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STANFORD / Engineering and Tech / ENGR 50 / What are the primary bond types?

What are the primary bond types?

What are the primary bond types?

Description

School: Stanford University
Department: Engineering and Tech
Course: Introduction to Materials Science, Nanotechnology Emphasis
Professor: Professor sinclair
Term: Spring 2016
Tags: Nanotechnology, materials, and Science
Cost: 50
Name: ENGR50 Review Sheet of some topics
Description: Covers all topics based on lecture notes before ionic structures
Uploaded: 04/24/2016
6 Pages 169 Views 1 Unlocks
Reviews


Synthesis of Nanomaterials


What are the primary bond types?



I. Synthesis of nanoparticles

A. Chemical Reduction

1. Produced by liquid chemical methods

B. Chloroauric acid process:

1. After dissolving H[AuCl4], the solution is stirred while a reducing agent is added

2. Causes Au3+ atoms to be neutralized to Au atoms

3. The solution becomes supersaturated with pure Au atoms, so the gold precipitates in nanometer­sized particles

II. Synthesis of nanowires

A. Vapor­Liquid­Solid method (VLS)

1. Solid nanoparticles (i.e. Au) placed on substrate

2. A gas is introduced (such as Silane, SiH4), and decomposes into Si and H

3. Silicon dissolves through nanoparticle, forming a liquid, then produces a solid Si crystal on substrate as it further dissolves


What is vapor-­liquid-­solid method (vls)?



4. Gradually a nanowire forms, liquid nanoparticle always on the top of

growing nanowire; nanowire grows beneath the liquid nanoparticle

III. Synthesis of nanotubes

A. CNT synthesis and CNT growth If you want to learn more check out What is the deutsch word of to play the guitar?

1. Substrate is prepared with a layer of the metal catalyst nanoparticles 2. Two gases are added introduced into the reactor: process gas and a carbon containing gas

3. CNTs grow on the site of the metal catalyst

4. The carbon containing gas breaks apart at the surface of the catalyst particle, and the carbon is transported to the edges of particle to form the CNT

5. Catalyst particles can stay at the tip of growing CNT or remain at the CNT base (depends on type of catalyst and its adhesion with substrate)

IV. Thin films


What is the synthesis of nanotubes?



A. PVD

1. No chemical reactions involved

2. Involves a substrate and a target element

3. Electron beams are repeatedly shot at the target element in vacuum 4. This process is called ablation

5. Once supersaturated, the target element releases its atoms in gas state, hitting the substrate, then solidifying

6. Over time, layers and layers of these released atoms will build up on the substrate surface We also discuss several other topics like What is an area between supply curve and price curve?

B. CVD

1. Chemical reactions involved

2. However less common because of creation of byproducts, which may taint the purity of the thin film

V. Multiple thin films

A. Photoresists and photolithography

1. Copper is placed on substrate

2. Add a photoresist, a light­sensitive polymer

3. Protecting some areas with a mask, UV light is shined

4. The photoresist is now prepared for etching

B. Etching

1. After etching, more copper is deposited

2. The remaining photoresist is removed, leaving behind a channel in the copper

Bonding

I. Primary bond types If you want to learn more check out What is the study of systems in state of constant motion at rest and constant velocity?

A. Metallic

1. Lowered energy state, allows force between metallic atoms to bond 2. Gives rise to freely flowing electrons, which is responsible for metals’ conductivity

3. Orderly, stately structure of atoms, very strong bond

B. Covalent

1. Sharing of electrons

2. Tetrahedral­shaped bonds

3. Stable balance of repulsive and attractive forces between

atoms/molecules

C. Ionic

1. Electrons given off and electrons taken away between bonded elements 2. Always involves a metal and a nonmetal, typically an alkali metal and group 7 element If you want to learn more check out Who was the leader of the french revolution?

3. Metals: usually electropositive (give out electron), called cation

4. Nonmetals: usually electronegative (take electron), called anion

5. Anion usually larger than cation radius

II. Secondary bond types

A. Van der Waals

1. Molecules, atoms are fluctuating dipoles

2. Small attraction between molecule to molecule

3. Example: Neon, methane

4. Low melting point

B. Hydrogen

1. Permanent dipole

2. Attraction between negative and positive dipole of the molecule

3. Example: water (H2O)

4. Usually involve C=O and OH groups in polymers

Interatomic Potential and Forces

I. Interatomic forces

A. Attractive

1. Equation given by Coulomb’s Law of attraction

2. F_attract = (z_1 * z_2 * e^2) / (4 * PI * Epsilon * a^2)

3. From the partial derivative of the interatomic potential, it is the term with the positive sign

B. Repulsive

1. Always has the form ­B/(a^m) Don't forget about the age old question of Define acceleration.

2. From the partial derivative of the interatomic potential, it is the term with the negative sign

II. Interatomic potential

A. Energy of the bond between atoms

B. Has the form: A/(a^m) ­ B/(a^n), where A, B, m, and n are given constants C. Lower temperature materials have a lower Vmax than higher temperature materials

D. Increasing the temperature causes higher amplitude of vibration between atoms (note frequency is still constant)

III. Important relationships to remember

A. At a0, where the distance between the atoms is equal to the sum of their radii, the interatomic net force is always 0.

