Transition metals Flashcards

Question

What is haemoglobin?

Answer 1

It is a protein found in blood that helps to transport oxygen around the body. It contain iron ions that are hexacoordinated with six coordinate bonds in an octahedral structure.

Answer 2

Four of the coordinate bonds come from a single multidentate ligand. Four nitrogen atoms from the same molecule coordinate around iron to form a circle.

Answer 3

The haemoglobin swaps its water ligand for a carbon monoxide ligand, forming carboxyhemoglobin. This is bad news because carbon monoxide is a strong ligand and doesn't readily exchange with oxygen or water ligands, meaning the haemoglobin can't transport oxygen anymore. Carbon monoxide poisoning starves the organs of oxygen, it can cause headaches, dizziness, unconsciousness and even death if it is not treated.

Answer 4

- in the lungs where oxygen concentration is high, an oxygen molecule substitutes the water ligand and bonds coordinately to the iron ion to form oxyhemoglobin, which is carried around the body in the blood. - When the oxyhemoglobin gets to a place where oxygen is needed, the oxygen molecule is exchanged for a water molecule. The haemoglobin then returns to the lungs and the whole process starts again.

Answer 5

If the two odd ligands are opposite each other it is the trans isomer. If they are next to each other it is the cis isomer.

Answer 6

optical isomerism is a type of stereoisomerism.

Answer 7

It is an optical isomer that can be used as an anti-cancer drug.

Answer 8

Normally the 3d orbitals of transition element ions all have the same energy. But when ligands come along and bond to the ions, some of the orbitals gain energy. This splits the 3d orbitals into two different energy levels.

Answer 9

Electrons tend to ocupy lower orbitals in the ground state. To jump up to the higher orbitals they need energy equal to the energy gap. They get this energy from visible light.

Answer 10

delta E = hv = hc/wavelength of light absorbed h - Planck's constant v - frequency of light absorbed c - speed of light

Answer 11

- the central metal ion - the metal ion's oxidation state - the ligands - the coordinate number

Answer 12

When visible light hits a transition metal ion, some frequencies are absorbed when electrons jump up to the higher orbitals. The frequencies absorbed depend on the size of the energy gap. The rest of the frequencies are transmitted or reflected. These transmitted or reflected frequencies combine to make the complement of the colour absorbed which is the colour you see.

Answer 13

No electrons will jump, so no energy will be absorbed. Therefore, the compound will look white or colourless.

Answer 14

1) white light is shone through a filter, whic his chosen to only tel through the cooour of light that is absorbed by the sample. 2) the light passes through the sample to a colorimeter, which calculates how much light was absorbed by the sample 3) The more concentrated a coloured solution is, the more light it will absorb. So you can use this measurement to work out the concentration of a solution of transition metal ions. 4) you can use a calibration curve to work out the unknown concentration

Answer 15

Measure the absorbance of known concentrations of solutions and plotting the results on a graph. Once you've done this, you can measure the absorbance of your sample and read its concentration off the graph.

Answer 16

It is where one ligand can be swapped for another, it usually causes a colour change.

Answer 17

If the ligands are of similar size and the same charge, then the co-ordination number of the complex ion doesn't change, and neither does the shape.

Answer 18

If the ligands are of different sizes there is a change of coordination number and a change of shape.

Answer 19

H2O, NH3, Cl-

Answer 20

H2, NCH2, CH2, NH2, C2O42-

Answer 21

It is harder to reverse when the new complex ion is much more stable than the old one.

Answer 22

If they form stronger bonds with the central metal ion than the old ligands did, the change is less easy to reverse. Like CN- forms stronger then H2O to fe3+

Answer 23

Multidentate ligands form more stable complexes than monodentate ligands, so a change is very hard to reverse.

Answer 24

Whe a ligand exchange reaction ocucurs, co-ordinate bonds are broken and formed. THe strength of th bodns being broken is often very similar to the strength of the new bonds being made. So the enthalpy change for a ligand reaction is usually very small.

Answer 25

The increase in stability explains why multidentate ligands always form more stable complexes than monodentate: When monodentate ligands are substituted with bidentate ligands, the number of particles in solution increases, the more particles the greater the entropy. Reaction that increases entropy are more likely to occur.

