Transition Metals Flashcards

Question

Cis-Trans Isomerism

Answer 1

Cis= 90 trans = 180

Answer 2

Cis platin binds to the DNA in cancer cells and stops cell replication. The two Cl- ions on Cisplatin are substituted for two N atoms on adjacent Guanine bases.

Answer 3

This only occurs when complexes form with at least two bidentate ligands.

Answer 4

ROY light = low energy BIV light = high energy

Answer 5

- Transition Metal compounds are coloured because they have partially filled d-sub shells. - In a pure transition metal atom all d-orbitals are of equal energy. This is known as the ground state. - In transition metal compounds, the presence of other atoms causes the d-orbitals to have slightly different energies. - This enables electrons to be excited from one d orbital to another. - When an electron moves from a lower energy d orbital to a higher energy d orbital (called the excited state) the energy needed to make the transition is taken from white light. - That colour of the light which is absorbed is then missing from the light which reflects from the substance. The colours which are reflected give the compound its colour. The difference in energy between the d sub-shells can be called ∆E

Answer 6

Ti(IV) = 1s2 2s2 2p6 3s2 3p6 Ti(III) = 1s2 2s2 2p6 3s2 3p6 3d1 Ti(III) has a d electron that can be excited to a higher level Absorbs one colour of light from white light Ti(IV) has no d electron so no electron transition with energy equal to that of visible light

Answer 7

One colour of light is absorbed The remaining colour are transmitted (through the solution)

Answer 8

∆E = hv ∆E = Change in Energy h = Planck Constant 6.626x10-34 v = Frequency of Light (Hz) ∆E = h x c/λ ∆E = Change in Energy h = Planck Constant 6.626x10-34 c = speed of light λ = wavelenght of Light

Answer 9

If a transition metal compound has LARGE ∆E between d sub-shells: - High energy light (i.e. Violet, Indigo and Blue) - will be absorbed to excite the electrons - Red, Orange, Yellow light will be reflected - This means the compound will look Red/Orange

Answer 10

If a transition metal compound has SMALL ∆E between d sub-shells: - Low energy light (i.e. Red, Orange, Yellow) - will be absorbed to excite the electrons - Blue, Indigo, Violet) will be reflected by the compound - This means the compound will look Blue/Purple

Answer 11

LOCo - Change in ligands - Change in oxidation state of the metal - Change in the coordination number of the complex

Answer 12

Suggest why it is important that the container for each sample has the same dimensions. - Absorption depends on path length Suggest why the coloured filter is used. - To select the colour that is most strongly absorbed ( blue solution = red filter Suggest one reason why a colorimetric method might be chosen in preference to titration. - Quicker to analyse extracted samples than by titration / uses smaller volumes of solution 1.0 = All the light passed through the sample, and none was absorbed.

Answer 13

- measure absorbance of known conc. - plot graph of absorbance v conc. - read value of conc. for measured absorbance for this graph

Answer 14

- A substance which increases the rate of a chemical reaction - Without being used up

Answer 15

- Provide an alternative reaction pathway - With a lower activation energy

Answer 16

where the catalyst is in a different phase to the reactants

Answer 17

where the catalyst is in the same phase as the reactants

Answer 18

The catalyst is usually a solid, and the reactants are either liquids or gases (1) Reactants adsorb onto the surface of the catalyst on an active site. (2) Reaction occurs on the surface of the catalyst (3) Products desorb from the surface if the catalyst

Answer 19

- Increase the surface area - Spread the catalyst over an inert support medium.

Answer 20

- Impurities can block the active sites. - This prevents the reactants from adsorbing - Purifying the reactants is the best way to prevent poisoning.

Answer 21

(1) Making Ammonia in the Haber Process Catalysed by solid IRON N2(g) + 3H2(g) ⇌ 2NH3(g) (2) Making Sulfuric Acid in the Contact Process Catalysed by solid Vanadium (V) Oxide – V2O5 Step 1 – Sulfur Dioxide is oxidised to Sulfur Trioxide - SO2(g) + V2O5(s) ⇌ SO3(g) + V2O4(s) Step 2 – The Vanadium (IV) Oxide is then converted back to Vanadium (V) Oxide with oxygen. - 2V2O4(s) + O2(g) ⇌ 2V2O5(s) This means the overall reaction is: 2SO2(g) + O2(g) → 2SO3(g) Sulfuric Acid is then formed by reacting SO3 with H2O. (3) Manufacture of Methanol Firstly this reaction forms a mixture known as synthesis gas CH4(g) + H2O(g) → CO(g) + 3H2(g) Then the following reaction is catalysed by solid Chromium (III) Oxide – Cr2O3 CO(g) + 2H2(g) → CH3OH(g)

Answer 22

only exist as 1 oxidation state

Answer 23

Peroxodisulfate ions oxidise iodide ions to iodine. The uncatalyzed reaction has a high Ea as the two negative ions repel each other: S2O82-(aq) + 2I-(aq) → 2SO42-(aq) + I2(aq) This reaction is catalysed by Fe2+ ions in a two-step process. Step 1: S2O82-(aq) + 2Fe2+(aq) → 2SO42-(aq) + 2Fe3+(aq) Step 2: 2I-(aq) + 2Fe3+(aq) → I2(aq) + 2Fe2+(aq) In the second step the Fe2+ catalyst is regenerated. Fe3+ could also be used as a catalyst, as step 2 could occur before step 1.

Answer 24

where one of the products of the reaction actually catalyses the reaction as it proceeds further

Answer 25

An example of this is the oxidation of ethanedioic acid by manganate (VII) ions. The uncatalyzed reaction has a high Ea as the two negative ions repel each other: 2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g) The Mn2+ ions produced can catalyse the reaction in a two step process as shown below. Initially the rate is slow, but as more catalyst is produced the rate increases. 1) In the first step the Mn2+ is oxidised to MnO4-. 4Mn2+(aq) + MnO4-(aq) + 8H+(aq) → 5Mn3+(aq) + 4H2O(l) 2) In the second step the Mn3+ is reduced by C2O42-. 2Mn3+(aq) + C2O42-(aq) → 2CO2(g) + 2Mn2+(aq) Because the MnO4- ions are a deep purple colour, the progress of this reaction can be tracked using a colorimeter. The reaction will go from purple to progressively clearer.

Answer 26

The rate starts off slow as there is no catalyst initially. The two negatively charged reactants collide with a very high Ea. Then as some Mn2+ is formed the rate increases as the reaction is being increasingly catalysed. The rate then decreases and levels off as the reactants get used up.