B. At a0, the interatomic potential V is always at a minimum

C. The partial derivative of the potential is the force

D. Finding the which distance gives the most net force: take the derivative of the expression relating force with the interatomic distance (F versus a) and set equal to zero

E. Elasticity modulus: directly related to interatomic force and potential

F. Steeper curve of graph: corresponds to higher Elasticity modulus

Crystal Structures:

I. FCC (front center)

A. Stacking scheme: ABCABC…

B. Symmetrical along all axes

C. Isotropic properties are due to the close­packed sheets’ ability to slide in many different directions without breakage or allowing crack propagation

D. Unit cell is a cube with atomic nuclei positioned at perfect 90 degree angles from each other If you want to learn more check out Is there an equivalence between the citizens and the state to goodness?

E. It is called Front centered because at each face of the cubic unit cell, there is always a atom “visible” in the center of the cube

F. Close­packed sheets of the crystal structure do not lie “on top of each other”, rather are stacked diagonally, angled at around 55 degrees

G. Total orientation: 12

H. FCC materials: High temperature Fe, Al, Cu, Ag, Au, Ni, Pd, Pt, High temperature Cobalt

I. Face diagonal relationship between a and r: sqrt(2) * a = 4 * r

J. Calculations

1. Number of atoms in a 10 nm particle

a) Formula: Volume of nanoparticle / Volume occupied by one atom

b) Volume occupied by one atom = a^3 / 4

c) Volume of spherical nanoparticle: 4/3 * PI * r^3 where r = a/2

2. Density of FCC unit cell

a) Can only consider the mass contained within the boundaries of the

defined unit cell

b) Therefore, there are only four atoms that are entirely contained

within any unit FCC cell

c) Total volume of the cubic cell is a^3, where a is the atomic radius

d) Formula: (4 atoms * atomic mass / volume contained) *

Avogadro’s number

e) Alternatively, (4 * amu / a^3) * 1.66 * 10 ^ ­24

f) Density units: gm/cc

II. BCC (body center)

A. Not made up of close­packed planes

B. Instead, there is just one atom in the center of the unit cell, with the other nuclei forming a cube around it

C. Number of atoms contained within a BCC unit cell: 2 (one in center, ⅛ * 8 atoms in each corner)

D. Materials: V, Cr, Mo, W

E. Instead of a 12­atom orientation, BCC has 8­atom orientation

F. This means that bonds are stronger ­> higher melting points than FCC and HCP G. Diagonal plane of BCC unit cell: relationship of r and a given by 4r = sqrt(3) * a III. HCP (Hexagonal Close Packed)

A. Stacking scheme: ABAB…

B. Note that it can also be ACAC, BABA, or CACA, depending on orientation of crystal

C. Unit cell is a parallelopiped, a parallelogram with perfect 120 degree top angle D. Each atom in the B layer is connected to three atoms on top and three atoms on the bottom (six bonds in total with A layer atoms)

E. Total coordination: 12 atoms (6+3+3)

F. HCP materials: Be, Mg, Zn, Cd, Ti, Zr, low­temperature Cobalt

G. Anisotropic properties based on how the close­packed planes can only move in one direction

H. Important HCP c to a ratio: c/a = sqrt(8/3), where a = 2 * atomic radius

Atomic Packing Factor

I. General Formula

A. APF = Volume occupied by spherical atoms / Total Volume

B. Notice how the APF does not depend on the size of atom ­ in both FCC and BCC, the radius ends up canceling out (shown below)

II. FCC

A. APF = (4 atoms * (4/3) * PI * r^3) / a^3

B. Substituting a = 4r/sqrt(2) using the FCC a to r relationship and simplifying: 1. APF = (sqrt(2) * PI )/ 6

III. BCC

A. APF = (2 atoms * (4/3) * PI * r^3) / a^3

B. Substituting a with the BCC a to r relationship and simplifying:

1. APF = (sqrt(3) * PI) / 8

IV. Calculating Volume change using APF (BCC to FCC or vice versa) A. Volume change = APF_ FCC ­ APF_BCC

A note on steels:

High temp Fe: nonmagnetic, FCC (gamma iron)

Low temp Fe: BCC, magnetic (alpha iron)

Regular Steel: BCC, structure, Fe + Carbon + alloys

Alloys:

I. Random crystal structures

A. Substitutional alloys

B. e.g. Brass (Cu, Zn) and Bronze (Cu, Sn)

1. The larger Zn atoms cause strain, which strengthens the copper

2. In both cases, the atoms are randomly “scattered” in close packed planes II. Interstitial Alloys

A. e.g. Steel

1. The added element has a smaller radius than the host atom

2. In steel, carbon is much smaller than Fe

III. Ordered alloys

A. Atoms in regular positions, not random

B. Ex: Nitinol (NiTi), crystal structure is the CsCl or B2 structure

1. Has one Titanium atom in the center of cube, all the other Ni atoms

located on cube corners

C. Unit cell structures are uniform

D. Formation of ordered compounds, which strengthen host metal

1. AlCu ­> CuAl2 + AlCu1

a) AlCu is FCC

b) CuAl2 forms an ordered compound

2. Fe(C) ­> Fe3C + FeC1

a) Fe(C) is FCC

b) Fe3C is the ordered compound in this reaction

3. NiAl ­> Ni3Al + NiAl1

a) NiAl is FCC

b) Ni3Al is a L12 crystal structure, which has Al atoms on the corners

of the cubic cell and the Ni atoms in the center of all the faces

Magnetic Properties:

I. Paramagnetism

II. Diamagnetism

III. Ferromagnetism

A. Strongest magnetization, the other two are weak

B. Especially important are Fe (BCC), Ni (FCC), and Co (HCP)

Domains:

Hard magnets: need to remove field, domains stay the same (e.g. steel)

Soft magnet: can be magnetized with exposure to magnetic field, domains change due to process called magnetic hysteresis

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