5- Classification & Taxonomy

The Five Kingdoms

There are 5 kingdoms in the classification system. Organisms are classified according to their evolutionary relationships (their phylogeny).

Phylogeny is the study of the evolutionary history of organisms, and gives us an insight as to how to group them and their extinct relatives. The base hierarchy in the classification system is the Kindom.

Generally, we can order the Kingdoms by increasing complexity. To help remember the names of the kingdoms, I was taught:

Pretty Polly Finds Parrots Attractive – Prokaryote, Protoctista, Fungi, Plantae, Animalia.

	Prokaryotes	Protoctista	Fungi	Plantae	Animalia
Cell Structure	Unicellular; no membrane bound organelles	Eukaryotes, Unicellular & Multicellular	Eukaryotes, Unicellular & Multicellular (Yeast)	Eukaryotic, Multicellular; Large Vacuoles	Eukaryotic, Multicellular
Cell Wall	Murein	Sometimes Polysaccharide	Chitin	Cellulose	None
Nutrition	Autotrophic, Aerobic Heterotrophic	Autotrophic, Hetrotrophic	Heterotrophic	Autotrophic (Photosynthetic)	Heterotrophic, Digestive System
Reproduction	Binary Fission	Fission	Spores	Seeds/Spores, Some asexual while some sexual	Develop from embryo
Example	Bacteria	Algae, Protozoa	Penicillin	Mosses, Ferns	Humans, Animals

NOTE

Q. What’s a photosynthetic organism?

A. An organism that gets its energy by absorbing light.

Q. What’s a autotrophic organism?

A. An organism which gets it’s energy from light (photosynthesis) or from chemical interaction (chemosynthesis).

Q. What’s a heterotrophic organism?

A. An organism that relies on complex organic matter for food.

Remember that 4 of the 5 kingdoms feature Eukaryotes! Only Prokaryotae contains Prokaryotes (no surprise there!).

Taxonomy (Breaking it down)

We break down organisms into a total of 7 hierarchical classes (including Kingdom above). That’s a lot of possible choices for organisms, and is know as Taxonomy, or Alpha Taxonomy.

The 7 levels are Kingdom, Phylum, Class, Order, Family, Genus and Species. You could remember this as:

King Penguins Climb Over Frozen Grassy Slopes

Here’s an example of two organisms and their taxonomy:

	Humans	Large White Butterfly
Kingdom	Animalia	Animalia
Phylum	Chordata	Arthropoda
Class	Mammalia	Insecta
Order	Primates	Lepidoptera
Family	Hominidae	Pieridae
Genus	Homo	Pieris
Species	sapiens	brassica

As you can see, humans are sapiens of the Genus Homo. AKA Homo sapiens (I bet you’ve heard that before!).

The only similarity between these two examples is that they are both in the Animalia kingdom. This means they share a great number of common traits, and so actually tells us a lot about the organisms.

It is also worth bearing in mind that Protoctista is often the ‘Other’ category where organisms who have no clear Kingdom are put. For example, Slime Moulds have fungi characteristics, yet are not quite suitable for classification in the Fungi Kingdom.

The Species

Species is the final tier on the taxonomy hierarchy; and is a group of organisms with similar traits. These include:

• Morphology (The outside appearance of an organism, including shape, colour, structure and pattern)
• Physiology (The way in which an organisms works, by looking at it’s biochemical, mechanical and physics functions)
• Behaviour

BUT most importantly, we can class two organisms as the same species if they can naturally breed together and produce fertile offspring.

The fertility point is an important one, as there are several organisms that can breed together, but produce a sterile offspring which cannot breed any further – such as a horse and a zebra which can produce a hybrid. This hybrid is sterile, so we know what the horse and the zebra are different species.

The Chemistry of Acid Chlorides, ROCl

Acid Chlorides:

Highly Reactive Carboxylic Acid Derivatives such as Acid Chlorides can be easily formed:

Easy preparation of acid halides from Carboxylic Acids

Acid Chlorides are:

• Highly reactive functional groups.
• Mainly involved in nucleophilic substitution reactions.
• Have identical reactions to acid bromides and acid anhydrides (so I will only focus on the Chlorides).

Flash animation showing the Step-by-step mechanism of the formation of an Acid Chloride from Thionyl Chloride and a Carboxylic Acid. Click to launch.

