Question

Molybdenum (At. wt. = \(96 \mathrm{~g} \mathrm{~mol}^{-1}\) ) crystallizes as bce crystal. If density of crystal is \(10.3 \mathrm{~g} / \mathrm{cm}^{3}\), then radius of Mo atom is (use \(N_{A}=6 \times 10^{23}\) ): (a) \(111 \mathrm{pm}\) (b) \(314 \mathrm{pm}\) (c) \(135.96 \mathrm{pm}\) (d) none of these

Short answer

The radius of Mo atom is approximately 135.96 pm.

Step-by-step solution

Understand the Body-Centered Cubic (BCC) Structure

In a body-centered cubic (BCC) structure, there are 2 atoms per unit cell. One atom is at the corner of the cube, which is shared with eight other cells. The second atom is at the body center of the cube, which isn't shared. Each corner atom contributes 1/8 of an atom to the cell, leading to 1 full atom from corners and 1 full atom from the center.

Calculate the Volume of the Unit Cell

Using the density formula \( \rho = \frac{mass}{volume} \), we can calculate the volume of the unit cell. Rearrange the formula to find volume: \( V = \frac{mass}{\rho} \). The mass of the unit cell is the atomic weight of Molybdenum times the number of atoms per unit cell divided by Avogadro's number: \( m = \frac{96 \mathrm{~g/mol} \times 2}{6 \times 10^{23}} \). Then, plug in the density of the crystal to find the volume: \( V = \frac{m}{10.3 \mathrm{~g/cm}^3} \).

Determine the Edge Length of the Unit Cell

After calculating the volume of the unit cell, determine the edge length (a) of the cube since the volume (V) is equal to \( a^3 \).

Calculate the Radius of Mo Atom in the BCC Structure

In a BCC crystal, the body diagonal (connecting opposite corners of the cube through the center) is equal to \( 4r \), where r is the atomic radius. The body diagonal can also be expressed using the edge length as \( \sqrt{3}a \). Equate these to solve for r: \( 4r = \sqrt{3}a \).

Perform the Calculations

Calculate the mass per unit cell: \( m = \frac{96 \times 2}{6 \times 10^{23}} \). Then, calculate the volume of the unit cell: \( V = \frac{m}{10.3} \). Next, find the edge length: \( a = \sqrt[3]{V} \). Finally, solve for the atomic radius using \( r = \frac{\sqrt{3}a}{4} \). Convert this radius into picometres to match the answer choices.

Key concepts

Understanding the Body-Centered Cubic (BCC) Structure

The body-centered cubic (BCC) structure is a type of crystal lattice that is common among metals, including elements like Molybdenum. It is called 'body-centered' because in addition to the atoms at the corners of the cube, there is one atom right at the center of the cube. This results in a total of 2 atoms per unit cell.

BCC is distinct in how atoms are arranged and touch each other - corner atoms do not touch each other directly; only the center atom and corner atoms touch. Each corner atom is part of 8 different unit cells, so only 1/8 of its volume is actually within a given cell. This contributes a total of 1 atom from the corners (1/8 * 8) and 1 from the center, simplifying the count of atoms within the BCC unit cell to 2.

Unit Cell Volume Calculation

To calculate the unit cell volume in a BCC structure, such as that of Molybdenum, we can use the relationship between mass, volume, and density with the formula: \( \rho = \frac{mass}{volume} \). Knowing the density and the mass of the unit cell (obtained by multiplying the atomic weight by the number of atoms per unit cell and dividing by Avogadro's number), you can rearrange the formula to find the volume, \( V = \frac{mass}{\rho} \).

Understanding this formula is crucial because it connects the macroscopic properties (like density) to the microscopic ones (like unit cell volume), a powerful concept that integrates physical measurements with atomic-scale phenomena.

Determining the Atomic Radius

The atomic radius, especially in a BCC crystal structure, is less straightforward to determine than in simpler structures. In BCC, atoms along the body diagonal are separated by a distance of \( 4r \) where \( r \) is the atomic radius. Since the body diagonal also equals \( \sqrt{3}a \) (where \( a \) is the edge length of the cube), these expressions can be set equal to each other to solve for \( r \).

This calculation is an elegant example of how geometric and algebraic relationships can be combined to find microscopic features of materials, bridging the gap between abstract mathematical concepts and tangible physical properties.

Application of Avogadro's Number

A critical concept in the realm of chemistry and physics is Avogadro's number (\( N_A \)), which is the number of entities (atoms, ions, molecules) in one mole of a substance. Its application in our calculations allows us to convert between the mass of atoms in grams (macroscopic) and the number of atoms (microscopic).

By applying Avogadro's number, we can determine the mass of atoms within a single unit cell, which is a stepping stone to calculate many other characteristics of materials, such as their density and atomic radii. The understanding and correct application of Avogadro's number are fundamental in the field of material science and in solving complex problems involving material properties.