Category Archives: Chemistry

Bases, Acid Base Reactions and Equilibriums

So we’ve heard of Ka/pKa, the acid dissociation constant – but what is there for bases? There is such a thing as a base dissociation constant Kb but more commonly the reverse of Ka is used.

A base would normally be indicated by a high pKa – meaning there is very little dissociated H+. Kb is simply the opposite as shown below.

Base Dissociation Constant

Base Dissociation Constant

So if we were to look at Ammonia (NH3):

Kb for Ammonia

Kb for Ammonia (NH3)

But…seeing as we aren’t using Kb/pKb we simply need to rearrange our base dissociation formula to fit into Ka/pKa:

Ka of Ammonia (NH3)

Ka of Ammonia (NH3)

From this you may notice it’s the other way round for the Ka for acids, as there the top line holds the base NH3 while in acids the acid is found on the bottom line (eg HCl).

Acid Base Reactions – and Equilibrium

Acid base reactions have an equilibrium, and to calculate it we simply combine the Ka of the acids with the Ka of the bases.

Consider the reaction between HCl and NaOH, producing NaCl and H2O. Their Ka equations would be as follows:…

Acids and Bases – pKa, Equilibrium Constant and logs

The best place to start is at acids – and as we know acids can give up a proton to become deprotonated.

Acid and Conjugate Base

Acid and Conjugate Base (& charges)

The acid will dissociate into a proton (H+) and cunjugate base (A-). Note that this is in equilibrium – there is a mixture of both sides of the equation present. The constant for this equilibium (the acid dissociation constant, Ka) tells us the position of the equilibrium. Ka is the concentration of the products over the concentration of the reagents:

Ka = Products Concentrations / Reagent Concentrations

Ka = Products Concentrations / Reagent Concentrations

therefore, for Hydrochloric Acid (HCl):

Acid Dissociation Constant for HCl

Acid Dissociation Constant for HCl

Simple, huh? What Ka tells us is a numeric value for the strength of an acid in solution. The larger the value, the smaller the extent of dissociation.

To illustrate, a strong acid like HCl has a Ka value of 1×10^7, which clearly shows a large bias towards products. Acetic acid on the other hand has a Ka value of 1.7×10^-5 – which strongly favours reagents.

So, what’s all this ‘log’ stuff?

We use logs to convert long numbers into a user friendly scale – as the numbers we often get are on a huge scale (see HCl and Acetic Acid above). To do this we put p into our acid dissociation constant Ka.

pKa & Log of the Concentrations

pKa & Log of the Concentrations

Simply, p = – log, so the result is the logarithm of negative Ka.

Going back to HCl & Acetic Acid:

HCl Ka = 1×10^7; – log Ka = -7, therefore pKa = -7.

Acetic Acid Ka = 1.7×10^-5; – log Ka = -4.76, therefore pKa = 4.76.

So stronger acids have lower pKa‘s (or have higher Ka‘s).

We can easily convert back into Ka:

Converting between pKa and Ka

Converting between pKa and Ka

pH and pKa

If the pH of a solution = the pKa, then the acid is in equilibrium – it is half dissociated. This scale goes either way – if pH is less than pKa then it’s mainly protonated acid; if pH is more than pKa it’s mainly deprotonated.

pH and pKa - Equilibrium & Protonation

pH and pKa - Equilibrium & Protonation

In the next post I’ll look at the equilibrium constant for bases & for acid base reactions.

Rates of Reactions, Chemical Kinetics & Orders

NOTE: Formatting needed–

One of the most important things to note with chemical reactions is that the molar concentrations of the substrates are often proportional to the rate of the reaction.

So if we take the simplest rate constant for an equation:

A + B –> C

We could might find the rate law to be:

Rate = k[A][B]

The coefficient k is called the rate constant and is dependant on temperature – this is independant of the concentrations of the substrates; so the larger the value of k, the faster the rate of the reaction. Also important is that the units of k will convert the product of the concentrations into a rate – so change in concentration per unit of time, often expressed as

While temperature increases increase the rate constant and rate of reaction in most cases, reactions with a large activation energy will have small rate constants as considerable temperature rises may be required for the reaction to occur at all.

Consider this theoretical example:

Rate = k[A][B] where k = 5 dm3 mol^-1; [A] = 1; [B] = 2

Therefore Rate = 5 dm3.mol-1.s-1 x 2 mol^^-6

The units all cancel to leave us with a rate of: 10

So the units for k in that example were dm3.mol-1.s-1. In another rate law, eg: Rate = k[A] we would find the units for k to be simply s-1.

Once we know the rate law and rate constant for that reaction we can go on to predict the reaction rate for any concentration of substrates.

– The Order of a Reaction

Reactions can usually be defined as either zero order (0), first order (1) or second order (2). The order of a reagent or the overall reaction depends on the effect varying the concentrations of substrates has on the rate of the reaction. So:

  • Zero Order – rate is not related to reactant A – rate is proportional to [A]0
  • First Order – rate is doubled as concentration of reagent B doubles – rate is proportional to [B]1
  • Second Order – rate is quadrupled as concentartion of reagent C doubles – rate is proportional to [C]2

Combining the above information, rate is proportional to [A]0[B]1[C]2 – therefore Rate = k[B]1[C]2 – so the reaction is 3rd order ( 1+2=3). Third order tells us the reaction is made of several parts.

