Problem 149 - A Complete Resource Book in Chemistry for JEE Main

Question

Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of \(4.29 \AA\). The radius of sodium atom is approximately (a) \(1.86 \AA\) (b) \(3.22 \AA\) (c) \(5.72 \AA\) (d) \(0.93 \AA\)

Step-by-Step Solution

Answer

The radius of the sodium atom is approximately \(1.86 \AA\).

Step 1: Understanding the BCC structure

In a body-centered cubic (BCC) lattice, atoms are located at each corner of the cube and one atom is in the center of the cube. The edge of the cube is given as \(4.29 \AA\). The atoms are in contact along the cube's body diagonal.

Step 2: Calculating the body diagonal

In a BCC unit cell, only the body diagonal is relevant for calculating the radius of an atom. If the edge length of the cube is \(a\), the body diagonal can be calculated using the formula \(\sqrt{3}a\). So, the body diagonal of this unit cell is \(\sqrt{3} \times 4.29 \AA\).

Step 3: Calculate the atom-to-atom distance along the diagonal

In a BCC structure, the body diagonal passes through two radii at the corners and the entire diameter of the atom in the center. The formula for the body diagonal is \(4r = \sqrt{3}a\), where \(r\) is the radius of the atom. Substituting \(a = 4.29 \AA\), the distance is \(4r = \sqrt{3} \times 4.29 \AA\).

Step 4: Solving for the radius

Now, we solve for the radius \(r\). Rearrange the equation from the previous step to find \(r = \frac{\sqrt{3} \times 4.29}{4}\). Calculate this to find the radius in \(\AA\).

Step 5: Final Calculation

After calculating \(r = \frac{\sqrt{3} \times 4.29}{4} \approx 1.855 \AA\). Therefore, the closest option is \(1.86 \AA\).

Key Concepts

Sodium Metal

Sodium metal is a soft, silvery-white metal that belongs to the alkali metal group in the periodic table. It is known for its high reactivity, particularly with water, to produce hydrogen gas and sodium hydroxide. This reactivity is due to the one valence electron in the outer shell which sodium easily loses to form positive ions. This makes sodium a good conductor of electricity. It is widely used in various applications, such as in street lights for producing a bright yellow-orange light, and as a heat exchanger in nuclear reactors.

• Belongs to the alkali metals group.
• Highly reactive with other substances.
• Commonly used in street lighting and industrial applications.

Understanding the properties of sodium metal, including its reactivity and applications, is key to comprehending the environment of its crystal lattice structure, which dictates many of its physical characteristics. The crystalline structure allows us to calculate atomic-scale properties like the atomic radius based on known measurements of the lattice.

Atomic Radius Calculation

The atomic radius is a measure of the size of an atom, typically the distance from the center of the nucleus to the outer electron shell. In the context of a crystalline structure, such as a body-centered cubic (BCC) lattice, the atomic radius can be determined using geometric considerations.
In a BCC structure, atoms are placed at each corner of a cube, with one atom in the center. The edge length of this cube helps determine the body diagonal, through which calculations are made to find the atomic radius. The formula used in a BCC lattice to relate the edge length \(a\) and the radius \(r\) of the atom is derived from the equation for the body diagonal:

• The body diagonal = \(\sqrt{3}a\).
• The relationship \(4r = \sqrt{3}a\) is used to calculate \(r\).

By solving the equation \(r = \frac{\sqrt{3} \times a}{4}\), where \(a\) is the edge length, we find the atomic radius. This type of calculation is critical for understanding how closely packed the atoms are within a metal, influencing its density and other properties.

Crystal Lattice Structures

Crystal lattice structures describe the ordered arrangement of atoms, ions, or molecules in a crystal. One common type of crystal lattice is the body-centered cubic (BCC) structure, which is the focus here.
In a BCC lattice, each cube has atoms at the eight corners and an additional atom in the center of the cube, differing from other structures like face-centered cubic (FCC) or simple cubic lattices. This arrangement impacts the material's properties, as it determines how atoms touch each other and how they are spaced.

• In BCC, there are 2 atoms per unit cell: 8 corners times \(\frac{1}{8}\) per atom, plus 1 body-centered atom.
• The BCC structure results in efficient packing, albeit slightly less than FCC structures.

These lattice arrangements are fundamental to understanding various physical properties, such as how metals bend, conduct heat and electricity, or respond to stress. Recognizing these structures helps in manipulating material properties for specific engineering and technological applications.