Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Sodium metal crystallizes in body centred cubic lattice with the cell edge, \(a=4.29 \AA\). What is the radius of sodium atom?

Step-by-Step Solution

Answer

Radius of sodium atom is 1.86 Å.

Step 1: Identify the Lattice Type

Sodium crystallizes in a body-centered cubic (BCC) lattice, which is a specific arrangement of atoms within the crystal.

Step 2: Understand the Relationship in BCC

In a BCC lattice, one atom is at each corner of the cube, and one atom is in the center. The distance between opposite corners of the cube (the body diagonal) is equal to 4 times the radius of the atom. So the relationship is given by: \(\sqrt{3}a = 4r\), where \(a\) is the cell edge length, and \(r\) is the radius of the atom.

Step 3: Substitute the Cell Edge Length

We know the cell edge length \(a = 4.29 \ \text{Å}\). Substitute this value into the body diagonal formula: \(\sqrt{3}(4.29) = 4r\).

Step 4: Calculate the Radius

Solve for \(r\):\[\sqrt{3}(4.29) = 4r \7.43 = 4r \r = \frac{7.43}{4} \r = 1.8575 \ \text{Å}\]

Step 5: Conclusion

The radius of a sodium atom in this BCC lattice is approximately \(1.86 \ \text{Å}\).

Key Concepts

Crystal Structure

In crystallography, the crystal structure refers to the orderly geometric spatial arrangement of atoms in the crystalline solids. A body-centered cubic (BCC) lattice is one of the common crystal structures. In a BCC structure, each cubic unit cell has:

• Atoms at each of its eight corners.
• An additional atom positioned at the center of the cube.

This structure arrangement leads to specific properties of the crystal. For instance, in metals like sodium, this compact arrangement affects how atoms pack together and their bonding energies. Understanding the crystal structure helps determine other properties of the material, like density and atomic radius. The BCC crystal structures are known for being densely packed, offering more efficient packing than some other configurations like simple cubic structures.

Atomic Radius Calculation

The calculation of atomic radius in a BCC lattice utilizes geometric relationships based on its spatial configuration. The key relationship to remember in a BCC lattice involves the body diagonal, which connects two opposite corners of the cube through its center atom. This diagonal is mathematically expressed as \[\sqrt{3}a = 4r\]where:

• \(a\) is the cell edge length.
• \(r\) is the radius of the atom.

The formula shows that the body diagonal is four times the atomic radius. Substituting the known cell edge length helps find the atomic radius. For example, for sodium with a cell edge of \(4.29 \, \text{Å}\), the radius can be calculated by solving \[\sqrt{3}(4.29) = 4r\]resulting in a radius of approximately \(1.86 \, \text{Å}\). This method is crucial for determining the size of atoms in crystalline structures.

Lattice Parameters

Lattice parameters are the set of values that describe the size and shape of the unit cell of a crystal lattice. These parameters are fundamental when you are working with crystal structures because they directly influence the physical and chemical properties of the material. In the context of a cubic system, particularly a body-centered cubic (BCC) lattice:

• The lattice parameter \(a\) represents the unit cell edge length.
• The angles between edges in a cubic system are always \(90^\circ\).
• The parameter \(a\) directly influences the atomic radius and packing efficiency.

Understanding lattice parameters helps in deriving other significant crystal properties such as density and spacing between planes in the lattice. As in the BCC lattice of sodium, knowing the lattice parameters allows us to calculate atomic dimensions and thus better grasp the material's bulk properties.