Metallic Bonding and Crystal Structures. Copyright © Houghton Mifflin Company. All rights...

Metallic Bonding and

Crystal Structures

Copyright © Houghton Mifflin Company. All rights reserved. 4 | 2

Imagine Forming a Metallic Solid Starting with the Isolated Atoms in a Regular Arrangement

(a) Atoms are far apart an non-interacting(b) Atoms are just touching(c) The atoms are tightly packed and closer than their sizeConclusion: Metal atoms lose their valance shell electrons to

the “electron sea” which acts to bind the atoms together.


The Unit Cells for the Crystal Lattices of Metallic Elements


The Crystal Structure of the Metallic Elements; The Semimetals Si and Ge; and the Noble Gases

in their Low-temperature, Solid Form


The crystal form to which any given element crystallizes is a balance of several factors

related to orbitals.

•The shape and occupation of orbitals. Example: Filled s and p orbitals are spherical and typically pack as spheres would. See noble gases, Group 1/2 ions and the last transition metals Groups 11/12.

•Non spherical unfilled orbitals tend to result in more localized electrons and will alter the geometric shape.


A Collection of Spherical Objects Is Packed Differently(a) a Simple Square Array (b) a Close Pack Layer

The simple square in (a) extended to 3-D results in a “Simple Cubic” Structure


In a Simple Cubic Solid, Each Corner of the Cube Is Occupied by an Atom, with the Center of the Atom at the Corner Point, Space

Is Filled with These Cubes Stacked in All Three Directions


A Collection of Spherical Objects Is Packed Differently(a) a Simple Square Array (b) a Close Pack Layer

The close packing in (b) extended to 3-D results 2 possible structures depending on how they are stacked


In Hexagonal Packing, Cations in the Second Layer Sit in Triangular

Hollows of the First Layer


• The second layer atoms can sit directly over the atoms in the first layer—called an AA pattern.

• Or the second layer can sit over the holes in the first layer—called an AB pattern.

Closest-Packed Structures Second Layer


The determiner of the overall structure depends on the third layer.

•If the third layer sits directly over an atom of the first layer, it is hexagonal close packing or HCC

•If the third layer sits over the holes in the first layer, it forms a Face-Centered unit cell or FCC.


Unit cell differences, HCP and FCC


Hexagonal Closest-Packed Structures

ABA packing produces the hexagonal crystal form


ABC Packing produces a “Face-Centered” cubic cell, or fcc.


The Relationship of the Close Packed Planes with the Unit Cell in an FCC Lattice Is Shown by Standing the Unit Cell on One of its Corners with the Diagonally Opposite Corner Directly Above It


The Relationship of the Close Packed Planes with the Unit Cell in an FCC Lattice Is Shown by Standing the Unit Cell on

One of its Corners with the Diagonally Opposite Corner

Directly Above It


Three Representations of a Face-centered Unit Cell


The Body-Centered Cubic Unit Cell

This third-type often occurs due to localized orbitals rather than the spherical shapes. It also occurs in alloys when to atoms are of different sizes.


Problem:

The density of Calcium is 1.55 g/cm3 and it crystallizes into an FCC unit cell. From this compute the radius of Ca in this crystal form. The atomic radius of Ca is given to be 2.23 Angstroms(AW = 40.078)

(Answer: 1.97 Angstroms)


X-Ray Diffraction Analysis


Diffraction from a Crystal


Problem:

Scattering of a crystal of Iron, Fe, using X-rays of wavelength 154 pm produced a primary spot at an angle of 32.45o. Given the density and AW of Fe:

a.) Determine the unit cell structure of Fe.

b.) Compute the covalent radius of Fe.

c.) What % of the unit cell in the Fe structure is occupied by atoms? (Packing Efficiency)


Comparison of Common Crystal Structures

Material Properties


A Typical Stress-Strain Plot


The Elastic (Young’s) Modulus

Note: E as represented in Schultz is given as mass/dist2. F/A is a pressure unit, mass/disttime2. Schultz has divided by the gravitational constant to get her values. Young’s Modulus is generally given in pressure units.


The Elastic Modulus for Transition Metals and Calcium First Increases and Then Decreases Across the Fourth Period


The Atomic-scale Result of Application of a Stretching Force. Bonds Along the Length are

Elongated, while Bonds Across the Width are Compressed


Plastic Distortion Occurs When the Strain Becomes too Great and the Atoms Shift,

Adopting New Neighbors


1.) Wires for electrical transmission on power lines are made of copper. If the copper line has a diameter of 2 mm and is covered with ice during an ice storm to a thickness of 1 inch radius, how much will the power lines stretch given the following:

•Ecu = 13480 kg/mm2

•Poles are 40 ft apart.•Density of Ice, Dice = 0.918 g/cm3

(Answer = 0.0643 in)

2.) Mo has a much higher elastic modulus. a.) Compute the stretch for Mo if E = 36140 kg/mm2.

(Answer = 0.0240 in or about 1/3 as much)

b.) Give at least three reasons why Mo is not used for electrical wires.


The Valence Electron Configuration of Fourth-Metal Period

• As one moves through the d-block elements, electrons added to the d-orbitals are more localized and result in tighter bonds.