Answer 26

It is usually a redox reaction

Answer 27

Adding them to zinc metal in an acidic solution

Answer 28

2VO2 + to 2VO2+ Yellow to Blue 2VO2+ to 2V3+ Blue to Green 2V3+ to 2V2+ Green to Violet

Answer 29

The redox potential of an ion or atom tells you how easily it is reduced to a lower oxidation state. The larger the redox potential, the less stable the ion will be, and so the more likely it is to be reduced.

Answer 30

It can vary depending on the environment it is in: pH - some ions need H+ ions to be reduced while some release OH- ions into solution. In general they are larger in higher oxidation states. ligands - standard electrode potentials are measured in aqueous solution, so any aqueous ions will be surrounded by water ligands. Different ligands may make the redox potential larger or smaller depending on how well they bind to the metal ion in a particular oxidation state.

Answer 31

It is easily reduced as the standard electrode potential is large. so silver ions are easily reduced to silver metal.

Answer 32

Tollen's reagent uses this reduction reaction to distinguish between aldehydes and ketones. It is prepared by adding just enough ammonia solution to silver nitrate solution to form a colourless solution containing the complex ion [Ag(NH3)2]+ When added to aldehydes, Tollen's reagent reacts to give a silver mirror on the inside of the test tube. The aldehyde is oxidised to a carboxylic acid, and the Ag+ ions are reduced to silver metal.

Answer 33

Transition metals have variable oxidation states which means they are often present in either the oxidising or reducing agents. Their colour changes also make them useful in titrations as it is easy to spot the end point.

Answer 34

Oxidising and reducing agents can be used in titrations to find out the concentration of the other.

Answer 35

1. First you measure out a quantity of a reducing agent using a pipette, and put it in a conical flask. 2. Using a measuring cylinder, you add about 20cm3 of dilute sulphuric acid to the flask, this is an excess, so you don't have to be too exact 3. Now you add the oxidising agent to the reducing agent using a burette, swirling the conical flask as you do so. 4. The oxidising agent that you add reacts with the reducing agent. This reaction will continue until all of the reducing agent is used up. The very next drop you add to the flask will give the mixture the colour of the oxidising agent. 5. Stop when the mixture in the flask just becomes tainted with the color of the oxidising agent and record the volume of the oxidising agent added. This is the rough titration 6. Now you do some accurate titrations. You need to do a few until you get two or more readings that are within 0.1 cm3 of each other.

Answer 36

The main oxidising agent used is aqueous potassium manganate, which contains purple manganate VII ions. Strong acidic conditions are needed for the ions to be reduced.

Answer 37

Transition metals and their compounds make good catalysts because they can change oxidation states by gaining or losing electrons within their d orbitals. This means they can transfer electrons to speed up reactions.

Answer 38

Vanadium oxide is able to oxidise SO2 to SO3 because it can be reduced to vanadium oxide. It is then oxidised back to vanadium oxide by oxygen ready to start over again.

Answer 39

It is a catalyst that is one that is in a different phase from the reactants

Answer 40

Support mediums are often used to make the area of a catalyst as large as possible. They help to minimise the cost of the reaction, because only a small coating of catalyst is needed to provide a large surface area.

Answer 41

It is iron

Answer 42

It is Vanadium (V) oxide

Answer 43

THey often work by adsorbing reactants onto active sites located on their surfaces. Impurities in the reaction mixture may also bind to the catalysts surface and block reactants from being adsorbed. This process is called catalyst poisioning.

Answer 44

It reduces the surface area of the catalyst avaliable to the reactants, slowing down the reaction. Catalyst poisioning increases the cost of a chemical process becuase less product can be made in a certian time or with a certain amount of energy. The catalyst may even need replacing or regenerating which costs money.

Answer 45

Sulfur poisons the iron catalyst in the Haber process. The hydrogen in the Haber process is produced from methane. The methane is obtained from natural gas, which contains impurities, including sulphur compounds. Any sulphur that is not removed is adsorbed onto the iron, forming iron sulfide, and stopping the iron from catalyzing the reaction efficiently.

Answer 46

They are in the same physical state as the reactants. They work by combining with the reactants to form an intermediate species which then reacts to form the products and reform the catalyst.

Answer 47

IT has two humps in it, corresponding to the two steps in the reactionn. The activation energy needed to form the intermediate is lower than that needed to make the products directly from the reactants.

Answer 48

The redox reaction between iodide ions and peroxodisulfate ions takes place annoyingly slowly because both ions are negatively charged. The ions repel each other, so it's unlikely they'll collide and react.