Acid Chlorides undergo a fair number of useful reactions. Below is a table illustrating them:

Paths to common, useful products of Acid Chlorides

The mechanism for all of these substitution reactions begins with the addition of Nu- or :NuH to the δ+ carbon atom of the carbonyl. This then creates an tetrahedral intermediate which then collapses to eject the chlorine (Cl-). The only difference with :NuH is an additional step where a base (such as pyridine) removes the H+ from the nucleophile.

Nucleophilic Substitution using Nu- and :NuH on Acid Chloride

Acid Chlorides can be converted into Ketones using organocopper reagents such as Me2CuLi and Ph2CuLi. This can be extremely useful in increasing chain length, amongst other things. The reason we used organocopper reagents instead of Grignard reagents (which we already know work) is down to how far the reaction goes. Grignard reagents are capable of converting Ketones into tertiary alcohols, and so tend to follow this route to completion.

The reactions involving Hydride ions are all run using weaker sources of H- than LiAlH4 (which would normally be the obvious choice). This is because the LiAlH4 will continue the conversion from an Aldehyde to a primary alcohol.

Addition of Aromatic Rings (Friedel Crafts Acylation):

Aromatic rings have no direct route for attack. They are poor nucleophiles (due to their stability) and as such require the Acid Chloride to be activated (made into a better electrophile) so they can be pulled in.

This activation can be achieved by using a Lewis acid such as AlCl3 or FeBr3. This reaction type is know as a Friedel Crafts Acylation. The animation below shows the mechanism and reaction scheme for this activation, and joining.

Friedel Crafts Acylation Mechanism - Addition of an Aromatic Ring to form Ketone. Click to launch animation.

If you’re wondering why the product does not reach further, simply consider the properties of the carbonyl group. The carbonyl group has electron withdrawing properties and as such reduces the available of electrons in the aromatic ring…requiring stonger conditions to instigate a second acylation reaction.

The Carbonyl, >C=O

One of the most important functional groups is the Carbonyl group.

Source: Wikipedia. Shows C=O and two additional atoms.

A couple of points about the carbonyl group:

• It is Planar (flat).
• Bond angles are 120 degrees.
• The Carbon = Oxygen double bond is the result of overlapping Pi and s orbitals.
• Both the Oxygen and Carbon atoms are sp2 hybridised.
• Oxygen has 2 lone pairs of electrons not involved in bonding.
• Oxygen is electronegative relative to Carbon and therefore the bond is polarised.

There are 2 ways to represent the polarisation of the carbonyl. Delta-notation to show partial charges, or Resonance forms to show the individual structures which contribute to the bonding sturture.

Resonance Forms

Reactivity:

There are three main loci of reactivity – with electrophiles, nucleophiles and bases.

• Reactions with Electrophiles, E+
  

  The Oxygen atom is electron rich and interacts with the Electrophile. One lone pair is used to form a new Sigma bond.

• Reactions with Nucleophiles, Nu-
  

  The carbon atom is electron deficient and so attracts the nucleophile.

• Reactions with Bases, BASE-
  

  A hydrogen on a neighbouring carbon is removed by strong bases, creating a resonance stabilised anion.

The reactivity of carbonyl compounds is influenced by the atoms attached.

Nucleophilic Substitution Reactions – Sn1 & Sn2 Stereochemistry

- Nuclephilic Substitution Reactions

The viability of nucleophilic substitution over a single bond is determined by the bond polarity. A nucleophile (Nu-) will attack the δ+ atom in a polar bond and replace the existing δ- atom.

A good example of this is the haloalkanes, where the halogens are more electronegative than the Carbon atom. As the the halogen has a higher affinity for -ve charge, the bonding electrons are found closer to the halogen than the carbon, shifting the dipole charges in the molecule.

Nuleophilic Substitution of Iodine with Cyanide in Iodomethane

As you can see the nucleophile (which likes +ve charge) attacks the δ+ carbon atom, and this essentially severs the C-I bond, releasing I-.

There are a large number of other suitable nucleophiles, including the following. I’ve included products too, excuse the lack of correct punctuation. This allows conversion of an Alkyl Halide into many different compounds.