• Past the 1/2 –filled point, the orbitals are becoming more spherical resulting in weaker bonding and a lower elastic modulus.


The Energy of a Typical Bond

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

7

8

9

10Softer MaterialStiffer Material

Comparison of Potentials for Strong and Weak Bonds

Distance (Angstroms)

Pot

enti

al E

nerg

y (e

V)


The Dark Line Represents the Variation in Energy as Two Rows of Atoms Slide over Each Other. In a Close Pack Structure (a) the Variation Is Quite Smooth. In a

Body-centered Cubic Structure (b), the Larger Spacing Between Atoms Means that there Is a Larger Energy

Barrier to Get from One Position to the Next.


A Close Pack Plane (FCC or HCP) and a BCC Lattice

A bcc structure has atoms that are farther apart and thus are harder to shift.


Slip Planes of the BCC, FCC, and HCP Crystal Lattices


It Takes a Great Deal of Effort to Pull All the Teeth of a Zipper Apart at Once, Pulling Apart a Pair of Teeth at a Time Requires Much Less EffortThus when plastic deformation occurs, it’s due to moving a few atoms at a time. This is done through defects.


(a) A Single Missing Atom in an Otherwise Regular Array is an Example of a Kind of Point Defect Called a Vacancy Defect.

(b) A Row of Atoms Where the Regular Crystalline Array Is Disrupted Is a Line Defect.

(c ) An Area Where Two Crystalline Arrays Meet but are out of Register with Each Other Is a Grain Boundary

Defects


Bubble Raft Depicting Defects


Find the Defects (a.k.a. Where’s Waldo?)


A “doped” surface


A Line Defect in a Simple Cubic Lattice


Successive ”Snapshots" of a Line Defect Rippling Through a Crystal Lattice


As the defect ripples through the solid lattice, it is difficult for it to cross the gap at the grain boundary.Thus a metal can be hardened by introducing grain boundaries.


Conductivity and Resistivity These two are inverse relationship to each other. Conductivity is presented for elements on your periodic chart.

It is important to recognize that electrons do not “move” through a metal. Rather, the push each other over one at a time.

In order to conduct, an electron has to have a place to go easily.


Factors Affecting Conductivity

Most metals have open and available orbitals that can exchange electrons.



In ½ filled orbitals, a moving electron must pair up and this requires more energy.



In filled orbitals, electrons must be able to use other orbitals such as p orbitals and those must be close in energy.

•Most Group 12 metals have low conductivity for this reason. They can conduct only because p-orbitals are close enough in energy to help.•Hg is very resistant because of interesting quantum relativistic effects. “f-orbital contraction” and “s-orbital isolation”.


Resistivity of First Transition Series Elements Decreases Smoothly from the Start of the Series

to the End, with the Notable Exception of Mn.


Resistivity Shows a General Decrease Across All Three Transition Series

Note the dramatic drop in resistivity down Group 7. Why?


½ Filled Elements Decrease in Resistivity down a GroupFrom Chapter 2 – Notice how energy levels get closer together as the period number increases. This makes additional orbitals available for electron movement than Mn. (We will see that orbital mixing will account for this even better.)


Resistivity of Transition Elements

Phase Transitions


Phase TransitionsPhase transitions occur when energy supplied allows different movement or separation between atoms.

•In a solid, atoms are bound in place and can move by vibrating.•In a liquid, sufficient energy is available to overcome the bonds holding the atoms in place but is insufficient to allow them to break away completely. •In a gas, atoms have gained sufficient energy to have escaped from the bulk material.•A phase transition can also occur by atoms changing their bonding and relationships to their neighbors.


(a) Below 13.6° C, the Stable Phase of Tin, has a Tetrahedral Arrangement of Atoms. (b) The Unit Cell of the Diamond Structure Is Related to the Face-centered Cubic Unit Cell by the Addition of an Atom Between Every Other Corner Atom and the Three Adjacent Face Center Atoms


Above 13.6° C , Tin Is Metallic and has Eight Nearest Neighbors in a Tetragonal Crystal


The Phases of Tin Plotted on a Temperature Line


Adding a Pressure Axis to the Temperature Line Generates a Phase Diagram: A Plot of the

Temperature and Pressure Range over which a Given Phase Is Stable


Alloys

• Alloys are mixtures of different metals or the dissolving of one metal into another.

• The atoms arrange together and generally produce very different properties based on the concentrations of each.

• Two major types of alloys are:– Substitutional – Here atoms of one metal are substituted in

the crystal by the atoms of another metal– Interstitial – Here, atoms fit in the “interstitial” spaces within

the unit cell.


Selected Alloys


Two Elements form a Substitutional Alloy over a Large Compositional Range if they are Similar in

Size, Crystal Structure, and Electronegativity


Copper-Nickel Solutions

The compatibility of metals to form alloys over a wide temperature range are guided by

Hume-Rothery rules.


