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Vanadium crystallizes with a body-centered cubic unit cell. The volume of the unit cell is \(0.0278 \mathrm{~nm}^{3}\). What is its atomic radius?

Short Answer

Expert verified
Answer: The atomic radius of vanadium in the BCC unit cell is approximately 0.131 nm.

Step by step solution

01

Find the lattice constant (a) of the BCC unit cell

The volume of a cubic unit cell is given by \(V = a^3\), where 'a' is the lattice constant. We can solve for 'a', given the volume: $$ a = \sqrt[3]{V} $$ Substitute the given volume, \(V = 0.0278 \mathrm{~nm}^{3}\): $$ a = \sqrt[3]{0.0278} $$ Calculate the lattice constant: $$ a \approx 0.303 \mathrm{~nm} $$
02

Determine the relation between atomic radius and lattice constant for BCC unit cell

In a BCC unit cell, atoms are located at the corners and the center of the cube. If we consider a body diagonal of the unit cell, there are two atomic radii (corner and center atoms) and one lattice constant (edge) on a straight line. We obtain a relation between the atomic radius and the lattice constant using the Pythagorean theorem for a right angle triangle formed using half of the body diagonal, half of the edge (lattice constant), and one atomic radius as its sides: $$ \left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 = (2r)^2 $$ Simplifying, we get: $$ \frac{a^2}{2} = 4r^2 $$
03

Calculate the atomic radius using the lattice constant and relation derived in Step 2

Using the derived relation and the calculated lattice constant, we can find the atomic radius 'r': $$ 4r^2 = \frac{a^2}{2} $$ Plug in the lattice constant value, \(a \approx 0.303 \mathrm{~nm}\): $$ 4r^2 = \frac{(0.303)^2}{2} $$ Solve for 'r': $$ r = \sqrt{\frac{(0.303)^2}{8}} $$ Compute the atomic radius: $$ r \approx 0.131 \mathrm{~nm} $$ The atomic radius of vanadium in the BCC unit cell is approximately \(0.131 \mathrm{~nm}\).

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Key Concepts

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

Body-Centered Cubic (BCC)
The body-centered cubic (BCC) structure is a crucial concept in crystallography that describes one of the ways in which atoms arrange themselves in a crystal lattice. In this structure, every unit cell has atoms at each of the eight corners of a cube, as well as one atom at the very center of the cube. This is a frequently observed arrangement in metals like vanadium, iron, and tungsten.
  • Atoms at the Corners: Each corner atom in the BCC structure is shared by eight neighboring unit cells, meaning it is not exclusive to one unit cell but rather a "borrowed" atom.
  • Central Atom: The central atom is entirely within the unit cell, contributing fully to the structure of that cell.
  • Coordination Number: The coordination number in a BCC structure is 8, meaning each atom is in direct contact with 8 other atoms.
Because of this arrangement, the BCC structure is less densely packed compared to the face-centered cubic structure, but it offers significant strength and hardness to metals that adopt this formation. Understanding these properties lays the foundation for calculating atomic positions and eventually deriving atomic radii.
Lattice Constant
The lattice constant is a useful term when examining crystal structures such as the BCC. It is defined as the edge length of the cubic unit cell in which the atoms are arranged. Calculating the lattice constant provides essential information about the scale of the crystal lattice, and it is a stepping stone towards identifying properties like the atomic radius.
To find the lattice constant when the unit cell volume is given, you use the cube root formula for volume:
  • Formula: \( a = \sqrt[3]{V} \)
  • Volume Substitution: By substituting the given volume of \( 0.0278 \mathrm{~nm}^{3} \), it calculates as follows:
  • Calculation: \( a \approx 0.303 \mathrm{~nm} \)
Measuring the lattice constant allows scientists and engineers to understand and predict how atoms will behave in a material, informing everything from electrical conductivity to strength.
Pythagorean Theorem in Crystallography
The Pythagorean Theorem is not just limited to calculating the hypotenuse of a triangle; it has significant applications in crystallography as well. For a BCC unit cell, it helps us find relationships between the lattice constant and the atomic radius.
Consider a body diagonal within the cube, which stretches from one corner through the center to the farthest opposite corner. In the context of the BCC structure, this body diagonal involves part of the cube's geometry and the arrangement of atoms across it.
Geometric Application:
  • The body diagonal consists of a relation involving two half-lattice constants and two atomic radii.
  • Formula Derivation: The diagonal length (from geometry) is \((2r)\), which aligns with the body diagonal formula by the Pythagorean theorem: \(\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 = (2r)^2 \).
  • By rearranging terms, you see how it simplifies to \(\frac{a^2}{2} = 4r^2 \), leading you to calculate the atomic radius.
This deduction allows for a clear understanding of how the physical dimensions of a BCC unit cell relate to the atoms within it, making it possible to estimate things like the atomic radius, as seen in the exercise with vanadium.

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Most popular questions from this chapter

Chloroform, \(\mathrm{CHCl}_{3}\), was once used as an anesthetic. In spy movies it is the liquid put in handkerchiefs to render victims unconscious. Its vapor pressure is \(197 \mathrm{~mm} \mathrm{Hg}\) at \(23^{\circ} \mathrm{C}\) and \(448 \mathrm{~mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). Estimate its (a) heat of vaporization. (b) normal boiling point.

The vapor pressure of bromine, \(\mathrm{Br}_{2}(l)\) at \(25^{\circ} \mathrm{C}\) is \(228 \mathrm{~mm} \mathrm{Hg}\). (a) How many grams of bromine will be present as a gas if liquid bromine is poured into an evacuated 2.00-L flask at \(25^{\circ} \mathrm{C} ?\) (b) If \(2.00 \mathrm{~g}\) of bromine is used, what is the final pressure in the flask? Will there be liquid in the flask? (c) If \(2.00 \mathrm{~g}\) of bromine is put into an evacuated \(750.0\) -mL flask at \(25^{\circ} \mathrm{C}\), what is the final pressure in the flask? Will there be any liquid in the flask?

In which of the following processes is it necessary to break covalent bonds as opposed to simply overcoming intermolecular forces? (a) subliming dry ice (b) vaporizing chloroform \(\left(\mathrm{CHCl}_{3}\right)\) (c) decomposing water into \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) (d) changing chlorine molecules into chlorine atoms

What are the strongest attractive forces that must be overcome to (a) melt ice? (b) sublime bromine? (c) boil chloroform (CHCl \(_{3}\) )? (d) vaporize benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) ?

Methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), has a normal boiling point of \(64.7^{\circ} \mathrm{C}\) and has a vapor pressure of \(203 \mathrm{~mm} \mathrm{Hg}\) at \(35^{\circ} \mathrm{C}\). Estimate (a) its heat of vaporization \(\left(\Delta H_{\text {vap }}\right)\). (b) its vapor pressure at \(40.0^{\circ} \mathrm{C}\).

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