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Note: From 2008, there will no longer be a choice. Most of the stuff here will be found in Unit 5: Equilibria, Energetics and Elements.
In the final section of Chemistry, assessed alongside Trends and Patterns, you have a choice: Biochemistry or Transition Metals. If you do biology, you'll be familiar with most of the Biochemistry, although it will present everything with a chemical twist. If you're just a chemist, you'll be familiar with Transition Metals from the final section of Trends and Patterns. On this site, we'll just look at Transition Metals.
Before we go any further, just what is a transition element? To recap Trends and Patterns, "a transition element is an element with one or more ions that have partially filled d oritals."
Remember the 3d-block of the periodic table:
All these are transition elements, other than Zn and Sc. Those two aren't transition elements because they only form one ion each.
Zn forms the Zn2+ ion with electronic configuration: 1s22s22p63s23p63d10. The d orbital is full and therefore, it isn't a transition element.
Sc forms the Sc3+ion. Its electronic configuration is: 1s22s22p63s23p63d0. The d orbital is empty and so it isn't a transition element.
Remember that Cr and Cu have weird electron configurations. As it's more stable, their 4d sub shell only contains one electron.
As the eight transition metals can form more than one ion, this means they can also have a variety of oxidation states. Iron, for example, can either be Fe3+ or Fe2+. The variety of oxidation states means that they are also good for redox reactions.
If iron goes from Fe2+ to Fe3+, it loses an electron and so is oxidised. If the reverse happens, it is oxidised.
We always write out half-equations in terms of reduction. So, iron's half equation is:
The ease with which transition elements are reduced varies. Cu2+ is very easy to reduce, while V2+ ions are much more difficult to reduce.
But to be more scientific than using 'harder ' and 'easier', we can calculate
a value for the ease of reduction. This is called the electrode potential for a reduction. In short hand, electrode potential is described as E. The electrode potentials are calculated experimentally under standard conditions,
so we say E
In order to measure electrode potential, we need to set up a half-cell.
To measure Fe2+ + 2e- Fe, we would need a rod of iron, placed in a 1 mol dm-3 solution of Fe2+:
Now we have a half cell, we need to attach it to another half cell, via a voltmeter, to create an electrochemical cell.
We always attach it to a H+/H2 half-cell, just because this is the 'standard' cell of comparison that chemists have agreed on. It looks like this:
Because it uses a gas and a solution, the gas has to be bubbled in using a glass 'bell'. The platinum electrode connects the gas to the solution.
We attach the two half cells together with a voltmeter and a salt bridge to form an electrochemical cell as shown below:
The standard electrode potential for a half-cell can be defined as the voltage measured under standard conditions when the half-cell is incorporated into an electrochemical cell with the other half-cell being the standard hydrogen electrode. The standard conditions are:
Electrons flow from one side to the other across the voltmeter, which records the electrode potential, while ions flow between the two half-cells through the salt bridge. The salt bridge can be something as simple as a piece of filter paper soaked in potassium nitrate.
The voltage recorded on the voltmeter varies depending on how well a half cell reduces in comparison to the H+/H2 half-cell. For Fe/Fe2+, the voltmeter records -0.44. This means the electrons are flowing like this:
The Fe/Fe2+ is the negative terminus, meaning electrons flow from there to the H+/H2 half-cell. Iron is losing electrons (oxidising) and hydrogen is gaining electrons (reducing):
If we look at a Cu/Cu2+ connected to the H+/H2 half-cell, however, the recording is positive: +0.34 V. This means electrons are flowing in the opposite direction, from the hydrogen half-cell, to the copper one. Here, Cu2+ is gaining electrons and is being reduced, while H2 is losing electrons and is being oxidised:
This means, quite simply, that Cu2+ has a much greater tendency to gain electrons that Fe2+. So, the more positive the E
When all the reactants are 'used up' (have been oxidised or reduced), an electrochemical cell stops working and 'goes flat'.
We've seen how to set up a half cell with a solid and an ion, and with a gas and an ion, but what about two ions? This uses two 1 mol dm-3 solutions of the ions, with a Pt electrode allowing electrical contact with the solutions. For example, if we made a Cu2+/Cu+ half cell, it would look like this:
Sometimes reductions need H+ ions in order to occur, such as MnO4- + 8H+ + 5e- Mn2+ + 4H2O. For this, we simply use a 1 mol dm-3 solution of H+ ions in addition to the other solutions.
