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Chapter 12: Problem 106

One of the most important enzymes in the worldnitrogenase, the plant protein that catalyzes nitrogen fixation contains active clusters of iron, sulfur, and molybdenum atoms. Crystalline molybdenum (Mo) has a body-centered cubic unit cell \(\left(d\right.\) of \(\left.\mathrm{Mo}=10.28 \mathrm{~g} / \mathrm{cm}^{3}\right) .\) (a) Determine the edge length of the unit cell. (b) Calculate the atomic radius of Mo.

Short Answer

Expert verified
Edge length (a) calculated first, then atomic radius (r) using BCC relations.

Step by step solution

01

- Understand the properties of the crystalline structure

In a body-centered cubic (BCC) unit cell, there is one atom at each corner and one atom at the center of the cube. For a BCC structure, the number of atoms in each unit cell is 2.
02

- Determine the molar mass of Molybdenum

The molar mass of Molybdenum (Mo) is given as approximately 95.95 g/mol.
03

- Calculate the number of atoms per unit volume

Using the density \(d = 10.28 \, \) g/cm³ and the molar mass, the number of atoms per cubic centimeter can be found using the formula: \[ n = \frac{{d \times N_A}}{M} \] where \(N_A\) is Avogadro's number \(6.022 \times 10^{23} mol^{-1}\).
04

- Calculate the mass of one unit cell

Using the density, determine the mass of one unit cell. First, recall that density \(d\) is defined as mass per unit volume. \[ V_\text{cell} = \frac{2 \times \text{molar mass of Mo}}{d \times N_A} \]
05

- Calculate the volume of the unit cell

Since we have the mass of the unit cell and the density, the volume \(V_\text{cell}\) of the unit cell can be calculated as: \[ V_\text{cell} = \frac{2 \times 95.95}{10.28 \times 6.022 \times 10^{23}} \]
06

- Calculate the edge length

The edge length \(a\) can be found from the volume using the formula for the volume of a cube: \[ a^3 = V_\text{cell} \] Solving for \(a\): \[ a = \root{3}\frac{2 \times 95.95}{10.28 \times 6.022 \times 10^{23}} \]
07

- Determining the atomic radius

For a BCC unit cell, the relationship between the edge length \(a\) and the atomic radius \(r\) is: \[ 4r = \root{3}a \] Solving for \(r\): \[ r = \frac{a \root{3}}{4} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nitrogenase Enzyme
The nitrogenase enzyme is a crucial protein for life on Earth. It helps certain plants convert nitrogen gas from the atmosphere into a form that plants can use, like ammonia. This process is known as nitrogen fixation. Without nitrogenase, plants wouldn't be able to absorb the nitrogen they need for growth.

Nitrogenase contains clusters of iron (Fe), sulfur (S), and molybdenum (Mo) atoms. These clusters are essential for the enzyme's ability to facilitate the chemical reactions involved in nitrogen fixation. Understanding the properties of molybdenum, one of these critical atoms, helps us grasp how the enzyme functions at a molecular level.
Molybdenum Atomic Radius
The atomic radius of molybdenum is a key parameter in understanding how atoms are arranged within materials. In the context of the nitrogenase enzyme, molybdenum plays a pivotal role. To calculate the atomic radius, we first need to understand its crystalline structure.

Molybdenum adopts a body-centered cubic (BCC) crystalline structure. From this structure, we can derive that each unit cell contains 2 molybdenum atoms. By calculating the edge length of the unit cell and using it in further formulas, we can find the atomic radius. The atomic radius is crucial in determining the enzyme's active site configuration and its overall function.
Crystalline Structure Calculations
Crystalline structures define how atoms are packed together in a material. For molybdenum, a body-centered cubic unit cell is considered. This means there is an atom at each corner and one in the center of the cube.

To calculate properties such as the unit cell's edge length and atomic radius, we use specific formulas. For instance, the density and molar mass help determine how many atoms exist in a given volume. These calculations involve Avogadro's number, which is a constant representing the number of atoms in a mole. Understanding these calculations helps us model the physical properties of materials accurately.
Unit Cell Edge Length
The edge length of a unit cell is a fundamental measurement in crystallography. It represents the distance between the corners of the cube in the unit cell. To find this for molybdenum, we start with the given density and molar mass.

By calculating the mass of one unit cell and its volume, we can determine the edge length. This is done using the formula for the volume of a cube. Finally, knowing that molybdenum has a BCC structure, we can relate this edge length to the atomic radius using the appropriate equations. Knowing the edge length helps in further understanding the geometric packing and spatial arrangement of atoms in the cell.

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Most popular questions from this chapter

Mercury (Hg) vapor is toxic and readily absorbed from the lungs. At \(20 .^{\circ} \mathrm{C},\) mercury \(\left(\Delta H_{\mathrm{vap}}=59.1 \mathrm{~kJ} / \mathrm{mol}\right)\) has a vapor pressure of \(1.20 \times 10^{-3}\) torr, which is high enough to be hazardous. To reduce the danger to workers in processing plants, \(\mathrm{Hg}\) is cooled to lower its vapor pressure. At what temperature would the vapor pressure of \(\mathrm{Hg}\) be at the safer level of \(5.0 \times 10^{-5}\) torr?

The repeat unit of the polystyrene of a coffee cup has the formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{2}\). If the molar mass of the polymer is \(3.5 \times 10^{5} \mathrm{~g} / \mathrm{mol},\) what is the degree of polymerization?

Rank the following in order of increasing surface tension at a given temperature, and explain your ranking: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

Classify each of the following as a conductor, an insulator, or a semiconductor: (a) carbon (graphite); (b) sulfur; (c) platinum.

One way of purifying gaseous \(\mathrm{H}_{2}\) is to pass it under high pressure through the holes of a metal's crystal structure. Palladium, which adopts a cubic closest packed structure, absorbs more \(\mathrm{H}_{2}\) than any other element and is one of the metals used for this purpose. How the metal and \(\mathrm{H}_{2}\) interact is unclear, but it is estimated that the density of absorbed \(\mathrm{H}_{2}\) approaches that of liquid hydrogen \((70.8 \mathrm{~g} / \mathrm{L}) .\) What volume (in \(\mathrm{L}\) ) of gaseous \(\mathrm{H}_{2}\) (at \(\mathrm{STP}\) ) can be packed into the spaces of \(1 \mathrm{dm}^{3}\) of palladium metal?
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