Question

In diamond, each carbon atom is bonded to four other carbon atoms tetrahedrally. Alternate tetrahedral void occupied by carbon atoms. The number of carbon atoms per unit cell is : (a) 4 (b) 6 (c) 8 (d) 12

Short answer

The number of carbon atoms per unit cell in a diamond is 8.

Step-by-step solution

Understanding the structure of diamond

Recognize that in a diamond, the carbon atoms are arranged in a tetrahedral lattice structure. This implies that each carbon atom is at the center of a tetrahedron formed by four other carbon atoms. Also, note that in a diamond cubic structure, half of the tetrahedral voids are occupied by carbon atoms.

Counting the carbon atoms in different locations

In a unit cell of diamond, there are 8 carbon atoms at the corners, each shared by 8 unit cells, and 6 face-centered atoms, each shared by 2 unit cells, and 4 inside the unit cell, fully belonging to it. The equation to find the total number of atoms per unit cell is: Total number of atoms = (number of corner atoms × 1/8) + (number of face-centered atoms × 1/2) + number of inside atoms.

Calculating the number of carbon atoms per unit cell

Use the formula from Step 2 to calculate the number of carbon atoms: (8 corners × 1/8 per corner atom) + (6 face-centered × 1/2 per face-centered atom) + (4 inside × 1 per inside atom) = 1 + 3 + 4 = 8.

Key concepts

Tetrahedral Lattice Structure

In the world of crystallography and materials science, the structure of a diamond is a shining example of a tetrahedral lattice structure. Imagine a 3D network where each point of connection is a carbon atom, and from that atom, four bonds extend out to join with four other carbon atoms. This creates a pyramidal shape, known as a tetrahedron.
Now picture a vast collection of these tetrahedra, interlocking with perfect symmetry to form a crystal lattice. That's the heart of diamond's unique and impressive properties - its unparalleled hardness and brilliant sparkle. In diamonds, every carbon atom is bonded in such a way that it forms part of multiple tetrahedra, each one sharing a corner with the next.
This intricate and robust arrangement of atoms gives diamond its incredible strength and ability to refract light in a way that produces that mesmerizing fire we associate with these precious stones. Understanding the tetrahedral lattice structure isn't just fascinating from a gemological perspective, but it's also critical for students preparing for competitive exams like JEE, as it intertwines with concepts of chemistry and physics.

Unit Cell Calculation

Calculating the number of atoms within a unit cell is a task that often perplexes students, but it's an important skill in fields like solid-state physics and materials science. So, how do we tackle this in the context of a diamond's unit cell? Let's start by simplifying our approach.
Consider the diamond cubic structure as a complex puzzle, where carbon atoms are located at different positions—corners, face centers, and completely inside the unit cell. Remember, the carbon atoms at the corners and face centers are like shared pieces of the puzzle belonging to multiple unit cells, while those completely inside are unique to their respective unit cells.
To find the total number of carbon atoms within one unit cell, we use a thoughtful counting strategy. For the corner atoms, since an atom at a corner is shared by eight surrounding cells, we only count an eighth of each. For the face-centered atoms, which are shared by two cells, we count half. And the atoms fully inside the unit cell? We count those in full. This careful accounting allows us to arrive at the accurate number of carbon atoms in a unit cell, a calculation essential for students mastering the concepts of crystal structures in physical chemistry, especially for competitive exams like the JEE.

Physical Chemistry for JEE

Physical chemistry explores the physical properties and transformations of materials, a critical portion of syllabi for competitive exams like the Joint Entrance Examination (JEE) in India. When delving into the physical chemistry for JEE, students are not only introduced to theoretical concepts but also to practical applications that lay the framework for advanced scientific and engineering disciplines.
In this context, understanding the crystal structure of substances like diamond is pivotal. It not only deepens a student's comprehension of atomic arrangement and bonding but also prepares them for complex problem-solving on topics such as unit cell calculations—key to cracking the JEE. Furthermore, these core principles lay the foundation for students to analyze and predict the behavior of different materials, an invaluable skill in research and applied sciences.
Therefore, integration of exercises that involve three-dimensional problem-solving, symmetrical analysis, and calculation strategies is crucial for aspirants. Not only do they need to engage with these concepts theoretically, but practicing with meticulous, step-by-step solutions paves their way to achieving excellence in subjects like physical chemistry.