Starting Material	Reacts With	Produces	AKA
RX	PPh3	RP+Ph3X-	Phophodium Salt
RX	R’S-	RSR’	Thioester
RX	Na2S	RSH	Thiol
RX	NC-	RCN	Nitrile
RX	HCC-	HCCR	Alkyne
RX	MeCOO-Ag+	MeCOOR	Ester
RX	EtO-	ROEt	Ether
RX	HO-	ROH	Alcohol
RX	N3-	RN3	Alkyl Azide
RN3	H2 / PdC	RNH2	Amide
RX	NH3	RNH3+X-	Ammonium Salt
RNH3+X-	HO-	RNH2	Amide

The ones in GREY at the bottom are the amide chain – there are two routes to an Amide. One is through an alkyl azide and the other through an ammonium salt.

*******************************************************************

A special note: Hydride (H-) reducing agents such as LiAlH4 can be used as sources of nucleophilic hydride ions which will replace the halogen group. This allows conversion of an alkyl halide into an alkane.

Nucleophilic Substitution of Halide with Hydride via LiAlH4

*******************************************************************

A stable molecule is a good leaving group, such that H2O is better than HO-. In haloalkanes, reactivity goes from RI > RF (travelling down the group).

This can be explained better when we look at the basicity and nucleophilicity of the atom/molecule. Note that while Nucleophilicity is a kinetic property determining the rate of reaction with a Carbon atom (how fast the reaction progresses), Basicity is a thermodynamic property determining an atom/ion/molecul’s ability to accept a proton.

Nucleophilicity (rate of reaction): NC- > I- > RO- > HP- > Br- > Cl- > ROH > H2O
Basicity (ability to accept proton): RO- > HO- > NC- > H2O > ROH > Cl- > Br- > I-

- The Reaction Pathway – Sn1 and Sn2 Reactions

There are two main pathways that a nucleophilic substitution reaction can follow:

Sn1 (Substitution, Nucleophilic, Unimolecular):

• Substrate ionises to form a planar intermediate carbocation in the rate determining step.
• The intermediate cation then rapidly reacts with the nucleophile. This means there are two transition states.
• This is a 1st order reaction as rate = k[substrate]. It is a unimolecular process.
• Favoured in polar solvents – this aids ionisation.
• Favoured Tertiary > Secondary > Primary as the two state process allows access to the carbon centre without steric hindrance (see Sn2 below).

Sn1 Substitution

Sn1 creates a racemic product (an equal amount of left and right enantiomers) which as a result is optically inactive. This means it will not rotate polarised light.

Sn2 (Substitution, Nucleophilic, Bimolecular):

• Reaction occurs completely within one transition state.
• This is a second order reaction as rate = k[substrate][nucleophile]
• Reaction favoured in polar aprotic solvents (solvents which have high polarity but cannot dissociate a H+) such as DMF (Dimethylformamide) and DMSO (Dimethyl Sulfoxide).
• Steric hindrance slows or stops reaction progression in tertiary systems as steric crowding stops attack by the nucleophile (aka there isn’t room!) and tertiary cations are quite stable. In this case we would expect Sn1. Note that the “back route” must be clear else the reaction will proceed by Sn1.
• Favoured Primary > Secondary > Tertiary.

Sn2 Substitution

Sn2 creates a product with an inverted stereo structure to that of the substrate. Essentially the Nucleophile attaches to the opposite side from the leaving group, inverting the molecule’s original stereochemistry.

- Alcohols

Alcohols are extremely important for synthesising new molecules:

Synthesising new molecules from Alcohols (In this case Propanol)

It is especially useful when you consider that we can already use the Alkyl Halide table from above to form a variety of molecules that way.

Hybridisation – Mixing Up Orbitals with sp, sp2, sp3

Essentially, hybridisation is the mixing of standard atomic orbitals to form new orbitals – which can be used to describe bonding in molecules.

Most importantly we have sp3, sp2 and sp hybridisation.

sp3 Hybridisation in Methane (CH4):

The best way I can describe sp3 hybridisation is in Methane (also the most basic choice!). This is simplified for expression. Remember that Carbon has 6 electrons.

• In methane (CH4), 1 Carbon binds with 4 Hydrogens. The carbon atom itself has only 2 electrons available for bonding in the 2p subshell.
  

  Carbon - Ground (normal) electron states. 1s2, 2s2, 2p1 2p1.

