Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

Original post by ayvaak

Posted from TSR Mobile

okay that's really weird lol...I'm not sure why that is :/

Reply 381

OP

Original post by ayvaak

Posted from TSR Mobile

Don't worry about it it's very very complicated and is way beyond our spec. It's to do with pi donor ligands if you really want to know (I did some research) but it's well beyond the level of our understanding for us to properly understand it

Reply 382

Aimen.

7

Original post by Don Pedro K.

Guys can someone tell me if this could be called 2-bromo-1-methylbenzene?

On wikipedia it said 1-bromo-2-methylbenzene but I don't understand why bromo is on the first carbon?

Reply 383

ayvaak

2

Original post by Aimen.

It could just be looking and finding the highest Mr group attached to the hexacarbon ring?

Posted from TSR Mobile

Reply 384

Aimen.

7

Original post by ayvaak

It could just be looking and finding the highest Mr group attached to the hexacarbon ring?

Posted from TSR Mobile

Not really sure, I'm just trying to learn it cuz once in a paper we were supposed to name an amino acid! so we start from carboxylic acid side!

Reply 385

Original post by Aimen.

Not really sure, I'm just trying to learn it cuz once in a paper we were supposed to name an amino acid! so we start from carboxylic acid side!

Yeah I just remember that -COOH takes priority above all and then go from there :smile:

Reply 386

OP

Original post by Don Pedro K.

Yeah I just remember that -COOH takes priority above all and then go from there :smile:

I would probably just go with instinct rather than trying explicitly to learn it, naming something is only ever gonna be a max of 1 mark and they often accept differing forms anyway so imo you're better off learning the stuff that can be big marks in the exam

Reply 387

Original post by samb1234

I would probably just go with instinct rather than trying explicitly to learn it, naming something is only ever gonna be a max of 1 mark and they often accept differing forms anyway so imo you're better off learning the stuff that can be big marks in the exam

Yeah but this is just something that I picked up along the way haha xD If you know that -COOH takes precedence, then that carbon atom will always be "number 1" so to speak.

Reply 388

lordoftheties

I can always identify the oxidizing or reducing agent when its an ion or molecule, but can't seem to do it when its a molecule. Can anyone guide me through the steps to doing this? Just like in the question below? Thanks
Screen Shot 2016-06-09 at 20.59.49.png

Reply 389

Spoderman14

1

With the ligand strength thing, it's to do with what type of bond(s) the ligand can form with the metal cation. Some like NH₃ can only form σ bonds with the cation, and then some which have electrons in p orbitals can donate those as π electrons too. There are also some ligands that have either vacant π* or d orbitals, which the metal cation can donate electrons to.
I found it really interesting to read up on some crystal field theory, it's a shame lots of interesting stuff gets left out in this course

Reply 390

Spoderman14

1

Original post by lordoftheties

I can always identify the oxidizing or reducing agent when its an ion or molecule, but can't seem to do it when its a molecule. Can anyone guide me through the steps to doing this? Just like in the question below? Thanks
Screen Shot 2016-06-09 at 20.59.49.png

Look at the oxidation number of sulfur wherever it appears, if it oxidises something, then it is itself reduced, so its oxidation number will decrease,

Reply 391

Original post by lordoftheties

I can always identify the oxidizing or reducing agent when its an ion or molecule, but can't seem to do it when its a molecule. Can anyone guide me through the steps to doing this? Just like in the question below? Thanks
Screen Shot 2016-06-09 at 20.59.49.png

You essentially need to look at how the oxidation state of sulfur changes in this case, since the oxidation states of hydrogen and oxygen are always +1 and -2 respectively (except of course for oxygen in the special cases when it is in hydrogen peroxide and something else if I remember correctly, where its oxidation state is -1 instead of -2).

To work out oxidation states, I like to use a lil' algebra. So let's take H2SO3 for example.

H2SO3 is a molecule with no charge, meaning that the oxidation states of all the elements which make up this molecule add to 0.

We know that the oxidation state of hydrogen is +1, and that of oxygen is -2. We have two hydrogen atoms and 3 oxygen atoms in this case. The unknown oxidation state is that of sulfur - let's name that "x".

So we can deduce:

2(1) + x + 3(-2) = 0.

2 + x -6 = 0.

x - 4 = 0 ----> Therefore, x = 4. So, the oxidation state of sulfur in H2SO3 is +4.

You do the same for the molecules/compounds containing sulfur on the product side to find if the oxidation state of sulfur increases or decreases.

We want to know in which scenario the sulfur (IV) acid is acting as an oxidising agent, which means that the acid itself will become reduced (i.e. the oxidation state of sulfur will DECREASE going from reactants to products).

Hope you're able to work out the answer based on this information :smile:

(edited 7 years ago)

Reply 392

CLGC98

6

So is the answer C? for the above ?

Reply 393

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Original post by CLGC98

So is the answer C? for the above ?

That's what I would have put, yes :smile:

Reply 394

gabby07

11

Hey y'all, sorry to bother you again, but I'm stuck on some multiple choice questions! :frown:

With the first one, I put C, and I can't for the life of me see why it's wrong lol. Surely E_cell for this reaction in C is positive? D is the right answer, and I see how that works in terms of giving E_cell> 0, but why can't it be C?

On the second one, I was unsure which answer to put down for part b). I was thinking green precipiate so I put B because the question mentioned Fe(II) but tbh I really have no idea why the answer is C.