– Measuring Rate & Integrated Rate Equations

0. Zero Order:

As a zero order reaction has a rate which is independant of any reagents, we can assume that Rate = k.

To identify a zero order reaction plot concentration of a reagent against time and you would see a straight line. The integrated rate equation is:

Integrated Rate Equation - Zero Order

Which means that the gradient (from y=mx+c) equals -k. This allows us to determine k from the graph.

Another feature of a zero order reaction is a decreasing half life as the reaction continues. The half life equation for zero order reactions is:

Half Life Equation for Zero Order Reaction

Where [A]0 is the initial concentration. Shows a decreasing half life as concentration falls.

1. First Order:

First order reactions have a rate proportional to the concentration of only one reagent. Any other reagents present will not affect the rate.

To identify a first order reaction plot In(concentration) against time to give a straight line. The integrated rate equation is:

Half Life Equation for Zero Order Reaction

Which means that as with zero order, k is the -ve of the gradient.

The half life of a first order reaction is constant thoughout the reaction:

Half Life Equation for First Order Reaction

This half life is dependant only on k as the half life remains constant regardless of concentration.

2.Second Order:

Second order reactions have a rate proportional either to 1 or 2 reagents (eg 2 x first order reagents or 1 x second order reagents).

To identify a second order reaction, plot 1/concentration against time to give a +ve straight line. The integrated rate equation is:

Integrated Rate Equation for Second Order Reactions

Which means that k = gradient (so the opposite of what we find in zero and first order reactions).

The half life of a second order reaction increases throughout the reaction:

Half Life of a Second Order Reaction

Shows an increasing half life with decreasing concentration.

2(1). Psuedo First Order:

Psuedo first order approximation is used when carrying out some second order reactions. It is useful as it is difficult to effectively control the concentrations of more than one reagent at the same time, and the psuedo technique simply places one reagent in excess at a constant level; essentially limiting the reaction rate the other reagent (you only control the concentration of one reagent).

Rate Order of a Psuedo First Order Reaction

The equation above illustrates that by putting [B] in excess we have essentially removed it from the rate reaction, allowing us to calculate the psuedo rate constant k‘.

Preparation & Reactions of Aldehydes and Ketones, RHO & ROR'

A couple of key points:

  • Aldehydes and Ketones both contain a carbonyl group, but are also less reactive than acid chlorides.
  • They do NOT react with organocopper reagents and weak hydride donors (as these weak reagents are involved in their own synthesis).
  • The reactions are addition rather than substitution as there is no leaving group.
  • They have one less bond to an electronegative atom than acid chlorides (no chlorine!).
Aldehyde & Ketone

Aldehyde & Ketone

They can be formed through reduction of Acid Chloride:

Aldehyde & ketone synthesised with Bu3SnH and R'2CuLi

Aldehyde & ketone synthesised with Bu3SnH and R

If an aromatic ring is being substituted then we must use friedel crafts acylation.

For Acid Chloride to Aldehyde we use Bu3SnH as a source of weak Hydride ions which displace a Cl-. We do not use a more obvious source such as LiAlH4 as this will result in the over reduction of the aldehyde into a primary alcohol.

For Acid Chloride to Ketone we use R’2CuLi as a source of nucleophilic R’ group.

and via reactions with Alcohols:

Simply, Primary alcohols lead to Aldehydes and secondary alcohols lead to Ketones when reacted with PCC. This is oxidation.

Aldehyde & Ketone synthesised from Alcohols

Aldehyde & Ketone synthesised from Alcohols

and finally with Alkanes:

Alkanes are just as simple as alcohols – just add O3 then PPh3 for an easy reaction!

Simple alkenes lead to aldehydes and more complex lead to ketones.

Aldehyde  Ketone synthesised from Alkenes

Aldehyde Ketone synthesised from Alkenes

Synthesis Summary:

In short:

From Acid Chloride to Aldehyde – Bu3SnH (as a source of H-)
From Acid Chloride to Ketone – R2CuLi (as a source of R)

From Alcohol to Aldehyde/Ketone – PCC
From Alkene to Aldehyde/Ketone – O3 then PPh3

– Reactions with Carbon Nucleophiles and Hydride Donors

As mentioned earlier, aldehydes and ketones do not react with weak hydride donors (eh Bu3SnH) or organocopper reagents (eg R2CuLi) – they need more powerful reagents.

These come in the form of Grignard reagents (eg RMgBr) and powerful halide donors (eg LiAlH4).


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

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.
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

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.

NuH on Acid Chloride

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.

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.

Wikipedia. Shows C=O and two additional atoms.

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

Resonance Forms


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.

    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.

    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.

    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.

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 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

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 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 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)

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.

    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.

    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.

    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.

    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.

A molecule of Ethene.

The electron configuration in carbon starts the same:

Carbon - Ground (normal) state of electrons


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

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

but the resulting spread is different:

Carbon hybridisation in Ethene (sp2)

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.

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.

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

Ethyne - sp Hybridisation

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


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

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

Metal Elements of the Periodic Table - Structures


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)


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.


  • 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.


  • 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.


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


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
      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.