Brittle intermetallic compound makes ultrastrong low-density steel with large ductility

Sang-Heon Kim,1, Hansoo Kim1, & Nack J. Kim1, Affiliations

Nature Volume: 518, Pages:77–79 Date published:(05 February 2015)

• Recently, high-aluminium low-density steels have been actively studied as a means of increasing the specific strength of an alloy by reducing its density3, 4, 5. But with increasing aluminium content a problem is encountered: brittle intermetallic compounds can form in the resulting alloys, leading to poor ductility. Here we show that an FeAl-type brittle but hard intermetallic compound (B2) can be effectively used as a strengthening second phase in high-aluminium low-density steel, while alleviating its harmful effect on ductility by controlling its morphology and dispersion. The specific tensile strength and ductility of the developed steel improve on those of the lightest and strongest metallic materials known, titanium alloys


A Simulated Temperature Line for a Mixture of Copper and Nickel


The Copper-Nickel Phase Diagram Indicates that Copper and Nickel Form a Homogeneous Solid Solution over the Entire

Composition Range. This results because Cu and Ni form a good Hume-Rothery Match


Question: A 40% Ni solution at 1400 oC is cooled.At what temperature does solid form and what is the composition?

Solid first appears at just under 1300oC. At that temperature, the solid is about 54% Ni and, thus, 46% Cu. The liquid is still 40% Ni


Question: A 40% Ni solution at 1400 oC is cooled to 1250oC .What are the compositions of the liquid and solid phases?

At 1250oC, the liquid is about 30% Ni and the solid is about 44% Ni.


The Sn-Pb Phase Diagram Is an Example of the Type of Behavior Found Among Elements with a

Poor Hume-Rothery Match


Parts of the Phase Diagram.


Reading the phase diagram.Question: What is the phase and composition of this alloy at 250oC for:

a.) 20% Sn b.) 70 % Sn ?

At 250oC the solution is a liquid-solid mixture at 20% Sn and liquid at 70% Sn


Reading the phase diagram.Question: What are the phases and compositions of a 40% Sn alloy as it is cooled from a.) 300oC to b.) 200oC to c.) 150oC as indicated by the points

drawn in?


Reading the phase diagram.Answer a.) At 300oC, the alloy is all liquid that is 40% Sn.


Reading the phase diagram.Answer b.) At 200oC, the alloy is a liquid/solid mixture.

The composition is found using a tie-line. At 250oC the solid is about 18% Sn and the liquid is about 58% Sn.


Reading the phase diagram.Answer c.) At 150oC, the alloy is a solidified mixture of two different compositions.

The compositions are found using a tie-line here as well.At 150oC there is a Sn rich solid of about 97% Sn/3% Pb and a Pb rich solid of

about 10% Sn/90% Pb..


Reading the phase diagram.Finally we can determine how much of the alloy is the 98 % Sn and how much is the 10% Sn. The fraction of the 10% Sn is the length of segment “a” divided by the total

length a + b. This gives about (40-10)/(98-10) x 100 = 34 %. So the solid phase is seen to be 34 % Pb rich alloy and 66 % Sn rich alloy.


The Cooling Rate Determines the Morphology of the Solid Formed

• We saw that as the alloy is cooled, two different forms of solid are produced that have different compositions.

• If the solid is cooled very rapidly, then the diffusion that occurs during slow cooling is not allowed to happen.

• The results is a very different alloy structure.

• Fast cooling produces many grain boundaries which we saw earlier increase the strength.


The Cooling Rate Determines the Morphology of the Solid Formed


A Portion of the Phase Diagram for Iron and Carbon

Rapid cooling is utilized in the manufacturing of steel. Rapid cooling is done from the Austentite phase to produce a less regular phase called Martensite.


Interstitial Alloys

If two materials do not match according to the Hume-Rothery rules, then they typically do not form substitutional alloys. If there is a disparity in size, then one atom might fit into the empty spaces in a crystal.

Often this stresses the crystal structure and adds to the strength. Such is the case with steel where carbon atoms are in the interstitial spaces.


(a) Two Unit Cells of the Iron Lattice are Shown with the Interstitial Site Indicated. (b) In steel, Carbon Occupies Some of the Interstitial Sites.

Is the space indicated by the site large enough to accommodate a carbon atom?


Question: What is the distance between the center atoms and also between the corner atoms? To fit, carbon must have a radius less than these dimensions.

Answer: The dimensions of the space are 0.362 Angstroms between side atoms and 1.481 Angstroms between corner atoms along the diagonal


The Nitinol – Memory Wire

The alloy of Nickel and Titanium produces a structure that alters in structure with heat and possesses a “memory” effect.

In the cool phase, Nitinol is easily bent but upon gentle warming, returns to it’s original shape.

Applications are limitless.


Nitinol Stent placed to keep blood flow through a major artery.

Open Journal of Modern NeurosurgeryVol.2 No.3(2012), Article ID:21174,5 pages


The Nitinol Unit Cell Showing Structure of the High-temperature Austenite Phase


The Low-temperature Phase of Nitinol Is Characterized by a Skewed Unit Cell, Compressed

Along Some Directions More than Others


Compressing and Tilting a Square Produces a Parallelogram


Portion of Ni-Ti Phase Diagram

Metallic Bonding and Crystal Structures. Copyright © Houghton Mifflin Company. All rights...

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