Problem 111 - A Complete Resource Book in Chemistry for JEE Main

Question

Gold (Au) crystallizes in cubic close packed (FCC). The atomic radius of gold is \(144 \mathrm{pm}\) and the atomic mass of \(\mathrm{Au}=197.0 \mathrm{amu}\). The density of Au is (a) \(19.4 \mathrm{~g} \mathrm{~cm}^{-3}\) (b) \(194 \mathrm{~g} \mathrm{~cm}^{-3}\) (c) \(39.4 \mathrm{~g} \mathrm{~cm}^{-3}\) (d) \(0.194 \mathrm{~g} \mathrm{~cm}^{-3}\)

Step-by-Step Solution

Answer

The density of gold is (a) 19.4 g/cm³.

Step 1: Understand the Structure

Gold crystallizes in a face-centered cubic (FCC) structure. In this structure, there are four atoms per unit cell.

Step 2: Relate Atomic Radius to Unit Cell Parameter

For FCC, the relation between atomic radius (r) and edge length (a) is given by: \(a = 2\sqrt{2}r\).Substituting, \(a = 2\sqrt{2} \times 144 \, \text{pm}\).

Step 3: Calculate the Edge Length

Calculate \(a\) using the atomic radius:\[a = 2\sqrt{2} \times 144 = 407.02 \, \text{pm}\] \(407.02 \, \text{pm} = 4.0702 \, \text{Å}\).

Step 4: Convert Edge Length to Volume

Convert Å to cm: \(4.0702 \, \text{Å} = 4.0702 \times 10^{-8} \, \text{cm}\).Calculate the volume of the unit cell: \(V = a^3 = (4.0702 \times 10^{-8} \, \text{cm})^3\).

Step 5: Calculate the Unit Cell Volume

\[V = 67.55 \times 10^{-24} \, \text{cm}^3\]

Step 6: Calculate Mass of Gold Atoms in Unit Cell

Mass of one gold atom: \(197 \, \text{amu},\) where \(1 \, \text{amu} = 1.66054 \times 10^{-24} \, \text{g}\).Mass of 4 atoms in the unit cell:\[M = 4 \times 197 \times 1.66054 \times 10^{-24} = 1.31 \times 10^{-21} \, \text{g}\]

Step 7: Determine Density

Density \(\rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}\)\[\rho = \frac{1.31 \times 10^{-21} \, \text{g}}{67.55 \times 10^{-24} \, \text{cm}^3} = 19.4 \, \text{g/cm}^3\]

Step 8: Select the Correct Answer

The calculated density matches option (a), which is \(19.4 \, \text{g/cm}^3\).

Key Concepts

Cubic Close Packed Structure

The cubic close packed (CCP) structure is a way in which atoms can be arranged in a crystal. It's also known as face-centered cubic (FCC) due to its unique geometrical arrangement. In a CCP structure:

• Atoms are packed closely and efficiently, resulting in a high density.
• Each atom is surrounded by 12 other atoms, forming a repeating pattern.
• This arrangement minimizes empty space and optimizes space utilization.

Understanding CCP is essential for comprehending how metals like gold and silver crystallize. The high packing efficiency (about 74%) contributes to the properties of metals, such as their density and structural stability.

Atomic Radius

The atomic radius is the distance from the center of an atom's nucleus to the outer boundary of its electrons. It's a crucial factor in determining how atoms will bond and interact in solid structures. When considering the cubic close packed (FCC) and other structures:

• The atomic radius helps establish the overall size of the crystal lattice.
• In FCC structures, the radius aids in determining the edge length, using the formula: \[a = 2\sqrt{2}r\]

For gold, with an atomic radius of 144 pm, this is pivotal not only for calculating edge lengths but also for predicting how atoms pack together within a unit cell, influencing the material's physical characteristics.

Unit Cell

A unit cell is the smallest repetitive pattern within a crystal lattice that shows the entire system's symmetry and structure. In the context of the face-centered cubic (FCC) structure:

• It contains 4 atoms per unit cell.
• The edges of the cell are defined by the formula \(a = 2\sqrt{2}r\).
• Calculating the unit cell's volume entails using the edge length and cubing it, \(V = a^3\).

The unit cell serves as a building block, allowing scientists to calculate various properties of materials, such as density, by understanding how atoms are arranged at the microscopic level. This forms the backbone of studying crystalline materials like metals.

Face-Centered Cubic

A face-centered cubic (FCC) structure is a type of crystal structure where atoms are located at each of the corners and the centers of all the cube faces. This setup ensures:

• Each of the corner atoms is shared among eight adjoining cubes.
• Face-centered atoms are shared between two adjacent unit cells.

Due to these shared positions, there are effectively four whole atoms in each FCC unit cell. This structure significantly impacts the material properties:

• It offers high packing efficiency, leading to the impressive density observed in metals like gold.
• FCC structures are present in many ductile metals, contributing to their malleability and thermal conductivity.

Understanding the FCC arrangement gives insight into the behavior and characteristics of materials, which is crucial for applications in engineering and material science.