Problem 19 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the \(\mathrm{C}-\mathrm{C}\) bond length (in angstroms) in diamond? (a) \(5.2\) (b) \(2.0\) (c) \(1.54\) (d) \(3.35\)

Step-by-Step Solution

Answer

The C-C bond length in diamond is 1.54 angstroms (option c).

Step 1: Identify the Bond Length Characteristic in Diamond

Diamond is a crystalline form of carbon, where each carbon atom forms a tetrahedral bond with four other carbon atoms. This tetrahedral bonding is characterized by \\( \mathrm{sp}^3 \) hybridization. Understanding this bonding geometry helps us identify the typical bond length for a C-C bond in diamond.

Step 2: Recall Common Bond Lengths

In general, the C-C bond length for a single bond (typical in sp3 hybridization) in organic compounds is around 1.54 angstroms. This is a well-known standard value for the C-C bond in diamond and other compounds with similar bonding.

Step 3: Compare with Given Options

Given the options: (a) 5.2 angstroms - significantly longer than typical bond lengths for C-C single bonds. (b) 2.0 angstroms - longer than the standard C-C bond length in diamond. (c) 1.54 angstroms - matches the typical bond length in diamond. (d) 3.35 angstroms - much longer than diamond’s characteristic bond length. Option (c) corresponds well with the known C-C bond length in diamond.

Step 4: Validate the Correct Option

The known C-C bond length for diamond being 1.54 angstroms matches perfectly with option (c). This measurement aligns with the structure and rigidity of the diamond lattice, confirming it as the standard C-C single bond length in diamond.

Key Concepts

Tetrahedral Bonding

Tetrahedral bonding is a fundamental concept in the structure of diamond. Each carbon atom in a diamond is connected to four other carbon atoms. This forms a three-dimensional shape called a tetrahedron. Imagine a triangular pyramid where each vertex represents a carbon atom, and each edge represents a bond between these atoms.
This bonding arrangement is incredibly strong and stable, contributing to diamond's renowned hardness. It's why diamonds can cut through almost any other material. In this tetrahedral pattern, the angles between each pair of bonds are exactly 109.5 degrees. This specific angle helps ensure that the atoms are as far apart from each other as possible, minimizing repulsion and maximizing bond strength.
Understanding this structure helps us appreciate why diamonds are so tough and why the bonding environment directly impacts its physical properties, including its unique C-C bond length.

sp3 Hybridization

To grasp the essence of tetrahedral bonding in diamond, it's essential to understand \( sp^3 \) hybridization. In chemistry, hybridization is a concept where atomic orbitals mix to form new hybrid orbitals. These new orbitals are the ones that actually determine how atoms bond in a molecule.
In \( sp^3 \) hybridization, one 's' orbital combines with three 'p' orbitals from the electron shell of a carbon atom. This combination forms four equivalent \( sp^3 \) hybrid orbitals. Each of these orbitals holds one electron and extends outward in space, giving the setup needed for forming four equivalent single bonds in the tetrahedral geometry.
This is crucial for diamond, where each carbon is bonded to four others, achieving stability and a balance preeminent in its crystal structure. The formation of these \( sp^3 \) hybrid orbitals ensures that all bonds are equal in length and strength, contributing to the structural integrity and ideal properties of diamond.

Diamond Structure

The diamond structure is one of the most iconic among crystalline materials. It is characterized by a rigid network where every carbon atom is tetrahedrally bonded to four other carbon atoms. This arrangement repeats in three dimensions, creating a lattice that appears as an endless configuration of interconnected tetrahedrons.
This network's orderliness is not only aesthetic but also functional. It results in diamond's hallmark properties: extreme hardness, high melting point, and excellent optical clarity.
In terms of bond lengths, each C-C bond is 1.54 angstroms long, a precise measurement reflecting the strong covalent bonding within the lattice. This structure makes diamond not only a gem of great visual beauty but also a material of significant industrial value.

• Hardness: Diamond can cut and shape other hard materials.
• Optical Clarity: Valued highly in jewelry for its brilliance.
• Thermal Conductivity: Excellent heat conductor, useful in electronics.

These properties all stem from the \( sp^3 \) hybridization and tetrahedral bonding inherent in the diamond structure.