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 \(\mathrm{Au}\) 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 approximately \(19.4 \mathrm{g/cm^3}\), matching option (a).

Step 1: Understand the Structure

Gold crystallizes in a Face-Centered Cubic (FCC) structure. An FCC unit cell contains 4 atoms and has a cubic shape.

Step 2: Relate Atomic Radius to Edge Length

The relationship between atomic radius \( r \) and the edge length \( a \) of an FCC crystal is \( a = 2\sqrt{2}r \). Given \( r = 144 \mathrm{pm} \), convert it to cm: \( 144 \mathrm{pm} = 144 \times 10^{-10} \mathrm{cm} \). Substituting this into the edge length formula gives: \( a = 2\sqrt{2} \times 144 \times 10^{-10} \mathrm{cm} \).

Step 3: Calculate the Edge Length

Perform the calculation to find \( a \): \[ a = 2 \times 1.414 \times 144 \times 10^{-10} \mathrm{cm} \approx 4.08 \times 10^{-8} \mathrm{cm} \]

Step 4: Calculate the Volume of the Unit Cell

The volume of a cubic unit cell \( V \) is \( a^3 \). Use \( a \) calculated in Step 3:\[ V = (4.08 \times 10^{-8} \mathrm{cm})^3 \approx 6.79 \times 10^{-23} \mathrm{cm}^3 \]

Step 5: Determine Mass of the Unit Cell

The mass of one gold atom is the atomic mass divided by Avogadro's number: \( \text{Mass of one Au atom} = \frac{197.0}{6.022 \times 10^{23}} \mathrm{g} \approx 3.27 \times 10^{-22} \mathrm{g} \). Since an FCC cell has 4 atoms: \( \text{Mass of unit cell} = 4 \times 3.27 \times 10^{-22} \mathrm{g} = 1.31 \times 10^{-21} \mathrm{g} \).

Step 6: Calculate the Density of Gold

Density \( \rho \) is defined as mass divided by volume. Using values from steps 4 and 5: \[ \rho = \frac{1.31 \times 10^{-21} \mathrm{g}}{6.79 \times 10^{-23} \mathrm{cm}^3} \approx 19.3 \mathrm{g/cm^3} \]

Step 7: Select the Closest Option

Compare the calculated density \( 19.3 \mathrm{g/cm^3} \) with given options, the closest match is: (a) \(19.4 \mathrm{g/cm^3} \).

Key Concepts

Face-centered Cubic Structure

The face-centered cubic (FCC) structure, also known as cubic close packing (CCP), is a common crystal arrangement in metals like gold. In this type of structure, atoms are positioned at each of the cube's corners and the centers of all the cube's faces. This arrangement results in each unit cell hosting 4 atoms.

The FCC structure is renowned for its high packing efficiency. This means that the spheres (representing atoms) are packed closely together to minimize empty space, achieving a packing efficiency of about 74%. As a result, this structure contributes to the high density and stability observed in metals like gold.

Atomic Radius

The atomic radius is a measure of the size of an atom's electron cloud, and for gold, it is given as 144 picometers (pm). The atomic radius plays a crucial role in determining the properties of the crystal lattice. In the context of an FCC structure, the atomic radius is used in calculating the edge length of the unit cell.

The formula to relate atomic radius to the edge length of an FCC unit cell is: \[ a = 2\sqrt{2}r \] where \( r \) is the atomic radius. For gold, substituting the given atomic radius into this formula helps compute the size of the unit cell, which is vital for further density calculations.

Density of Gold

Density is a physical property that describes the mass of a substance per unit volume. To calculate the density of gold in a crystal structure, one must know both the mass and the volume of its unit cell. The mass is determined by the number of atoms in the unit cell and the mass of each atom, while the volume is derived from the cube of the unit cell's edge length.

In this context, the density formula is: \[ \rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} \]After picking an FCC structure and using calculations, we find gold's density to be approximately 19.3 g/cm³, closely matching the given option.

Unit Cell Volume

The unit cell is the smallest repeating unit in a crystal structure, and its volume is pivotal for density calculations. In a cubic structure like the FCC, the volume \( V \) is calculated using the cube of the edge length \( a \):\[ V = a^3 \]

After determining the edge length from the atomic radius, we can compute the unit cell volume, which is necessary for calculating the crystal's density. For gold, this calculation reveals a volume of approximately \( 6.79 \times 10^{-23} \mathrm{cm}^3 \), tying into the overall density solution of the precious metal.