The following table gives the electrode potentials of a number of different reductions. You don't have to learn them (I hope!).
As well as predicting how good a reaction is at reducing, E
For example, we might have the following:
The question might ask you to calculate the value on the voltmeter. If we look
at the E
The difference between the two half-cell values will give us the overall cell voltage. We always subtract the least positive from the most positive, so: +0.8 - (-0.76) = +156 V. The more positive Ag+/Ag will be at the positive pole and Zn/Zn2+ at the negative:
The question might also ask you to work out which reaction is oxidising and which is reducing, and to give the overall equation.
As we know, the more positive an E
In order to write a balanced equation, the electrons need to cancel out, so we times the Ag+/Ag or divide Zn2+/Zn by two:
A word of warning with E
In real life, the conditions that half-cells are under are unlikely to be standard.
If we make an electrochemical cell without standard conditions, we measure the E value. The E value can vary depending on the conditions:
It's because the equilibrium of the half-equation is being disrupted, and we all know Le Chatalier will move the equilibrium position in order to oppose the change.
The Fe2+/Fe3+ half-equation, for example is Fe3+ + e- Fe2+ and the E
If we were to increase the Fe3+ concentration, the E might equal +0.85 V instead, becoming more positive. If we decreased the Fe3+ concentration, E might equal +0.73 V, less positive.
If we increased the Fe2+ concentration instead, the E might decrease to +0.70 V. If we decreased the Fe2+ concentration, the E might increase to +0.85 V.
If two half-cells have E
Because the Cl2/Cl- half-cell is more positive, the MnO2 cannot oxidise it. However, if we were to add lots of concentrated HCl, the concentration of H+ would increase, as would the concentration of Cl-. This would mean the E of MnO2/Mn2+ would increase, and the E of Cl2/Cl- would decrease. They might now look like this:
As the E of MnO4-/Mn2+ is more positive, it would be reduce, while Cl2/Cl- oxidised. The balanced equation would be:
That's about everything you need to know for Electrode Potential. Make sure you can draw all three types of half-cells, and know which way the electrons flow, and you should be fine!
Ligands and Complexes
As we already know from Trends and Patterns, transition elements can form complex ions.
A complex ion is a central transition metal surrounded by ions or molecules called ligands, and a ligand has at least one lone pair, which it donates to the transition metal to form coordinating, dative bonds.
There are three common shapes formed when complex ions form: octahedral, tetrahedral, and square planar.
In all the examples above, the ligands are monodentate, which means they only donate one pair of electrons. However, some molecules can donate more than one pair. If they can donate two, they are called bidentate, and more than two are polydentate.
[Ni(en)3]2+ is an example with bidentate ligands:
Ahh, old cis and trans isomerism... You thought you'd escaped it when at last you finished Chains, Rings and Spectroscopy, but it's back again with complex ions.
Remember that stereoisomers have the same structure, but take on different spatial arrangement. Cis-trans can occur with octahedral or square planar shapes, but not tetrahedrals.
In octahedral complexes, this occurs when we have four of one monodentate and two of another, such as in [CoCl2(NH3)4]+.
There are two different ways of drawing the complex ions: where the Cl-Co-Cl bond is 90o and where the Cl-Co-Cl bond is 180o. This forms the cis and trans stereoisomers:
Optical isomers are two molecules that are non-superimposable mirror images of each other. Sometimes, optical isomerism occurs with octahedral molecules:
Make sure you're able to draw this!
That's it for this little section... nothing too bad, hmm?
Most things that are coloured absorb certain wavelengths in the spectrum of white light and transmit (allow to pass through) or reflect others. The transmitted or reflected light hits our eyes and we see the colour this makes.
Transition metals are the same. If we have a green solution of Fe2+ ions, it appears green because it adsorbs all light other than the green transmitted.
But just why are certain colours adsorbed and others transmitted? It's all about the electrons in the 3d subshell...
The 3d subshell contains five separate electron orbitals and each is able to contain two electrons...so ten electrons in total. The five orbitals look like this:
The five orbitals all have the same energy and are called degenerate.
