2 | intechemistry
October 13, 2010 at 10:07 pm

pleasure.

8 | intechemistry
November 5, 2010 at 12:09 am

The reaction is actually pretty complicated. In reality this reaction is very messy: Lots of things are produced and lots of reactions happen. I think it’s OK to write it like this as a kind of summary of the overall reaction:

8HI(g) + H2SO4(l) –> H2S(g) + 4H2O(l) + 4I2(s)
Ref: New Understanding Chemistry for A-level. Lister and Renshaw 3rd Edn, p211.

But you can write other equations like equations that yield sulfur if you insist. It’s not likely they will ask you for the HI reaction (or NaI or KI) reaction with sulfur for the reason above – it’s complicated. They are far more likely to ask you for equations for the HCl (or NaCl or KCl) reaction and HBr (or NaBr or KBr) reaction.

28 | intechemistry
December 23, 2010 at 3:44 pm

Facer means anhydrous Copper(II) compounds e.g. CuSO4(s). We are more used to seeing and using CuSO4.5H2O(s) {incidentally, sometimes called bluestone}.

CuSO4.5H2O(s) contains water LIGANDS. There are no LIGANDS in anhydrous CuSO4. Its the ligands that cause the d-subshell to split which consequently allows d-d transitions to occur which in turn is responsible for the colour.

Remove the ligands and Cu(II)’s ligands and the d-subshell is no longer split and so d-d transitions cannot occur.

If you heat CuSO4.5H2O(s) witha Bunsen flame, the bolue colour gradually turns white as the water is driven off. Over time it goes blue again as the watervapor in the air rehydrates the anhydrous compound.

Hope that helps.

30 | intechemistry
December 27, 2010 at 8:16 pm

Salam.

A few factors, the most important being:
Whether the process has deltaS(tot) > 0

The above can be broken down into smaller considerateions e.g.
1) The size and M-L bond strength of the existing/previous complex
2) The rate of ligand exchange
3) The size and structire of the ‘incoming’ ligand and the outgoing ligand
4) The charge on the complex to that of the ligand may also be a factor, although I’ve never studied that relationship.
5) The presence of particular EM frequencies.
6) The existing occupancy of the t.metals d-subshell
7) The concentraion of the incoming ligands (eqm effect)

& probably a few more

34 | intechemistry
February 26, 2011 at 4:13 pm

I hope you and your roommate are no longer stuck. If you get stuck again try that ‘shakey leg’ thing

36 | intechemistry
February 26, 2011 at 5:27 pm

lol

38 | intechemistry
April 8, 2011 at 12:37 am

Xavier. Again, your progression through the simplified world of A-level chemistry has uncoverd another weakness in these A-level models. Amphoteric or not is way too ‘binary’, at a higher level of knowledge, the boundaries often become blurred sometimes. This A-level specification tries to just point out the ‘easily seen’ amphoteric species. Hence Cr3+ Zn2+ and Al3+ (last specification also included Pb and Sn ions, now dropped)

Have a look at this… http://intechemistry.files.wordpress.com/2010/09/for-xavier-8-april-20112.jpg
[ Source: Inorganic Chemistry (Second Edition)
by D.F. Schriver, P.W. Atkins, C.H. Langford. Pub Oxford.
ISBN: 0-19-8555396-X ]

40 | intechemistry
May 5, 2011 at 3:19 pm

Hello again ‘Chemistry Lover’ – great name by the way
Can I ask you to post your question in the appropriate place. I need some level of organisation or the usability of the blog will ‘degrade’. It takes my time to move comments around. Thanks

So, why 4 and not 6 NH3. There are many factors. 1) Entropy considerations (inc therefore, delta H changes) 2) geometry/space reasons 3) Occupancy of orbitals, 4) energy of those orbitals… and probably a number of other things too.

The reasons are too complex for me to quantify. You will not have to explain them, but will be expected to know what the product is. So here, it’s just a matter of learning the fact. I don’t like to say that, but on occasions it’s a must.

