A-level Chemistry Revision Squad!
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im in year 12
Original post by Jade Li
Are you guys in year 12 or 13? I'm in year 12 
Original post by samb1234
this isn't technically true. If you are talking about the kind of pressures we experience on earth than yes it will sublime, but at extremely high pressures it is possible to get liquid diamond, and there are a few predictions that liquid diamond may exist on other planets in our solar system. Also, although an interesting question this thread is probably better used for questions to which people can work out the answer to, rather than trivia or random guesses.
I was asking it as I thought it was an interesting answer, nothing more.
Original post by osayuki
I have a question from OCR May 2013 paper that I can't do. Any help is appreciated please.
iii) The F-B-F bond angle in BF3 is different from the F-B-F bond angle in H3NBF3. Complete the table to predict the F-B-F bond angle in BF3 and in H3NBF3.
I got 120 for BF3 which is right but I can't work out the H3NBF3 bond angle. The m.s. says is 109.5.
iv) The H-N-H bond angle in NH3 is 107. A student predicted that the bond angle in H3NBF3 is larger. Explain why the student might expect the H-N-H bond angle to be larger in H3NBF3 than in NH3.
I know that the lone pair in NH3 repels more than the bonded pair of electrons. Also N in NH3 has 3 bonding pairs and 1 pair of electrons.
But I don't understand why the m.s. said N in H3NBF3 has 4 bonding pairs and no lone pairs.
iii) The F-B-F bond angle in BF3 is different from the F-B-F bond angle in H3NBF3. Complete the table to predict the F-B-F bond angle in BF3 and in H3NBF3.
I got 120 for BF3 which is right but I can't work out the H3NBF3 bond angle. The m.s. says is 109.5.
iv) The H-N-H bond angle in NH3 is 107. A student predicted that the bond angle in H3NBF3 is larger. Explain why the student might expect the H-N-H bond angle to be larger in H3NBF3 than in NH3.
I know that the lone pair in NH3 repels more than the bonded pair of electrons. Also N in NH3 has 3 bonding pairs and 1 pair of electrons.
But I don't understand why the m.s. said N in H3NBF3 has 4 bonding pairs and no lone pairs.
ii) 120 is right as its a trigonal planar. The h3nbf3...
it would help if you saw the boron having 4 bonds, 3 sinlge bonds to each fluorine and 1 bond to nitrogen ( even though this is a dative/co ordinate bond) , and as you may know this is then a tetrahedral structure which has a bond anlge of 109.5, therfreo bond anlge is 109.5.
iv) when it is NH3 you are quite right it has one lone pair of elcetrons which cause more repelling, therefore a bond anlge of 107 between H-N-H, however once this undergoes dative bonding and shares both its electrons with boron , this no longer acts as a lone pair, it acts like a normal sinlge covalent bond. Therefore nh3 has 3 bonding pairs and 1 lone pair , so the lone pair repeels the bond pairs even more, causing angles of 107, however when the lone pair forms a bond with Boron, this now acts a normal bond pair ,and as you know if an element has 4 bond pairs and no electron pairs, anlge is 109.5
I hope this helped
Original post by Science_help
ii) 120 is right as its a trigonal planar. The h3nbf3...
it would help if you saw the boron having 4 bonds, 3 sinlge bonds to each fluorine and 1 bond to nitrogen ( even though this is a dative/co ordinate bond) , and as you may know this is then a tetrahedral structure which has a bond anlge of 109.5, therfreo bond anlge is 109.5.
iv) when it is NH3 you are quite right it has one lone pair of elcetrons which cause more repelling, therefore a bond anlge of 107 between H-N-H, however once this undergoes dative bonding and shares both its electrons with boron , this no longer acts as a lone pair, it acts like a normal sinlge covalent bond. Therefore nh3 has 3 bonding pairs and 1 lone pair , so the lone pair repeels the bond pairs even more, causing angles of 107, however when the lone pair forms a bond with Boron, this now acts a normal bond pair ,and as you know if an element has 4 bond pairs and no electron pairs, anlge is 109.5
I hope this helped
it would help if you saw the boron having 4 bonds, 3 sinlge bonds to each fluorine and 1 bond to nitrogen ( even though this is a dative/co ordinate bond) , and as you may know this is then a tetrahedral structure which has a bond anlge of 109.5, therfreo bond anlge is 109.5.
iv) when it is NH3 you are quite right it has one lone pair of elcetrons which cause more repelling, therefore a bond anlge of 107 between H-N-H, however once this undergoes dative bonding and shares both its electrons with boron , this no longer acts as a lone pair, it acts like a normal sinlge covalent bond. Therefore nh3 has 3 bonding pairs and 1 lone pair , so the lone pair repeels the bond pairs even more, causing angles of 107, however when the lone pair forms a bond with Boron, this now acts a normal bond pair ,and as you know if an element has 4 bond pairs and no electron pairs, anlge is 109.5
I hope this helped
Thank you so much
Original post by Science_help
I was asking it as I thought it was an interesting answer, nothing more.