With the third one, is it because benzene has more electrons, therefore greater london forces? I always thought hydrogen bonding was much stronger than the effect of london forces....?

And finally, the last one I don't see how the diazonium ion can decompose to give phenol, which is the answer. Can anyone explain this?

Thanks you guys so much! :smile:

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Reply 395

Original post by gabby07

Hey y'all, sorry to bother you again, but I'm stuck on some multiple choice questions! :frown:

With the first one, I put C, and I can't for the life of me see why it's wrong lol. Surely E_cell for this reaction in C is positive? D is the right answer, and I see how that works in terms of giving E_cell> 0, but why can't it be C?

On the second one, I was unsure which answer to put down for part b). I was thinking green precipiate so I put B because the question mentioned Fe(II) but tbh I really have no idea why the answer is C.

With the third one, is it because benzene has more electrons, therefore greater london forces? I always thought hydrogen bonding was much stronger than the effect of london forces....?

And finally, the last one I don't see how the diazonium ion can decompose to give phenol, which is the answer. Can anyone explain this?

Thanks you guys so much! :smile:

In answer to your first question regarding as to why it is D and not C, I think it is because in C, there is no redox reaction occurring; both the oxidation state of Co3+ and Cl2 decreases. That's the only reason I can think of :s!

I think with the second one, if insufficient acid is added, the green Fe2+ gets oxidised to Fe3+, which I think is a of a browny colour? Not too sure to be honest. Ask @samb1234 haha :smile:

For the third one, I think B is the only reasonable answer. A doesn't really explain high boiling temperature because boiling only overcomes intermolecular forces in molecules; the actual bonds of the individual molecules themselves are not affected. So that eliminates A. As for C, you don't get hydrogen bonds between benzene so that answer is wrong. D is also incorrect because again, the bonds within the molecule do not get broken so the strength of these bonds is irrelevant to this scenario. Hence, B seems like the only viable answer in this case!

As for why the diazonium ion decomposes into phenol, I have no clue; I just know that it does :s I'm sorry that I can't help you there xD

Reply 396

OP

Original post by Don Pedro K.

In answer to your first question regarding as to why it is D and not C, I think it is because in C, there is no redox reaction occurring; both the oxidation state of Co3+ and Cl2 decreases. That's the only reason I can think of :s!

I think with the second one, if insufficient acid is added, the green Fe2+ gets oxidised to Fe3+, which I think is a of a browny colour? Not too sure to be honest. Ask @samb1234 haha :smile:

For the third one, I think B is the only reasonable answer. A doesn't really explain high boiling temperature because boiling only overcomes intermolecular forces in molecules; the actual bonds of the individual molecules themselves are not affected. So that eliminates A. As for C, you don't get hydrogen bonds between benzene so that answer is wrong. D is also incorrect because again, the bonds within the molecule do not get broken so the strength of these bonds is irrelevant to this scenario. Hence, B seems like the only viable answer in this case!

As for why the diazonium ion decomposes into phenol, I have no clue; I just know that it does :s I'm sorry that I can't help you there xD

Original post by gabby07

Hey y'all, sorry to bother you again, but I'm stuck on some multiple choice questions! :frown:

With the first one, I put C, and I can't for the life of me see why it's wrong lol. Surely E_cell for this reaction in C is positive? D is the right answer, and I see how that works in terms of giving E_cell> 0, but why can't it be C?On the second one, I was unsure which answer to put down for part b). I was thinking green precipiate so I put B because the question mentioned Fe(II) but tbh I really have no idea why the answer is C.With the third one, is it because benzene has more electrons, therefore greater london forces? I always thought hydrogen bonding was much stronger than the effect of london forces....?And finally, the last one I don't see how the diazonium ion can decompose to give phenol, which is the answer. Can anyone explain this?Thanks you guys so much! :smile:

Yeah the second one is that. Q3 is due to the fact that benzene has delocalised electrons so instantaneous dipoles of much larger magnitudes than most other molecules can be set up. Diazonium doesn't really 'decompose' as such, they are just hugely susceptible to attacks from nucleophiles so all that happens is an oh from water subs onto the ring and nitrogen is given off

Reply 397

Original post by samb1234

Yeah the second one is that. Q3 is due to the fact that benzene has delocalised electrons so instantaneous dipoles of much larger magnitudes than most other molecules can be set up. Diazonium doesn't really 'decompose' as such, they are just hugely susceptible to attacks from nucleophiles so all that happens is an oh from water subs onto the ring and nitrogen is given off

Ahhh okay awesome :biggrin:

! I am starting to think whether you're just a degree level student trolling us by posing as an A2 student haha lol xD

For the second one, why is that insufficient acid brings about this effect?

Reply 398

OP

Original post by Don Pedro K.

Ahhh okay awesome :biggrin:

! I am starting to think whether you're just a degree level student trolling us by posing as an A2 student haha lol xD

For the second one, why is that insufficient acid brings about this effect?

oh sorry im actually wrong about the brown one - if you have insufficient acid you get the formation of managanese oxide which is brown, after first forming a complex with 4 waters and 2 hydroxides. I guess keeping the acid conc high means that the deprotonation is very unlikely to happen and hence you don't get the brownness. And thanks haha, chemguide tends to have more detail than you really need lol.

Reply 399