Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

At room temperature, sodium crystallizes in a BCC lattice with the cell edge (a) \(4.24 \AA\). Find the density of sodium. (Atomic wt of \(\mathrm{Na}=23\) ) (a) \(2.01 \mathrm{~g} / \mathrm{cm}^{3}\) (b) \(1.002 \mathrm{~g} / \mathrm{cm}^{3}\) (c) \(3.003 \mathrm{~g} / \mathrm{cm}^{3}\) (d) \(2.004 \mathrm{~g} / \mathrm{cm}^{3}\)

Step-by-Step Solution

Answer

(b) 1.002 g/cm³

Step 1: Understand the BCC Structure

In a body-centered cubic (BCC) lattice, there is one atom at each corner of the cube and one atom in the center. This configuration accounts for a total of 2 atoms per unit cell.

Step 2: Convert Lattice Parameter

The given cell edge is in angstroms. We need to convert it to centimeters for use in density calculations:\[4.24 \, \text{Å} = 4.24 \times 10^{-8} \, \text{cm}\]

Step 3: Calculate Volume of Unit Cell

The volume of the cubic unit cell is calculated using:\[V = a^3 = (4.24 \times 10^{-8} \, \text{cm})^3 = 7.617 \times 10^{-23} \, \text{cm}^3\]

Step 4: Find Mass of Atoms in Unit Cell

Sodium's atomic weight is given as 23 g/mol, and the number of atoms in one mole (Avogadro's number) is \(6.022 \times 10^{23}\). Therefore, the mass of 2 sodium atoms (as in one BCC unit cell) can be found as follows:\[\text{Mass of 2 Na atoms} = \frac{23 \times 2}{6.022 \times 10^{23}} = 7.637 \times 10^{-23} \, \text{g}\]

Step 5: Calculate Density

Finally, density is mass per unit volume. Thus, the density can be calculated as:\[\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{7.637 \times 10^{-23} \, \text{g}}{7.617 \times 10^{-23} \, \text{cm}^3} = 1.002 \, \text{g/cm}^3\]

Step 6: Select Correct Option

The calculated density of sodium in its BCC configuration is \(1.002 \, \text{g/cm}^3\). This corresponds to option (b).

Key Concepts

Density Calculation

Density is a fundamental concept in materials science and chemistry, representing how much mass is contained within a given volume. Typically, density is expressed in grams per cubic centimeter (g/cm³). To calculate density, you need two main components: the mass of the substance and its volume. In the case of a crystalline solid like sodium forming a body-centered cubic (BCC) structure, the key steps to calculate density involve understanding the atomic arrangement and using the given lattice parameters.

• First, you identify the number of atoms present in a particular unit cell configuration. For BCC structures, this is 2 atoms per unit cell.
• Then, you calculate the mass of these atoms based on the atomic weight and Avogadro's number.
• Finally, you use the volume of the unit cell, determined from the lattice parameters, to find density using the formula: \[\text{Density} = \frac{\text{Mass of atoms}}{\text{Volume of unit cell}}.\]

Thus, understanding the density involves grasping how these microscopic parameters culminate in measurable macroscopic properties.

Unit Cell Volume

The unit cell volume is an essential concept in crystallography, providing the dimensional space that a single repeatable pattern (unit cell) occupies in a crystal lattice. The volume of a unit cell is especially important when calculating the density, as it helps determine how much space is available for atoms within the structure. For a cubic lattice, such as the BCC structure of sodium, the unit cell volume is straightforward to calculate. You use the formula: \[V = a^3,\]where \(a\) is the cell edge length. Given in angstroms, the cell edge must be converted to a more usable unit like centimeters (1 angstrom = \(10^{-8}\) cm). The straightforward nature of the calculation hides the fascinating complexity of how these structures come together to form a solid. While a BCC unit cell may seem simple, its geometric properties critically define the material's overall density and physical characteristics.

Atomic Structure

The atomic structure of a material informs us about how atoms are arranged in space, which directly affects the material's physical properties. In the case of the body-centered cubic (BCC) lattice, atoms are tightly packed into a specific geometrical format that optimizes space within a crystalline structure.

• In a BCC lattice, each corner of the cube hosts one atom, and there is one additional atom at the center of the cube. This results in 2 atoms being accounted for per unit cell, adding to the unique characteristics of BCC arrangements.
• The specific arrangement of atoms not only influences density but also impacts properties such as hardness and melting point.
• Understanding atomic structure is pivotal in predicting material performance and behavior under various conditions. The BCC structure, common in metals like sodium, is efficient for occupying space within the lattice, impacting material properties on a macroscopic scale.

Sodium Crystallization

Sodium is an alkali metal known for its highly reactive nature and metallic bonding characteristics. At room temperature, sodium crystallizes in a body-centered cubic (BCC) structure, a common arrangement for metals.

• The BCC structure is notable for fewer atoms per unit cell compared to face-centered cubic (FCC) lattices, requiring only 2 atoms per unit cell.
• This specific arrangement contributes to sodium's relatively low density compared to other materials, as less mass is concentrated in the volume of the unit cell.
• Sodium's crystallization pattern affects various properties, including conductivity and reactivity. The BCC formation allows for easier slipping of atomic layers, giving sodium its malleability.
• In industrial applications, understanding sodium's crystallization behaviour is crucial for manipulating its properties for specific uses, such as in batteries or chemical synthesis.

The unique crystalline arrangement of sodium reflects the intriguing interplay between atomic structure and macroscopic properties.