 6 atoms of element B in the effective unit cell representation of the compound.
Ratio of A : B = Number of A atoms : Number of B atoms
Ratio of A : B = 8 : 6
Step 5: Simplify the Ratio to Find the Formula
To get the simplest whole-number ratio, we divide both numbers by their greatest common divisor, which is 2.
Simplified ratio of A : B = $\frac{8}{2} : \frac{6}{2}$
Simplified ratio of A : B = 4 : 3
This ratio represents the relative number of A and B atoms in the compound.
Step 6: Write the Formula of the Compound
The formula of the compound is written by representing the elements with their symbols and the simplified ratio as subscripts.
Formula = A$_{\text{ratio of A}}$ B$_{\text{ratio of B}}$
Formula = A$_4$ B$_3$
Thus, the formula of the compound formed by elements A and B is A$_4$B$_3$.
Summary of Calculations
| Component | Quantity | Calculation/Reason |
|---|---|---|
| Number of B atoms (hcp lattice) | 6 | Effective number of atoms per hcp unit cell |
| Total tetrahedral voids | 12 | 2 × Number of B atoms (2N) |
| Fraction of tetrahedral voids occupied by A | $\frac{2}{3}$ | Given in the question |
| Number of A atoms | 8 | $\frac{2}{3} \times 12$ |
| Ratio of A : B atoms | 8 : 6 | From number of A and B atoms |
| Simplified ratio of A : B atoms | 4 : 3 | Dividing by 2 |
| Compound Formula | A$_4$B$_3$ | Based on the simplified ratio |
Revision Table: Crystal Lattice Concepts
| Lattice Type | Effective Atoms per Unit Cell (N) | Number of Tetrahedral Voids | Number of Octahedral Voids |
|---|---|---|---|
| Simple Cubic (SC) | 1 | 0 | 0 |
| Body-Centered Cubic (BCC) | 2 | 0 | 0 |
| Face-Centered Cubic (FCC) / Cubic Close-Packed (ccp) | 4 | 8 (2N) | 4 (N) |
| Hexagonal Close-Packed (hcp) | 6 | 12 (2N) | 6 (N) |
Additional Information: Voids in Crystal Structures
When atoms pack together to form a crystal lattice, they don't completely fill the space. The empty spaces left between the spheres are called interstitial sites or voids. The two most common types in close-packed structures are tetrahedral and octahedral voids.
- Tetrahedral Voids: These voids are surrounded by four spheres arranged at the corners of a tetrahedron. There are two tetrahedral voids for each atom in a close-packed structure (hcp or ccp).
- Octahedral Voids: These voids are surrounded by six spheres arranged at the corners of an octahedron. There is one octahedral void for each atom in a close-packed structure (hcp or ccp).
Smaller atoms or ions can occupy these voids. The formula of ionic compounds or alloys formed by this mechanism depends on the stoichiometry of the host lattice atoms and the interstitial atoms/ions, as well as the fraction of voids occupied.
Important Questions from Crystalline Solid
How many anions surround a sodium ion in a crystal lattice of sodium chloride?
Among the following statements, choose the correct statements,
A. In Ionic solid, ions are the constituent particles.
B. Ionic solids are soft.
C. Ionic solid are electrical insulators in the solid state.
D. Ionic solid conduct electricity in molten state.
E. Ionic solid have low melting and boiling points.
Choose the correct answer from the options given below:
Which of the following statements best describes the characteristics of a crystalline solid?