42 | intechemistry
May 6, 2011 at 12:26 am

No problem.

Sometimes there isn’t much of a ‘clever’ way to learn. One suggestion I’d make is to list all the reactions down and try and group them according to common features, e.g. all one group in which all the complexes have 6 NH3′s Take the most heavily populated category as your DEFAULT case for all compounds. Then, look at the minor categories, and make them ‘exceptions’ to the rule. Do that for other minor groups too, e.g. those with 2 NH3 like Ag. Pretty quickly you will cover all the groups. Then, if necessary, think of some way in which to tie the groups together. I don’t have a specific trick to learn this one.

– If anyone does have any good tips here, I’d love to hear of it –

You might like to give this a quick read: http://hsc.csu.edu.au/study/memorisation.htm

44 | intechemistry
May 7, 2011 at 4:57 pm

Salam.
Problem: The energies involved in many aspects of t-metal complexes are very similar. To be able to fully predict exactly what happens in all cases, you need to know these energies and ‘sum’ them up.

A small energy difference relating to one aspect involved with complexes can affect the overall energy of the system giving the appearance of a different or unpredictable result.

Identification and evaluation of all those factors is beyond A-level. So, just get familiar with the ones specifically mentioned in the syllabus and memorise them.

Assume all have 6-coordination and just learn the small number of 4 coordinate ones (e.g. cis-platin) and the exact shape they have.

If they asked you something (or gave you info suggesting) that implies a 4-coordinate complex, and you had to state the shape, then as it’s not a specified learning requirement, then you are almost 100% certain to get marks if you said square planar or tetrahedral, but none if you said octahedral.

46 | intechemistry
May 12, 2011 at 1:24 am

Salam.

Yes, as a quick answer you are right.

The discussion below expands on that slightly.

The definition of a catalyst is that it’s not consumed in a reaction. In the case of t.metals, because the energy involved in them giving and taking electrons, is quite low, it’s easy to do and hence, it is easy to return to the original state as it was before the reaction.

For gp1 elements however, they give away e- very easily but it’s very hard for them to take electrons back again afterwards. They’re all around about -2.xx V each} The initial state of the gp1 element is unlikely to be regenerated, hence not a catalyst.

48 | intechemistry
May 15, 2011 at 10:41 pm

I’d be happy with that, but I suspect a question asking about that might require a bit more detail. Your answer would be appropriate for a question worth about one mark.

54 | intechemistry
June 6, 2011 at 5:55 am

@51, chocoviolicious1808, June 4, 2011 at 1:02 pm
I presume you meant {see bold O.S.}:
“anyway, in the preparation of chromium (II) ethanoate, is it the Zn(s) or H2 that reduces Cr from dichromate (VI) to Cr2+?”
If I am right that you made a simple typo, then state Zn as it’s the reagent used.

@52, chocoviolicious1808, June 5, 2011 at 9:33 pm
Use of ⇌ in redox EQUILIBRIUM is fine.
I’d only worry about it IF the answer scheme rejected “⇌” which I very much doubt it does.
→ might be (casually) used simply to show which direction the reaction goes in.

60 | intechemistry
June 12, 2011 at 2:56 pm

@ 55, emrys ong, June 6, 2011 at 1:58 pm (sorry I’ve got the answer ordering mixed up)

You asked:
i have one ques in mind. let say they ask us to find whether oxygen can oxidise Fe2+ to Fe3+ in acidic cond. how do i know which 1 to use?
½O2(g) + 2H+(aq) + 2e- ⇌ H2O(l) +1.23
O2(g) + 2H+(aq) + 2e- ⇌ H2O2(aq) +0.68
because they will give different feasibility

You would not use
O2(g) + 2H+(aq) + 2e- ⇌ H2O2(aq) +0.68
why?
Because Fe3+(aq) + e- ⇌ Fe2+(aq) +0.77 p17 data book
You have as your reactants Fe2+ and O2. so you must swap the Fe equation.
Fe2+(aq) ⇌ Fe3+(aq) + e- -0.77
Then you add it to the ‘oxygen’ equation (O2 must be on the LHS)
The E(std)cell would be -’ve hence not feasible.