Answer 49

But if Fe2+ ions are added, things speed up because each stage of the reaction involves a positive and negative ion, so there is no repulsion. - first, the Fe2+ ions are oxidised - The Fe3+ ions oxidise the I- ions.

Answer 50

Another example of a homogenous catalyst is Mn2+ in the reaction between C2O42- and MnO4-. It is an autocatalyst reaction because Mn2+ ios a produce of the reaction and acts as a catalyst for the reaction. This means that as the reaction progresses and the amount of the product increases, the reaction speeds up.

Answer 51

Autocatalysis is when a produce catalyses the reaction.

Answer 52

When transition metal compounds dissolve in water, the water molecules form coordinate bonds with the metal ions. This forms metal aqua complex ions, In general, six water molecules from coordinate bonds with each metal ion. The water molecules do this by donating a non-bonding pair of electrons from their oxygen.

Answer 53

Metal 3+ ions are pretty small but have a big charge so they have a high charge density. The metal 2+ ions have a much lower charge density. This makes the 3+ ions much more polarising than the 2+ ions. More polarising power means that htey attract electrons from the oxygen atoms of the cooridinated water molecules more strongly, weakening the OH bond. So it is more likely that a hydrogen ion will be released. And more hydrogen ions mean a more acidic solution.

Answer 54

In a solution containing metal aqua 2+ ions, there is a reaction between the metal aqua ion and the water, this is a hydrolysis or acidity reaction. The meal aqua 2+ ions release H+ ions, so an acidic solution is formed. There is only slight dissociation though so the solution is only weakly acidic. 3+ ions dissociate more than 2+ so form more acidic solutions.

Answer 55

Adding OH- ions to solutions of metal aqua ions produces insoluble metal hydroxides. 1) In water metal aqua 3+ ions such are iron or aluminium form an equilibrium. If you add more Oh- H3O+ is removed shifting the equilibrium to the right. 2) Now another equilibrium is set up in the solution. Again the OH- ions remove the H3O+ from the solution, pulling the equilibrium to the right. 3) This happens one last time and you are left with an insoluble uncharged metal hydroxide.

Answer 56

There are only two steps this time because only two of the water ligands need to be deprotonated to make the +2 complex uncharged.

Answer 57

with an acid it goes from Al(OH)3(H2O)3 to [Al(H2O)6]3+ +3H2O with a base it goes from Al(OH)3(H2O)3 to [Al(OH4]- +3H2O

Answer 58

- add ammonia solution - because hydroxide ions are formed in ammonia solution, adding a small amount of ammonia solution gives the same result as adding sodium hydroxide - In some cases, such as with Cu(OH)2(H2O)4, a further reaction happens if you add excess ammonia solution. The water and hydroxide ligands are displaced. This forms a charged complex which is soluble in water, so the precipitate dissolves.

Answer 59

Metal 2+ ions react with sodium carbonate to form insoluble metal carbonates. But metal 3+ ions are stronger acids so there is a higher concentration of hydronium ions in solution. Rather than displacing water from the metal ions, the carbonate ions react with them, removing them from the solution and shifting the equilibrium to form a different precipitate.

Answer 60

1) Measure out samples of the unknown metal ion solution into three separate test tubes 2) To the first test tube add sodium hydroxide solution dropwise, using a droppping pipette and record any changes you see. Then add more NaOH dropwise so that it is in excess. 3) To the second test tube, add ammonia soution dropwise, using a dropping pipette, and record any changes you see. Keep addin ammonia so that it is in excess. 4) To the third test tube, add sodium carbonate solution dropwise. Record your observations.

Answer 61

All four metal aqua ions will form precipitates with sodium carbonate. The solutions containing AL3+ or Fe3+ will also form bubbles as CO2 is formed.

Answer 62

All four metal aqua ions will form precipitates with ammonia, but only the copper hydroxide precipitate will dissolve in an excess of ammonia. This is because it undergoes a ligand exchange reaction with excess ammonia.

Answer 63

All four metal aqua ions will form precipitates with sodium hydroxide, but only the aluminium hydroxide precipitate will dissolve in an excess of sodium hydroxide. This is because it is amphoteric.

Answer 64

blue to blue precipitate

Answer 65

deep blue solution

Answer 66

green-blue precipitate

Answer 67

green precipitate

Answer 68

Green precipitate

Answer 69

White precipitate

Answer 70

colourelss solution

Answer 71

white precipitate

Answer 72

yellow to brown precipitate

Answer 73

brown precipitate