In an ion with only one electron in the 3d subshell, the electron can travel freely between the orbitals. If the atom or ion is isolated from others, its electrons can also move freely between the orbitals, even if there are as many as nine electrons.
However, molecules are rarely isolate, and so rarely degenerate. In an octahedral complex ion, for example, the six pairs of coordinating bonds from the ligand form on the 3dx2-y2 and 3dz2 axis. This causes the 3dx2-y2 and 3dz2 orbitals to have a higher energy than the other three orbitals, something known as d-orbital splitting.
Though the orbitals are still part of the same subshell, there is an energy gap, labelled ΔE, between the two groups of orbitals:
So where does colour come from? It's all about electron promotion. If a transition element has at least one electron in the lower d orbitals and at least one gap in the upper d orbitals, the electron can move from the lower to the upper orbitals - a process called promotion.
Electron promotion requires energy, which is provided by visible light. For an electron to be promoted, it needs to absorb a particular light: the light that provides the exact energy required to bridge the ΔE. The remaining light is transmitted, and this is the colour we see. Pretty cool, huh?
In the example above, green light is being absorbed by [Ti(H2O)6]3+ for electron promotion, so the solution appears the colour that red and blue make: a violet.
When a ligand displaces another in ligand substitution, the ΔH value changes, which explains the colour change.
Remember how in Trends and Patterns we said there were a few exceptions to the rule that transition elements are coloured? Now we know about electron promotion, we can explain why:
We also need to know that transition metal complexes are often used in paints. Titanium(iv) oxide is often used in white paint because of the colourless Ti4+ ions, and lots of coloured ions are used in coloured pigmentations. Something called Cu2+-phthalocyanine is a blue pigment called Monastral blue.
Many non-transition compounds in the world aren't coloured, despite the fact that their electrons could be promoted, like in transition metals. However, the energy gaps between other orbitals are too large to be satisfied by visible light. The promotion may occur under x-ray or uv light, but our eyes can't detect these anyway.
In order to analyse colour in a complex, we use a visible spectrometer. This separates the light into the spectrum and measures its intensity. You need to know:
A quantum is a tiny 'packet' of light energy. The size of one quantum depends on the frequency of the light (number of waves per second, measured in hertz) and Planck's constant, which equals 6.63x10-34 J s.To symbolise Plank's constant, we use the letter h, for frequency f and for a quantum e.
Visible light has a frequency of 6x1014 Hz. If we put this into the equation, we see a quantum of visible light is:
This may be a tiny amount of energy, but it could be enough for promotion. Electron promotion cannot use quanta (plural) that are too big or too small, but exactly the right size.
We need to know about four transition elements: vanadium, chromium, cobalt and copper.
Let's start with vanadium, discovered in 1801, and named after the Norse god 'Vandanis'. Vanadium a hard, steely gray and very resistant to corrosion. It exists in various oxidation states: +2, +3, +4 and +5, though at higher states it needs be be combined with oxygen, such as VO2+, VO2+ and VO3-.
Like all transition elements, it is good for redox reactions. Zinc can be used to reduce it from VO2+ to VO2+ to V3+ and finally to V2+, and at each stage, a different colour will be seen:
Vanadium is very important in industry because it catalyses the first step in the formation of sulphuric acid - one of the most important acids in Chemistry. The reaction is:
Unfortunately, this is very slow as the activation energy is so high. However, we can use V2O5 in two intermediate reactions:
Each of these steps has a much lower activation energy than the reaction without the catalyst. V2O5 can do this because it forms stable compounds in a range of oxidation states. As the V2O5 isn't being used up, but offering an alternate route, it is a catalyst.
Chromium was first discovered in 1790, and its name comes from the Greek word 'chroma', which means colour. Its silvery appearance has led to it being used on bike and car bumpers.
Chromium also forms a range of oxidation states - +2, +3, +4, +5 and +6 - and the +3 and +6 are the most common.
In the +6 state, chromium forms two types of oxyanions (negative ions that contain oxygen):Cr2O72- and CrO42-. These are usually found in potassium salts: potassium dichromate (K2Cr2O7) and potassium chromate(VI) (Kr2CrO4). The Cr2O72- ion is orange in solution and CrO42- is yellow in solution.