• In order for 4 hydrogens to bind there need to be 4 electrons available for bonding, which cannot be achieved at the moment. The pull of a hydrogen nucleus results in an electron being excited from the 2s subshell into the 2p subshell, where it is available for bonding.
  

  Carbon - An electron has been excited to the 2p orbital.

• This excitation changes the forces on the valence (bonding) electrons as the nucleus now exerts a stronger effective core portential upon them. This and other factors leads to the creation of a new ‘hybridised orbital’, called sp3.
  

  Carbon - Hybridisation forms sp3 orbital.

  This leaves 4 valence electrons which will each overlap with the s orbital of a Hydrogen to form a σ (sigma) bond. These hydrogens space themselves as far apart as possible, leading to the tetrahedral structure of methane.

  

  3D animation of methane. Produced on ChemSketch.

  Each of the bonds in the image above are σ-bonds.

  

  Methane Hybridisation. Shows the S orbits of H overlapping with sp3 orbitals of C. Note 2 electrons in each bond, one from carbon and one from hydrogen. Image by K. Aainsqatsi, released into public domain.

sp2 Hybridisation in Ethene (C2H4):

This is similar to sp3 hybridisation, except there are only 2 hydrogen nuclei pulling on the bonding electrons (which need an electron each) and the other 2 electrons are required for the π (pi) bond (double bond) between the two Carbons.

A molecule of Ethene.

The electron configuration in carbon starts the same:

Carbon - Ground (normal) state of electrons

then:

Carbon - An electron has been excited to the 2p orbital.

but the resulting spread is different:

Carbon hybridisation in Ethene (sp2)

Only 2 of the 2p orbitals are used in sp2 hybridisation; in contrast to the 3 used in sp3 hybridisation (you should be seeing where the numbers come from!).

This leaves us with 3 sp3-orbitals and 1 p-orbital to bond with. 2 of the sp3 orbitals are used for forming σ-bonds with the 2 hydrogens, while the remaining sp3 orbital binds with the other carbon to form a σ-bond and the p-orbital bonds with a p-orbital from the other carbon to form a π-bond.

Every double bond (regardless of what atoms it joins) consists of a π-bond and a σ-bond.

Shows seperate carbon atoms in sp2 hybridisation, then combined to form ethene.

sp Hybridisation in Ethyne (C2H2):

This can occur on an atom with a triple bond such as the alkynes. Ethyne is the simplest.

Ethyne. Triple bonded Carbon with 2 hydrogens.

In this case we only have 1 hydrogen attached to a carbon, and three bonds between each carbon.  That’s 1 hybridised bond between the carbon and hydrogen with another hybridised bond between the carbons. The other two p-orbitals form two more bonds between the carbons.

Ethyne - sp Hybridisation

This essentially means that the triple bond consists of 1 σ-bond and 2 π-bonds.

Summary:

Essentially, the hybridisation of the carbon atom is based on the number of bonds to other carbons or identical atoms.
sp3 = single bond
sp2 = double bond
sp = triple bond

Drawing Chemical Formulae on your PC

You will undoubtedly need to draw an equation out on your computer at some point, and there are several ways to do this.

1. Load up Paint and spend hours perfecting a drawing.
2. Use an Office suite and spend equally long hours trying to get all the lines in the right place.
3. Use a specialist application to do it in seconds.

Assuming you chose option 3, we’d like to introduce you to MDL ISIS DRAW 2.5. Best of all, it’s free for personal use!

A couple of the templates provided by ISIS DRAW

You must register first but you can then download ISIS draw through their site here. I believe they are working on replacing it as MDL has now become Symyx, so I am unsure whether it will stay free when they release the next version. The current version (just updated) now works better with Vista.

You could also use an application called ChemSketch. This can be downloaded from their website here.

These applications are very useful for drawing accurate molecules, checking them (for over bonded atoms etc) and naming. They can also generate 3D images of the molecules you make.

I’m not going to go into how to use them here, I’m just introducing you!

How about 3D?

There are quite a few notable molecular visualisation applications and I’ve just included some of the easier to use here.

You can also use the above programs to render in 3D, the easiest to use (I think) is ChemSketch. Simply draw a molecule in the standard view and click copy to 3D – done!

Another choice is Jmol. Jmol is a java applet which means it runs in your browser without installing (there is a stand-alone downloadable version also available). You can check that out here if you need added functuality over ChemSketch. There’s also a host of demonstrations and guides through the link. Like the rest of the applications on this page, it’s free!

