Problem 17 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In a face centred cubic lattice, atoms of A form the corner points and atoms of \(\mathrm{B}\) form the face centred points. If two atoms of \(\mathrm{A}\) are missing from the corner points, the formula of the ionic compound is (a) \(\mathrm{AB}_{3}\) (b) \(\mathrm{AB}_{4}\) (c) \(\mathrm{A}_{2} \mathrm{~B}_{5}\) (d) \(\mathrm{AB}_{2}\)

Step-by-Step Solution

Answer

The formula of the ionic compound is (b) \( AB_4 \).

Step 1: Calculate Total Atoms at Corners

In a face-centered cubic (FCC) lattice, there are 8 atoms at the corners of the cube. Since each corner atom is shared by 8 adjacent cubes, the contribution of corner atoms is \( \frac{1}{8} \) per cube. However, according to the problem, 2 of these corner atoms (A-atoms) are missing. Therefore, the effective number of A-atoms at the corners is \( 8 - 2 = 6 \), contributing \( 6 \times \frac{1}{8} = \frac{3}{4} \) atom per unit cell.

Step 2: Calculate Total Face-Centered Atoms

Each face of the cube in an FCC lattice has 1 atom at the center, and there are 6 faces. Atoms at the face are shared between 2 adjacent cubes, so each face-centered atom contributes \( \frac{1}{2} \). Therefore, the total contribution of face-centered atoms (B-atoms) is \( 6 \times \frac{1}{2} = 3 \) atoms per unit cell.

Step 3: Write the Empirical Formula of the Compound

Integrate the effective count from Step 1 for A-atoms and Step 2 for B-atoms to determine the empirical formula: \( A_{\frac{3}{4}}B_3 \). To remove fractions, multiply both terms by 4 to simplify the formula, resulting in \( A_3B_{12} \), which reduces to \( AB_4 \).

Step 4: Match with Options

Comparing the derived formula \( AB_4 \) with the given options, we see that option (b) \( AB_4 \) matches our derived formula. Thus, the formula of the ionic compound is \( AB_4 \).

Key Concepts

Ionic Compounds

Ionic compounds are formed by the electrostatic attraction between positively and negatively charged ions. These oppositely charged ions result from the transfer of electrons, often from a metal to a non-metal. This transfer helps achieve a stable electronic configuration, similar to noble gases.

In this example, we're looking at a structure where elements A and B form such a compound. When A and B bond, one likely loses electrons becoming a positive ion, and the other gains those electrons, becoming a negative ion. Together, this forms a neutral compound overall.

• The formation of ionic bonds generally results in a rigid crystal lattice.
• The stoichiometry, or balance of the ions, determines the empirical formula.

Understanding the ionic nature of compounds helps in aligning the involved elements to satisfy valency requirements.

Crystal Lattice Structures

Crystal lattice structures in ionic compounds are regular, repeating arrangements of ions. These structures ensure that every ion is surrounded by ions of the opposite charge. This orderly pattern maximizes attraction while minimizing repulsion.

The face-centered cubic (FCC) structure is one common type of crystal lattice.

• In FCC, atoms are shared between corners and faces of the cubic structure.
• It involves sharing these atoms across several adjacent cells, impacting the effective number of atoms.

For the given problem, element A is placed at the corners, and element B is positioned at the centers of the faces. The contribution of each atom needs to be counted correctly by considering shared positions.

Empirical Formula Calculation

Calculating the empirical formula involves determining the simplest whole-number ratio of ions in a compound. This ratio reflects the composition of equal numbers of positive and negative charges that lead to a stable arrangement.

• Begin by calculating each type of atom's contribution within the lattice.
• Consider the missing atoms or additional factors that may affect this count.
• Simplify the ratio to the smallest whole numbers to derive the formula.

In the exercise provided, A-atoms missing from corners affect their total number, while B-atoms at face centers provide another contribution to the count. By applying lattice sharing principles, we establish the formula based on their adjusted contributions. After simplifying the fractional amounts, multiplying through by a common factor helps eliminate fractions, arriving at a cleaner empirical formula.