Yeah as I said it is an interesting question, I didn't mean to come across quite like that. Anyway best of luck with AS (assuming your school still take) and let me know if I can help with anything as I did AS last year
Original post by Haddar
carry out experiments to study the solubility of simple molecules in differentsolvents
Part of the specification, posted a thread with no luck
Part of the specification, posted a thread with no luck
could be to do with the fact that polar molecules only dissolve in polar solvents (e.g. an ionic compound in water) whereas non-polar solvents tend to dissolve in non-polar solvents (e.g. hexane). This is to do with the large electro-negativity difference between two atoms of a molecule creating a slightly positive and negative dipole, and hence a polar molecule.
I'm taking the full A Level in the October/November session, CIE board. Does anyone know how the practical skills papers are? I mean, do I have to set apparatus and carry out experiments in front of an examiner? I have the Chemistry coursebook from Cambridge University Press, and the practical skills chapter has some exercises that only require to draw graphics with given data and write plans for experiments. If the questions are like this I'm ok, otherwise I'll have to start practicing my laboratory skills in real life
Original post by nutcase13
could be to do with the fact that polar molecules only dissolve in polar solvents (e.g. an ionic compound in water) whereas non-polar solvents tend to dissolve in non-polar solvents (e.g. hexane). This is to do with the large electro-negativity difference between two atoms of a molecule creating a slightly positive and negative dipole, and hence a polar molecule.
Thanks for the reply, and as a test all I have to do is add different solutes do solvents and check what dissolves in what?
Original post by Haddar
Thanks for the reply, and as a test all I have to do is add different solutes do solvents and check what dissolves in what?
Yeah basically you would just have test tubes with a few different types of solvent (e.g water, organic solvent etc) and record your observations when you add the compounds to each test tube
Original post by osayuki
I have a question from OCR May 2013 paper that I can't do. Any help is appreciated please.
iii) The F-B-F bond angle in BF3 is different from the F-B-F bond angle in H3NBF3. Complete the table to predict the F-B-F bond angle in BF3 and in H3NBF3.
I got 120 for BF3 which is right but I can't work out the H3NBF3 bond angle. The m.s. says is 109.5.
iv) The H-N-H bond angle in NH3 is 107. A student predicted that the bond angle in H3NBF3 is larger. Explain why the student might expect the H-N-H bond angle to be larger in H3NBF3 than in NH3.
I know that the lone pair in NH3 repels more than the bonded pair of electrons. Also N in NH3 has 3 bonding pairs and 1 pair of electrons.
But I don't understand why the m.s. said N in H3NBF3 has 4 bonding pairs and no lone pairs.
iii) The F-B-F bond angle in BF3 is different from the F-B-F bond angle in H3NBF3. Complete the table to predict the F-B-F bond angle in BF3 and in H3NBF3.
I got 120 for BF3 which is right but I can't work out the H3NBF3 bond angle. The m.s. says is 109.5.
iv) The H-N-H bond angle in NH3 is 107. A student predicted that the bond angle in H3NBF3 is larger. Explain why the student might expect the H-N-H bond angle to be larger in H3NBF3 than in NH3.
I know that the lone pair in NH3 repels more than the bonded pair of electrons. Also N in NH3 has 3 bonding pairs and 1 pair of electrons.
But I don't understand why the m.s. said N in H3NBF3 has 4 bonding pairs and no lone pairs.
Well the two central atoms in H3NBF3 are N and B there exists a dative covalent bond between the N and B as N has a lone pair and B accepts this so both these atoms now have 4 bonding regions of electrons thus all electron pairs repel and the shape is tetrahedral with a bond angle of 109.5.
For those of us doing OCR A, F324 and F325. I am in the process of making big revision sheets out of my notes, so far I have these three if any of you are interested:
Benzene reactions:
Identifying carbonyl compounds (I also included tests for primary/secondary/tirtiary alcohols and halides from AS because there was loads of empty space):
Rates, equilibrium and pH (didn't have enough space for buffers though):
Hopefully someone finds that helpful, any suggestions on what topics to do next?
Benzene reactions:
Spoiler
Identifying carbonyl compounds (I also included tests for primary/secondary/tirtiary alcohols and halides from AS because there was loads of empty space):
Spoiler
Rates, equilibrium and pH (didn't have enough space for buffers though):
Spoiler
Hopefully someone finds that helpful, any suggestions on what topics to do next?
