Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An element with molar mass \(2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}\) forms a cubic unit cell with edge length \(405 \mathrm{pm}\). If its density is \(2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\), the radius of the element is approximately \(\times 10^{-12} \mathrm{~m}\) (to the nearest integer).

Step-by-Step Solution

Answer

The approximate radius is 143 in \(\times 10^{-12} \text{ m}\).

Step 1: Convert Edge Length

First, we need to convert the edge length given in picometers to meters so that it is compatible with the units of density. The edge length is given as 405 pm. Converting picometers to meters:\[ 405 ext{ pm} = 405 \times 10^{-12} ext{ m} \]

Step 2: Calculate Volume of Unit Cell

Next, calculate the volume of the cubic unit cell using the formula for the volume of a cube, which is the edge length cubed.\[ V = a^3 \]Substituting the edge length:\[ V = (405 \times 10^{-12} ext{ m})^3 \approx 6.64 \times 10^{-29} ext{ m}^3 \]

Step 3: Calculate Mass and Moles in Unit Cell

Calculate the mass of the element using its density and volume of the unit cell. \[ \rho = \frac{m}{V} \] So, the mass of the unit cell is:\[ m = \rho \times V = (2.7 \times 10^{3} ext{ kg/m}^3) \times (6.64 \times 10^{-29} ext{ m}^3) \approx 1.79 \times 10^{-25} ext{ kg} \]

Step 4: Calculate the Number of Moles in a Unit Cell

Using the molar mass, find the number of moles in the unit cell. The molar mass is given as \(2.7 \times 10^{-2} ext{ kg/mol} \). The number of moles \(n\) is given by:\[ n = \frac{m}{M} \]Substituting the values:\[ n = \frac{1.79 \times 10^{-25}}{2.7 \times 10^{-2}} \approx 6.63 \times 10^{-24} ext{ mol} \]

Step 5: Determine the Number of Atoms per Unit Cell

Find out how many atoms are present in a unit cell by dividing the number of moles by Avogadro's number (\(6.022 \times 10^{23} ext{ mol}^{-1}\)).\[ \text{Atoms per unit cell} = n \times N_A = (6.63 \times 10^{-24} \text{ mol}) \times (6.022 \times 10^{23} ext{ mol}^{-1}) \approx 4 \]

Step 6: Identify the Structure Type

Since there are approximately 4 atoms per unit cell, this indicates a face-centered cubic (FCC) structure, where each atom touches another atom along the face diagonal.

Step 7: Calculate Atomic Radius for FCC

For FCC, the relationship between the edge length \(a\) and the atomic radius \(r\) is given by:\[ 4r = \sqrt{2} \times a \] Solving for \(r\):\[ r = \frac{\sqrt{2}}{4} \times a \approx \frac{1.414}{4} \times 405 \times 10^{-12} \text{ m} \approx 1.43 \times 10^{-10} \text{ m} \] Thus, the radius in the scale \(\times 10^{-12} \text{ m}\) is:\[ 1.43 \times 10^{2} \] which approximately equals 143 to the nearest integer.

Key Concepts

Atomic Radius Calculation

Understanding how to calculate the atomic radius is a fundamental concept in chemistry, particularly when dealing with solid-state structures like the cubic unit cell. In this scenario, we have a face-centered cubic (FCC) structure, which simplifies the radius calculation. When calculating the atomic radius in a cubic unit cell, it's important to understand that different arrangements of atoms will lead to different formulas. In a face-centered cubic structure, each of the eight corners of the cube has an atom, as well as one atom on each face of the cube. Thus, atoms are in contact along the face diagonal, not along the cube edges. To calculate the atomic radius for an FCC structure:

• The relationship between the edge length \(a\) and the atomic radius \(r\) is \(4r = \sqrt{2} \times a\).
• We solve for \(r\) by rearranging the formula to \(r = \frac{\sqrt{2}}{4} \times a\).

Understanding these relationships helps when visualizing the structure itself and aids in calculating important metrics like atomic radius.

Face-Centered Cubic Structure

The face-centered cubic (FCC) structure is one of the most common and stable arrangements of atoms in solid materials. It's a close packing arrangement where each cube has atoms on all its corners and one in the center of each face.In an FCC arrangement:

• The total number of atoms per unit cell is 4. This is calculated as follows: Each of the 8 corner atoms contributes \(\frac{1}{8}\) of an atom (since each corner atom is shared among 8 adjacent cells), and each of the 6 face-centered atoms contributes \(\frac{1}{2}\) of an atom (since face atoms are shared between two cells).
• The coordination number, or the number of nearest neighbors for each atom, is 12, which means each atom is closely packed with 12 other atoms.

Due to its symmetry and density, the FCC structure is prevalent in metals and other crystalline solids. Recognizing this structure helps predict and understand the physical properties of the material.

Molar Mass and Density Relation

The relationship between molar mass and density provides crucial insight into the properties and identity of a substance. In this context, the mass of a unit cell can be related to the density and volume of the cubic unit cell.Here's what you need to know:

• Molar Mass (M): It represents the mass of one mole of a substance, generally measured in grams per mole (g/mol) or kilograms per mole (kg/mol). In our exercise, the molar mass is given as \(2.7 \times 10^{-2}\) kg/mol.
• Density (\(\rho\)): This is the mass per unit volume of a material, which is provided as \(2.7 \times 10^{3}\, \text{kg/m}^3\) in the problem.
• Using these values along with the calculated volume of the unit cell, you can compute the mass of the unit cell and then, using molar mass, find the number of moles contained in a unit cell. Finally, with Avogadro's number, determine the number of atoms per unit cell.

Understanding how molar mass and density relate aids in calculating unknown properties of rectangular crystals and designing materials with specific desired density characteristics.