Problem 40 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The edge length of unit cell of a metal having molecular weight 75 \(\mathrm{g} / \mathrm{mol}\) is \(5 \AA\) which crystallizes in cubic lattice. If the density is \(2 \mathrm{~g} / \mathrm{cc}\) then find the radius of metal \(\operatorname{atom}\left(N_{A}=6 \times 10^{23}\right)\). Give the answer in \(\mathrm{pm}\)

Step-by-Step Solution

Answer

The radius of the metal atom is approximately 177 pm.

Step 1: Understanding Unit Cell and Density Formula

To find the radius of a metal atom in a cubic unit cell, we begin with the formula for density \( \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \), where \( Z \) is the number of atoms per unit cell, \( M \) is the molar mass, \( a \) is the edge length of the unit cell, and \( N_A \) is Avogadro's number.

Step 2: Solve for Number of Atoms per Unit Cell

Using the given density \( \rho = 2 \mathrm{~g/cc} \), molar mass \( M = 75 \mathrm{~g/mol} \), and \( a = 5 \times 10^{-8} \mathrm{~cm} \), we can rearrange the formula to solve for \( Z \): \( Z = \frac{\rho \cdot a^3 \cdot N_A}{M} \).

Step 3: Calculate Contribution of Cubic Types

For cubic systems such as simple cubic, body-centered cubic (BCC), and face-centered cubic (FCC), the values of \( Z \) are usually 1, 2, and 4, respectively. We test these by substituting them into the calculated formula for \( Z \) to match the density.

Step 4: Identify Crystal Structure

Calculate \( Z \) using parameters: \( Z = \frac{2 \cdot (5 \times 10^{-8})^3 \cdot 6 \times 10^{23}}{75} \approx 4 \), identifying the crystal structure as FCC.

Step 5: Find Atomic Radius for FCC

In an FCC structure, the atomic radius \( r \) is related to the edge length \( a \) by \( r = \frac{a\sqrt{2}}{4} \).

Step 6: Calculate Radius in Picometers

Using \( a = 5 \mathrm{~\AA} = 500 \mathrm{~pm} \), the radius \( r = \frac{500 \times \sqrt{2}}{4} \approx 177 \mathrm{~pm} \).

Key Concepts

density formula

Understanding the density formula is the first step towards solving problems related to unit cells in crystalline solids. Density (\(\rho\)) in the realm of crystallography is defined as mass per unit volume. The formula is expressed as \( \rho = \frac{Z \cdot M}{a^3 \cdot N_A}\).

• In this formula, \(Z\) represents the number of atoms per unit cell. Different structures have different standard values for \(Z\), which are determined by their packing arrangement.
• \(M\) is the molar mass of the compound, which is typically given in the problem statement in grams per mole (g/mol).
• \(a\) is the edge length of the unit cell, expressed in centimeters (cm) when matching units is necessary.
• \(N_A\) is Avogadro’s number – a key constant in chemistry.

This formula helps us link the density of a material to its structural and molecular attributes. Knowing this can help determine other properties, such as the type of cubic unit cell and the atomic radius.

Avogadro's number

Avogadro's number (\(N_A\)) is fundamental in chemistry and material science. It's the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. Set at roughly \( 6.022 \times 10^{23} \), this number allows chemists to convert between microscopic and macroscopic scales.
For example, in the context of a unit cell, Avogadro's number lets us calculate the number of atoms in a given quantity of material, based on its molar mass.
This conversion is critical when employing the density formula. It helps adjust theoretical models to reflect real-world observations and measurements. By incorporating \(N_A\), we reconcile elements of classical chemistry with modern physical measurements.

face-centered cubic (FCC)

The face-centered cubic (FCC) structure is one of the most efficient ways atoms can be packed in a solid. In this configuration, atoms are positioned at each corner and at the center of each face of the cube.

• In an FCC unit cell, the coordination number—which represents the number of nearest neighboring atoms—is 12.
• The number of atoms per unit cell, \(Z\), is 4 for FCC. This accounts for the shared atoms at the corners and faces.
• The atomic radius in FCC is related to the edge length, with the equation: \(r = \frac{a\sqrt{2}}{4}\).

This structure is prevalent in metals like aluminum, copper, and gold, contributing to their characteristic properties, such as high ductility and good electrical conductivity. The efficient arrangement in FCC provides stability and strength to the material.

atomic radius

The atomic radius is a measure extending from the nucleus of an atom to its outermost electron cloud boundary. In a crystalline solid, knowing the atomic radius is essential for understanding how closely the atoms pack together.
For a face-centered cubic (FCC) structure, the atomic radius is directly related to the edge length of the cubic unit cell. This relationship is described by the equation: \(r = \frac{a\sqrt{2}}{4}\).

• This equation reflects the geometric considerations of an FCC lattice, where the atoms touch along the face diagonal.
• In practical terms, if you know the edge length and the crystal structure type, you can easily compute the atomic radius.
• For conversions between units, remember that 1 Å (angstrom) is equivalent to 100 pm (picometers).

Understanding the atomic radius helps predict how atoms will interact and bond, influencing the material's properties and behaviors.