The final example is Polyview – you simply fill in a form here and it pumps out a very nice 3D image or animation that can be put into presentations etc easily.

For more information on any of this and more I would suggest checking out this site, there is a huge number of resources for Jmol and other tools.

Chemical & Mathematical Equations in MS Word & LateX

When writing chemical or mathematical equations in Microsoft Word it is often easiest to use the equation editor.

This editor allows you better formatting than entering a formula with the standard tools. It’s also supplied with most copies of Microsoft Office, which you will probably find installed on institution/corporate computer facilities.

I have prepared this using Microsoft Office 2007. If your Office looks different, you should be able to find Equation Editor in “Insert > Object > Microsoft Equation 3.0″. If you can’t find it at all then you will need to install it from your Microsoft Office CD (this site shows you how to install).

If you don’t have MS Office then you could use a free editor such as this LateX Equation Editor which will render your equations as images which you can then copy into your work. They have a working trial at the top of the page, click the center image to load it.

The equation editor contains quite a few pre-set equations which you can enter automatically and gives you access to all the symbols you will need.

Quite simply, find out more through practice! There are more in depth guides available on this tool, including this one for older versions of Office. A simple google search will reveal more information.

Close Packing of Atoms & Metallic Elements

Close packing of atoms is simply close packing of spheres. There is only one way to efficiently arrange circles in 2D…and we can visualise the stacking with 2D drawings…

…each sphere (seen from above) is surrounded by 6 other spheres. This is a single layer, the next step is to stack the layers.

The most efficient place to put the next spheres is in the depressions between each sphere on layer one. So, we have one of two options.

We can either have rows in A or rows in B. Due to the size of the spheres it is not possible to fill ALL holes (A and B).

In position A:

In position B:

The fact is that the two sketches above are mirror images. This means it doesn’t matter which holes the second layer fall into! The important thing here is the creation of a new hole…directly above the spheres in layer 1. So for layer 3, we either have a repeat of layer 1 or a new layer in a new position.

Making 3 different layers with a new position:

Matching layer 1:

There are no spheres/atoms in the same position on the first 3 layer spread, but in the second we can see that layer 3 is in the same position as layer 1.

So, we have several possible outcomes, including:

1. 1..2..3..1..2..3..1..2..3……. Three different layers, could be cubic close packing (ccp).
2. 1..2..1..2..1..2..1..2..1……. Two different layers, could be hexagonal close packing (hcp).

HCP and CCP are the simplest and most common close packing structures. Each atom is surrounded by 12 other atoms – giving both these structures co-ordination numbers of 12.

So while these are close packed, a body centered cubic arrangement would NOT be, as it only has a co-ordination number of 8. This can also be illustrated with packing densities, where HCP and CCP have a density of 74.1% while BCC (body centered cubic, I) has a density of only 68% (although it is still a common metallic structure).

- Structures of Metallic Elements

In 1883 William Barlow (the curator of the science museum) suggested that the metals would each take one of three structures. These were hcp (hexagonal close packing), ccp (cubic close packing) and bcc (body centered cubic structure – which is NOT close packing).

Note that both hcp and ccp follow a face centered structure and as such have the same packing density, of 74.1%.

This image shows the structures of the metals in the periodic table at room temperature & pressure. I have it correct as far as I know, if you see any problems let me know.

Metal Elements of the Periodic Table - Structures

Polymorphism:

Some metals hold different structures at different temperatures and pressures. This is called Polymorphism and can occur in any compound.

Iron (Fe) is one such metal, and at atmospheric pressure a simple change in temperature will change it’s structure:

Above 1809K – Liquid – (No crystal struture)
1809K – 1665K – Delta-Fe – Body Centered Cubic (bcc)
1665K – 1184K – γ-Fe – Face Centered Cubic (fcc)
Under 1184K – α-Fe – Body Centered Cubic (bcc)

Note – Polymorphism of elements is known as allotropy. An example of allotropy is how carbon forms as diamond, graphite, nanotubules etc.

- Interstitial Holes

Close packing leads to two different types of hole between layers. One has a co-ordination number of 4 and the other of 6 (the number of surrounding spheres/atoms).

These are Tetrahedral (Td) with the co-ordination number of 4 and Octahedral with the co-ordination number of 6.