Also, but this is a bit trivial, there was no indication or declaration that the reaction produced H2O2. So choosing the ‘oxygen’ equation where just H2O was formed is the usual thing to do in such cases.

from what I’ve seen, many students seem to think they can just chose some equation in which they ‘see oxygen’ and then just use that. But the question restricts you. You must have the O2 on the left and Fe2+ on the left. For a feasible reaction you must get E(std)cell to be +’ve

62 | intechemistry
June 12, 2011 at 3:05 pm

@ 56, emrys ong, June 6, 2011 at 2:03 pm

you asked:
Fe2+ –> Fe3+ + e-
MnO4- + 8H+ + 5e- –> Mn2+ + 4h2O
to detect the titration we need to see when there is permanent pink colour.
may i know why?

The permanent pink is due to a slight excess of KMnO4- from the last drop you added from the burette. It’s pale pink (and NOT deep purple) because the slight excess of KMnO4 is diluted in the large volume of liquid in the conical flask.

Do not say the pale pink colour is because of the Mn2+(aq) ions produced. Yes, Mn2+(aq) is pale pink but the intensity is so low compared to the diluted KMnO4- that it’s not the Mn2+ pale pink that your eyes use to see the end point. Also, the pale pink of Mn2+ would almost definitely be ‘masked’ by the Fe species, even if the Fe species were in low concentrations. Even if a colourless redox system is used instead of Fe, then as the titration goes on, the Mn2+ would be present from the beginning and increase in concentration as the titration proceeded. Hence there would be no sudden change/appearance/disappearance of colour even of your eyes were good enough to see it.

64 | intechemistry
June 12, 2011 at 3:18 pm

@ 61, chocoviolicious1808, June 12, 2011 at 3:02 pm, Re: answer #59.

Oh, this whole question is not really concerning the new specification? Cos if im not mistaken, there is one answer which says that it is becos 8 mol of H+ is needed to react with 1 mol MnO4-. Honestly, that answer is the easiest to understand , so i only understand that particular one.

Well, yeah, you’re quite right, that’s important too (and I did mention that near the end ) but your query was specific to oxidation by air, hence the focus of my answer was specific to that.

74 | intechemistry
June 23, 2011 at 10:20 am

The SHE is always put on the left according to the convention.
Facer is being naughty and I’m not sure why he is doing it, but I have no doubt he could justify it (i.e. maybe vis high resistance voltmeter is swapped around – just a guess, or maybe he knows which species is being oxidised e.g. one electrode being ‘eaten’ away hence oxidation, so mentally automatically corrected the sign.)

Putting the SHE on the left always gives the correct sign, + or -, for the standard reduction potential If you put the SHE on the other side, the magnitude will be the same, but the sign will be the wrong way around.

Example: Using the SHE on the left, in accordance with the convention, gives the Zn2+(aq) | Zn(s) half cell a reduction potential of -0.76 V
Putting the SHE on the right would give Zn2+(aq) | Zn(s) half cell a reduction potential of +0.76 V

Re: HNO3. The H+ in the SHE comes from H2(g) not a conventional acid.
The equation is H2(g) ⇌ 2H+(aq) + 2e-

2. In aqueous solution, most Cu(I) complexes are unstable and revert to Cu(II). Out of solution many Cu compounds revert to Cu(I) on heating from Cu(II). If you are talking about compounds or complexes then I am not too sure why this is the case, but it’s going to be the usual consideration of ionisation energies, lattice energies, polarisation, crystal lattice structure etc. In Cu compounds and complexes the empty orbitals in 4s and 3d get filled by the ligand, so in the complex, we don’t really have empty or partially empty d-subshells any more.

