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A2 OCR Chemistry B F334 - June 2015

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Original post by Jaz_ace
do we really have to know how to name ligands, havent seen a past paper question about it???


I've not seen a past paper question either. The spec says:

(s) explain and use the terms: ligand, complex/complex ion,ligand substitution, recalling the formulae of the followingexamples of complex ions from the chemistry of: iron:[Fe(H2O)6]2+, [Fe(H2O)6]3+; copper: [Cu(H2O)6]2+,[Cu(NH3)4]2+, [CuCl4]2-;
(t) describe the formation of complexes in terms of coordinate(dative) bonding between ligand and central metal ion;explain the terms bidentate and polydentate as applied toligands, exemplified by ethanedioate and EDTA4–;

So I'd say no you don't. Not sure why we've had to learn it but if that Biology exam was anything to go on, I'm learning it anyway!
Original post by kingcobra007
It forms 6 due to 6 lone pairs, it just has a +4 charge on the ion (not related to coordination number)


so EDTA4- forms 6 bonds due to 6 lone pairs and if so it all the time? the overall charge is -2 when it is with ca2+ i know that part.
Original post by Anonymous696061
Aww, thank you, & sure!

Acid-base titration:
1 - Fill your burette to zero with the solution of known conc (-in all the scenario is I've seen, this is usually the alkali).
2 - Using a pipette, transfer a known volume of your solution of unknown conc into a conical flask (-again, this is usually the acid from the scenarios I've seen!).
3 - Add in a few drops of suitable indicator, such as methyl orange or phenolphthalein. (-ONLY if it is a titration that requires indicators! Titrations with potassium permanganate do not require any indicators!)
4 - Start the titration, swirling the conical flask throughout; when you first see a colour change, immediately turn down the tap on the burette, so the solution of known conc is added drop-wise.
5 - When a permanent colour persists for 20 seconds, stop the titration & record the titre value in a table. (-For permanganate titrations, the end of the titration will be 'when a permanent pink'.)
6 - Repeat the titration until you have 2 concordant results (i.e. 2 titres within 0.1cm^3 of each other), then calculate the average of these titres.

There are 2 redox titrations in F334:
-Potassium manganate(VII) titrations - this an additional step of acidifying the solution of unknown conc, after transferring into the conical flask (step 2 above) -the rest of the steps are the same. (-KMnO4 is a strong oxidising agent, so can be used to find conc's of solutions containing Fe2+ ions or H2O2, through redox titrations.)

-Iodine-thiosulfate titrations - this is used to determine the conc of oxidising agents; it involves an additional step BEFORE carrying out the titration above: react a known vol of the oxidising agent with an excess of acidified potassium iodide solution. The iodine that is produced can then be titrate against sodium thiosulfate; also, towards the end of the reaction, add a few drops of starch solution. This will give an intense blue-black colour that will fade to clear at the end point.

Sorry this is so long, I hope it helps!


Thanks so much :smile:
Does anyone understand proton nmr? if so mind sharing??
Original post by sophieeabiii
Does anyone understand proton nmr? if so mind sharing??


Thats for f335 exam but do you still want me to explain? high resolution or low
Original post by MathsAddict
Thats for f335 exam but do you still want me to explain? high resolution or low

Both please! :wink:
Original post by MathsAddict
Love PM exams my brains absorbs so much info in the morning so i can look through experiments again.


LOL, for me I love pm when I need to do some last min revision😅, otherwise prefer am to get it over & done with!:P
Original post by mathsaddict
thats for f335 exam but do you still want me to explain? High resolution or low


yes please!
Original post by sophieeabiii
yes please!


How do i multi quote so i can answer to both of you.
Original post by jackhadrill
Both please! :wink:


how do i multi quote?
Original post by MathsAddict
How do i multi quote so i can answer to both of you.


I think you click the button with the 2 commas next to the reply button but i'm not sure
Drawing a blank here. How do we work out the charge of a complex ion again? Do the ligands not affect the overall charge or?
Original post by misteltain
Drawing a blank here. How do we work out the charge of a complex ion again? Do the ligands not affect the overall charge or?


Depends on if the ligand has a charge.

I.e.

Water is neutral so will not affect the overall charge (charge of the complex will equal the charge of the metal)
Halide ions have a 1- charge so will affect the overall charge
Original post by MathsAddict
how do i multi quote?


Original post by MathsAddict
How do i multi quote so i can answer to both of you.


Original post by jackhadrill
Both please! :wink:


Original post by sophieeabiii
yes please!


Atoms have a property called nuclear spin and this exhibited by Hydrogen protons. They resonate at a certain frequency called the lamar frequency which corrosponds to the radio waves.
Some hydrogen protons go against the magnetic field they are exposed to and have a higher energy than the ones aligned with the field.

Each hydrogen can be in a different chemical enviroment for example -HCH2-CH2-C0H

Here there are three different environments because the COH the h is next to an O that is unique in that region so thats one environment
The CH2 has a C-C bond next to it which both have different groups so that CH2 the two hydrogens here are in their own enviroment respectfully.
The HCH2 he hydrogen are in another enviroment again unique to them

Lets compare that to propanone and you can see the C=O is in the middle and that its symettrical so all 6 hydrogens are virtually in the same enviroment to prove this draw both molecules or use page 78 of the revise book.
Now some times you are given a chemical shift to absorption graphs and are told to work out the structure which can be tricky, so you need to use high resolution NMR.