More to come… **UPDATE ME**

Crystal Structure Studies

Crystalline solids are composed of regular repeating patterns over relatively long distances ( >1000 Å). This is known as having a long range order.

ONLY crystalline solids produce macroscopic crystals (visible crystals).

Crystals have flat faces which can vary in size from crystal to crystal. The angle between similar faces is constant, and they break (cleave) in preferred directions (a feature exploited by diamond cutters).

Crystal Structure Studies tell us:

• Chemical Characteristics
  • How are atoms connected?
  • What are the bond lengths and angles between them?
  • What does this say about the bonding?
• Inorganic Solids
  • Structural features which may lead to an important property.
  • How changing a metal coordination sphere may alter the property (eg. in superconductors and electrical materials).
• Organic Compounds
  • Which points of stereochemistry are important.
  • How solubility is affected by the way the molecules pack together.
  • How many ways the molecules can pack together.
• Solution vs Solid State Structure
  • Structures may differ between different phases.

A couple of things about Molecule Packing:

• Atoms, ions or molecules always try to pack into the lowest energy configuration.
• This configuration can then be repeated for a large number of units.
• As the configuration is repeated, a regular pattern forms and a lattice emerges through the crystalline as a whole.
• This pattern may interact with certain wavelengths of radiation and lead to diffraction (constructive interference) which provides a means of studying the pattern.

3D Lattices and Translations – Unit Cells:

We can use three translations (a, b and c) to illustrate distance between atoms, ions or molecules; and three angles between each of these translations (α, β and γ). These parameters define the size and shape of the unit cell.

*Note that angle α corresponds to the translation a, angle β to translation b…etc.*

The unit cell is:

• A small volume defined by 6 faces, consisting of 3 identical pairs.
• A small unit of the larger structure which is then repeated.

“The smallest repeating unit that shows the full symmetry of the crystal structure.”

There are 7 Crystal Systems

• Triclinic – a≠b≠c – α≠β≠γ (ALL Different!)
• Monoclinic – a≠b≠c – α=γ=90°, β
• Orthohombic – a≠b≠c – α=β=γ=90°
• Tetragonal – a=b≠c – α=β=γ=90°
• Hexagonal – a=b≠c – α=β=90°, γ=120° (Need a and c)
• Rhombohedral – a=b=c – α=β=γ≠90°
• Cubic – a=b=c – α=β=γ=90° (ALL Equal, Only need a)

Let’s look at some unit cells. Try and work out the number of atoms in each 2D cell:

You should have:
i) 1
ii) 1
iii) 2

Why? This is something we’ll explore a little more later, but for now consider this. i) has 4 dots (atoms, ions or molecules) in it’s shape. Each of these four dots can participate in a total of four unit cells.

Look at the red dot. It is shared between four unit cells and so only contributes 1/4 to the contents of that cell. The same can be said for each other dot at the corner of a cell, as we assume that it is only attributing 1/4.

So we know that i) has 4 corners, each worth 1/4. That totals 1!

ii) is just the same, but iii) is different. It has 4 corners each worth 1/4 BUT it also has a single dot in the middle that is not shared with any other cell…meaning it is worth a whole 1. So (4×1/4)+1 = 2.

- Bravais Lattices

Although there may be an infinite number of chemical structures, there are only 14 3D lattice types, known as Bravais lattices.

We’re just going to look at the cubic system to start with. Remember, a=b=c – α=β=γ=90°. In the cubic system there are 3 lattice types, P (primitive), I (body centered), and F (face centered).

• Primitive cubic cells contain only corner dots, which can be shared between 8 different unit cells, meaning each dot is worth 1/8. 8*(1/8)  is 1 lattice point.
• Body Centered (I) cubic cells have the primitive structure but feature a central dot, worth 1 (as it is exclusive to that cell.  This gives 8*(1/8) + 1 = 2 lattice points.
• Face Centered (F) cubic cells once again follow the primitive structure but this time feature a dot on each of the 6 faces. Each of these dots can be shared between two cells and so each is worth 1/2. This gives 8*(1/8) + 6*(1/2) = 1+3 = 4 lattice points.

So, when counting particles in a unit cell:

• An atom at a corner counts for 1/8
• An atom on an edge counts for 1/4
• An atom on a Face counts for 1/2
• An atom within the cell counts for 1

Technically this is only true for cells with 90 degree angles, but it does actually work for all cells.