If you are just talking about Cu ions only (like as if you are doing an ionisation energy reading) then Cu+1 is more stable than Cu+2, because less energy is needed to remove one electron (forming Cu+1) rather than losing 2e- to form Cu+2.

76 | intechemistry
June 23, 2011 at 2:55 pm

Emrys. I got my days mixed up, so i’ve removed my insane ramblings. Ignore me.

78 | intechemistry
June 23, 2011 at 10:08 pm

According to the Chem dictionary #2 in the sidebar links, WoC is “Water that is stoichiometrically bound in a crystal” i.e. the number of water molecules necessary for the adoption of the crystal structure. Some (usually most) of these water molecules might be bound to the metal (standard metal-ligand interaction). You can add (a little) more water but the crystal structure won’t change. Removing that water of crystallisation will change the crystal structure. And adding an excess of water will cause the ions to dissolve breaking down the crystal structure.

A compound is hydrated whenever water molecules are present with(in) the compound in question. So CuSO4 is anhydrous, CuSO4.H2O (unnatural) is hydrated. CuSO4.5H2O (natural) is hydrated. CuSO4(aq) is also hydrated. In a hydrated compound the water molecules could be ligands but not necessarily so – they may fit into the structure in other ways.

I read CuSO4.5H2O the pentahydrate (natural) has 4 H2O’s on the Cu and one H2O bonding to the sulfate.

80 | intechemistry
June 24, 2011 at 11:30 am

It is a bit
Just treat WoC as being like the metal-ligand bonds that you’ve seen before and any more water after a certain amount (5 for CuSO4) will start to ‘dissolve’ the crystals.

82 | intechemistry
January 3, 2012 at 10:14 pm

5.3.1 j
discuss the use of hydrogen and alcohol fuel cells as energy
sources, including the source of the hydrogen and alcohol, eg
used in space exploration, in electric cars
5.3.1.k
demonstrate an understanding of the principles of modern
breathalysers based on an ethanol fuel cell and compare this
to methods based on the use of IR and to the reduction of
chromium compounds.

I personally do not give notes on this (I leave it as a student project). But you could try the very very good http://www.drbateman.net/asa2sums/sum5.3/sum5.3.htm

84 | intechemistry
January 6, 2012 at 9:30 pm

That info is covered by what’s known as Core Practical 29
but i don’t have that.
See: http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/GCE_Chemistry_8CH01Practical_work_Mapping_Green.pdf for that reference.

If you could get a hold of CP29, it may help.

Sorry, that’s all I can do at this moment.

90 | intechemistry
January 26, 2012 at 9:49 pm

And a good day to u 2. You are very much on the right lines… they form complexes because T.metal ions have many vacant orbitals of an energy and shape similar to that of many nucleophiles. Also the formation of complexes leads to a increase in total entropy (the physical reason why EVERYTHING happens)

The geometrical arrangement of the replaceable ligands in cis-platin is such that the molecule can bond to both strands of DNA preventing transcription hence reproduction of the cell, hence the cell will die. In the other configuration, the cis-platin molecule isn’t the right fashion to lock both strands together.

Try here for more info that sounds more ‘medical’: http://www.netdoctor.co.uk/cancer/medicines/cisplatin.html
or google around e.g. How does cic-platin work.

92 | intechemistry
February 13, 2012 at 11:00 pm

A deprotonation is where a H+ (which is a proton) is removed from some substance by using a base. In this topic the substance is usually an aqua complex e.g. [Cu(H2O)6]2+(aq) and the base is a hydroxide ion OH- (which can be generated when ammonia is added to water). The -OH pulls a proton from the H2O ligand.

A ligand exchange is where one species with a coordinate bond to a metal ion is replaced by a different species. The other species makes it’s own coordinate bond to the metal ionE.g.the EDTA ligand can replace the H2O ligand.