They form an equilibrium:
So, if we use acidic conditions and flood the reaction with H+ ions, the equilibrium will try to oppose through Le Chatalier's Principal and move it to the LHS, producing lots of orange Cr2O72-.
If we use alkali solutions and flood with OH-, these will react with the H+ in the reaction to form water. To replenish these, and decrease the conc of water, the equilibrium will move to the RHS through Le Chatalier's Principal, thus producing lots of yellow CrO42-.
In acidic solutions, potassium dichromate is a very strong oxidising agent, reducing the chromium ions to Cr3+, which is very stable.
Although Cr3+ ions violet in solution, the colour change seen is orange to green. This is because, under high temperatures, Cr3+ undergoes ligand substitution, forming the green [Cr(H2O)6]3+ complex.
In 1913, a Sheffield chemist Harry Brearley was trying to find a new alloy of steel for gun barrels. He tried one alloy containing about 13% of chromium, but found it useless for the task. However, when he came to his scrap metal pile months later, he found this alloy hadn't rusted. On further investigation, he found that an alloy of 70% iron, 20% chromium and 10% nickel formed an ideal stainless steel.
The chromium and nickel oxidise quickly. Unlike flaky iron oxide, their oxides form a layer that is hard and impervious to air and water, which protects the iron from rusting.
If smaller amounts of chromium are added to steel, a much harder alloy of steel is formed, useful for tools.
Discovered in 1735, cobalt is a hard unreactive metal, white but with a slightly blue appearance. Its name comes from the German 'Kobald', which means 'goblin'.
Cobalt forms a range of compounds in the +2 and +3 oxidation states, though the +2 state is the most stable.
Most octahedral complexes of Co2+ are pink, such as [Co(H2O)6]2+, while most tetrahedral complexes are blue, like [CoCl4]2-. Anhydrous cobalt chloride paper can be used to detect water - it turns from blue to pink as the [Co(H2O)6]2+ complex forms.
Although Co3+ forms an aqueous complex, [Co(H2O)6]3+, which is blue, it is so easily reduced to [Co(H2O)6]2+ that no reactions occur with it. Many of the compounds that involve Co3+, like CoF3, immediately reduce when they come in contact with water. But in complexes with ligands other than water, Co3+ is stable.
For example, the reaction [Co(NH3)6]3+ + e- [Co(NH3)6]2+ has a low E
e- [CoH2O)6]2+, however, has an E
Finally, we arrive at copper, a metal that has been around since ancient times. The name comes from 'Cuprum' the Latin for Cyprus, where copper was found. A reddish-brown, soft, ductile metal, its ability to conduct electricity and heat is second only to silver.
Copper forms a wide range of stable compounds in the +2 state, but unlike most others, also forms compounds in the +1 state. While virtually all +2 compounds of Cu are blue, most compounds of +1 are white solids.
As aqueous solutions, Cu2+ is stable, but Cu+ isn't. If the two are mixed, disproportionate can occur, where both oxidation and reduction occur. Look at the half-equations:
The top equation is the most positive, so this will be reduced, forcing the other to be oxidised:
The overall equation looks like this:
Cu+ has been oxidised to Cu2+ and also reduced to Cu. Cu+ is not stable in solution because of this disproportionation.
Insoluble solids containing Cu+ are stable, however. Copper halides are insoluble and will not dissociate. To get them, Cu2+ is mixed with excess I- ions, reducing the Cu2+ to Cu+:
Then, the Cu+ reacts with an iodide ion to form the halide:
The solid is insoluble, so no disproportionation occurs. Although the solid is white, it appears brown due to the colour of the iodine formed at the same time.
There are a few alloys that you need to know:
In order to work out how much copper there is in bass, we use titration.
For example: It was found that the iodine reacted with 47.8 cm3 of 0.2 mol dm-3 sodium thiosulphate. Find the mass of copper in the alloy.
Using this, we can work out the moles of sodium thiosulphate:
We know from the equation that the moles of sodium thiosulphate equal the moles of copper.
Next, we just times by the mr of copper:
And with that, Transition Elements is complete! Let me know if you think anything is missing!