- Projection Drawings:

Although we could draw the 3D shape of a cell every time, it is easier to draw a projection – a top down sketch of the cell.

Each atom’s height within the cell is indicated as a fraction of the cell height. Note the cell heights in the drawing below.

Here are the projection drawings for each cubic lattice:

- Packing Densities:

To find the structure offering the maximum density of atoms, we can calculate the atomic volume and packing density per unit cell. This involves some fairly basic maths which I have recapped here. We’ll start with Primitive cubic cell.

Volume of cubes = a³ (a*a*a)

Volume of a Sphere/Atom = 4/3 x π x r³
(r = atomic radius)

Packing Density = n x p / a³
(n = number of lattice points per cell, p = Volume of Atom)

Primitive:

There is 1 lattice point per cell – and the atomic radius is half the distance between the closest atom points. In this case, r = 1/2a.

Therefore:

• Volume of atom = 4/3 x 3.142 x (1/2a)³ = 0.524a³

The Packing Density of these atoms is then simply the volume of one atom (which we have found to be 0.524 a³) divided by the volume of the cube (which is a³):

• Packing Density = (0.524a³) / a³ = 0.524

Body Centered:

There are 2 lattice points per cell, and in this case the radius is not simply 1/2a as the central atom is the closest point. More trigonometry!

The distance between the 2 corner points is √3a, as shown in the first image. Thus the distance between the two closest atoms is half of this…√3a/2. This means the atomic radii must be √3a/4.

Therefore:

• Volume of Atom = 4/3 x 3.142 x (√3a/4)³ = 0.34a³
• Packing Density = (2 x 0.34a³) / a³ = 0.68

(Note the doubling of the atomic volume in the packing density calculation – this is because Body Centered cubes contain 2 lattice points.)

Face Centered:

4 lattice points per cell, with more trigonometry.

The distance between a corner atom and the corner atom diagonally across the same face is √2a. This means the distance between a corner atom and a face atom is half of this, √2a/2. This means the atomic radii must be √2a/4.

Therefore:

• Volume of Atom = 4/3 x 3.142 x (√2a/4)³ = 0.185a³
• Packing Density = (4 x 0.185a³) / a³ = 0.74

Summary:

Primitive Density (P) = 0.524 = 52.4%

Body Density (I) = 0.68 = 68%

Face Density (F) = 0.741 = 74.1%

You can see the density increasing as you move down the list.

- Using this knowledge

1. Finding the metallic radius of α-tungsten. We know the unit cell is cubic (so we can use what’s above), a = 3.15Å, Atomic Mass = 183.85g mol^-1 and it’s measured density is 19.25g cm^-3 near room temperature and pressure. Step by step:
   1. We need to calculate the number of atoms in the unit cell (the number of lattice points). We know:
      Density = (Mass of 1 atom x Number of Atoms in Cell) / Volume of Unit Cell
      Therefore:
      19.2 = ((183.85/6.023×10^23)* x N) / (3.15×10^-8)*^3
      and Number of Atoms = 2 (rounded to leave an integer).

      We now know that there are 2 lattice points, or atoms per unit cell. Looking above we know that this unit cell is Body Centered.

      (* Finding the mass of 1 atom is done by taking the molar mass of an element and dividing it by Avogadro’s number, AKA the number of atoms per mole.)
      (** This value has been converted into Metres from Angstroms. 1Å = 1×10^8 Metres.)

   2. Calculate the metallic radius – we have learnt that the unit cell is body centered, so we already know the distances involved.
      The atomic radius in this lattice is √3a/4, and we know that a for α-tungsten = 3.15Å. Therefore the metallic radius is (√3 x 3.15)/4 = 1.36Å.

Some Basic Trigonometry

It seems that doing maths for as long as I can remember didn’t make it stick in my mind so I’m having a little refresh as I go. I’m starting with some basic trig.

- Pythagorus’ Theorum – Find the length of a third side on a Right angled Triangle.

Very simple. Say we have a right angled triangle with sides a, b and c – where c is the hypotenuse…

Crucially, a² + b² = c²

AKA the sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).

So how about this triangle?

5² + 9² = c²
25 + 81 = c²
c² = 106

Now square root 106…

c = ~10.30 (4sf) or √106 in surd form.