The low resolution NMR shows you the peaks and tells you how many hydrogens are present which is hard to use to construct a molecule from but is useful. A high NMR will break down this peak in to smaller peaks which tells you what environments it is in. Cant all be explained using texts i wish it was visual i hope that made some sense.

Also another point the little peaks in the High resolution NMR will indicate this useful information.

if you are next to a CH3 molecule it will have 4 little peaks to demonstrate that you are next to a CH3 molecule.
If you are next to a CH2 molecule it have 3 little peaks to demonstrate that you are next to a CH2 molecule
if it has no H then it will be one peak which means the neighbouring molecule has no H atoms.
(edited 8 years ago)
For recrystallisation questions when they ask us to make it clear how impurities are removed or what a suitable solvent is. How do we answer these questions because different mark schemes say different things !



Posted from TSR Mobile
Original post by misteltain
Drawing a blank here. How do we work out the charge of a complex ion again? Do the ligands not affect the overall charge or?


You need to know the charge of the ligand (e.g. Cl is 1-, NH4 is 1+, H2O is 0) and the metal ion (e.g. Na is 1+, Ca is 2+ and they should tell you if it's copper or iron as they have variable oxidation states). Then you add together all the charges (remember to multiply ligands if necessary) and that's your overall charge. For example:

[Cu(Cl)4]2-
Work out the charge on copper
Cl has a -1 charge and there are 4 of them so that's 4- altogether. The overall charge for the complex ion is 2- so there is a difference of 2+ missing from somewhere- the copper. Therefore the charge on the copper is 2+.
Original post by pinkgorilla
For recrystallisation questions when they ask us to make it clear how impurities are removed or what a suitable solvent is. How do we answer these questions because different mark schemes say different things !



Posted from TSR Mobile


Suitable solvent:
- Dissolves solute at high temperatures
- Dissolves (almost) no solute at lower temperatures

Impurities removed:
- Solvent dissolves solute and filtering removes insoluble impurities
Original post by ThePhoenixLament
You need to know the charge of the ligand (e.g. Cl is 1-, NH4 is 1+, H2O is 0) and the metal ion (e.g. Na is 1+, Ca is 2+ and they should tell you if it's copper or iron as they have variable oxidation states). Then you add together all the charges (remember to multiply ligands if necessary) and that's your overall charge. For example:

[Cu(Cl)4]2-
Work out the charge on copper
Cl has a -1 charge and there are 4 of them so that's 4- altogether. The overall charge for the complex ion is 2- so there is a difference of 2+ missing from somewhere- the copper. Therefore the charge on the copper is 2+.


Original post by jackhadrill
Depends on if the ligand has a charge.

I.e.

Water is neutral so will not affect the overall charge (charge of the complex will equal the charge of the metal)
Halide ions have a 1- charge so will affect the overall charge


Thanks to both of you <3
Original post by MathsAddict
Atoms have a property called nuclear spin and this exhibited by Hydrogen protons. They resonate at a certain frequency called the lamar frequency which corrosponds to the radio waves.
Some hydrogen protons go against the magnetic field they are exposed to and have a higher energy than the ones aligned with the field.

Each hydrogen can be in a different chemical enviroment for example -HCH2-CH2-C0H

Here there are three different environments because the COH the h is next to an O that is unique in that region so thats one environment
The CH2 has a C-C bond next to it which both have different groups so that CH2 the two hydrogens here are in their own enviroment respectfully.
The HCH2 he hydrogen are in another enviroment again unique to them

Lets compare that to propanone and you can see the C=O is in the middle and that its symettrical so all 6 hydrogens are virtually in the same enviroment to prove this draw both molecules or use page 78 of the revise book.
Now some times you are given a chemical shift to absorption graphs and are told to work out the structure which can be tricky, so you need to use high resolution NMR.

The low resolution NMR shows you the peaks and tells you how many hydrogens are present which is hard to use to construct a molecule from but is useful. A high NMR will break down this peak in to smaller peaks which tells you what environments it is in. Cant all be explained using texts i wish it was visual i hope that made some sense.

Also another point the little peaks in the High resolution NMR will indicate this useful information.

if you are next to a CH3 molecule it will have 4 little peaks to demonstrate that you are next to a CH3 molecule.
If you are next to a CH2 molecule it have 3 little peaks to demonstrate that you are next to a CH2 molecule
if it has no H then it will be one peak which means the neighbouring molecule has no H atoms.


Thank you so much that was quite helpful :smile:
Original post by chemistrykid123
Good good I'm glad, yes I bet they are great we had an actual amazing (and interesting) group last year, it does not feel as daunting now I know somebody going through with me, I know what you mean I have not been revising intensely as I did last year trying to go with the calm relaxed approach this time. You ready?


I'm just currently watching videos and slowly creating more brain cells in the process ( Hopefully it's enough). Are you ready?

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