Original post by samb1234
Yeah as I said it is an interesting question, I didn't mean to come across quite like that. Anyway best of luck with AS (assuming your school still take) and let me know if I can help with anything as I did AS last year
Its completely fine, I know you didn't mean it like that.
And yes my sixth form are still doing as levels, and thank you for the offer
Good luck in your A2 exams.
What other subjects are you taking? and what did you get in as ? what did you apply for ?
Spoiler
Original post by Science_help
Its completely fine, I know you didn't mean it like that.
And yes my sixth form are still doing as levels, and thank you for the offer
Good luck in your A2 exams.
What other subjects are you taking? and what did you get in as ? what did you apply for ?
And yes my sixth form are still doing as levels, and thank you for the offer
Good luck in your A2 exams.
What other subjects are you taking? and what did you get in as ? what did you apply for ?
Spoiler
Maths further maths physics chem, I'll PM you my results and unis
Original post by RMNDK
That's quite cruel to be honest. I mean you understand the concept and you were pretty much spot on.
The pH values are just one of those things you need to learn. For instance, NaOH is a strong base and so will have a pH of 13-14.
pH values vary across exam boards. You might also be expected to recall pH values for specific compounds.
But you can try remember these steps.
The pH values are just one of those things you need to learn. For instance, NaOH is a strong base and so will have a pH of 13-14.
pH values vary across exam boards. You might also be expected to recall pH values for specific compounds.
But you can try remember these steps.
1.
What is being titrated? i.e. what's in the beaker. That's the starting pH. If it's a strong acid you're going to have a low pH of 1-3. If it's a weak acid, it'll be 3-6. If it's a strong base, it'll be 12-14. If it's a weak base, it'll be 8-11.
2.
What is it being titrated against? i.e. what's in the burette. That's the thing that will change the pH. If it's a strong acid, the end will be 1-3. If it's a weak acid it'll be 3-6, and so on.
3.
Draw the vertical portion at the equivalent point and about 1.5-2 pH off the end values. What I mean by this is say your end pH is 13 (e.g. your titrant is NaOH). Draw the vertical portion up to 11 or 11.5, maybe 12.
4.
Combine the vertical portion with the two pH values by nice curves.
THANK YOU SO MUCH. Very helpful
hey guys, can someone please help me with the attactched pastpaper question ..
sorry about the bad quality .. Compound P is CH3CH2CH2CH3
so what i know is that the more the branching the lower the boiling point due ti less effective london attractions sooo,, how does compound P have a lower boiling point than compound s when S is more branched I dont understand..
sorry about the bad quality .. Compound P is CH3CH2CH2CH3
so what i know is that the more the branching the lower the boiling point due ti less effective london attractions sooo,, how does compound P have a lower boiling point than compound s when S is more branched I dont understand..
Original post by NoorL
hey guys, can someone please help me with the attactched pastpaper question ..
sorry about the bad quality .. Compound P is CH3CH2CH2CH3
so what i know is that the more the branching the lower the boiling point due ti less effective london attractions sooo,, how does compound P have a lower boiling point than compound s when S is more branched I dont understand..
sorry about the bad quality .. Compound P is CH3CH2CH2CH3
so what i know is that the more the branching the lower the boiling point due ti less effective london attractions sooo,, how does compound P have a lower boiling point than compound s when S is more branched I dont understand..
What is the answer in the mark scheme?
And P has a lower bpt because compound S has a halogen group attached, meaning the C-Br bond is polar which allows for dipole-dipole intermolecular bonds.
Duh.
(edited 8 years ago)
Original post by High Stakes
What is the answer in the mark scheme?
And P has a lower bpt because compound S has a halogen group attached, meaning the C-Br bond is polar which allows for dipole-dipole intermolecular bonds.
Duh.
And P has a lower bpt because compound S has a halogen group attached, meaning the C-Br bond is polar which allows for dipole-dipole intermolecular bonds.
Duh.
the answer is D.. annd thanks
Original post by High Stakes
What is the answer in the mark scheme?
And P has a lower bpt because compound S has a halogen group attached, meaning the C-Br bond is polar which allows for dipole-dipole intermolecular bonds.
Duh.
And P has a lower bpt because compound S has a halogen group attached, meaning the C-Br bond is polar which allows for dipole-dipole intermolecular bonds.
Duh.
butttttt dont london forcess have a higher effect on the boilingpoint than dipole diple
Original post by NoorL
butttttt dont london forcess have a higher effect on the boilingpoint than dipole diple
No...
In terms of increasing strength:
1) van der Waals / London forces.
2) Dipole-dipole forces.
3) Hydrogen bonding.
This is basic chem 101. Get these solid otherwise it's going to make you shakey at the higher stuff.
--
@RonnieRJ - How's chem going for you?
Are you with AQA too?Quick Reply
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