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Original post by MintCrepz
How on earth were we meant to know this at AS?



Original post by krisshP
It does not say in the spec that we need to know the systematic nAvq,e of NO2- ion, yet they asked that. Isn't that illegal?


I agree with you, I only know because i've done F334 and F335
Reply 1041
I i have to admit, i didn't this exam easy at all, but from past experience, there wasn't any particularly hard questions apart from the question at the end which was quite querky.

I reckon atleast 83 for the A, and may be up to 85. Remember they usually have to cap the percentage of people who achieve A's and i know a lot of people who seem like they got close to 100, or atleast 95.
Original post by anthonyrf
What did people say for the bonding question!? I said hydrogen bonds between molecules of propan-1ol and pentene had Id-Id and so on..


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Anyone? lol


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Reply 1043
Original post by anthonyrf
Exactly what I said!! lol

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Original post by MintCrepz
Yeah yeah :smile: sounds like we got similar answers !


Haha yup, seems like i've gotten similar to quite a few people on here! :wink:

Original post by juniorx
i said less co2 and the total enery was lower?


Yeah thats probably gonna be alright, defo the co2 bit

In fact for the C-F bond its more strong as its a smaller atom right :P

I found the paper decent but a lot of my friends hated it haha, so I think boundaries will be about 65-70ish for an A?
(edited 10 years ago)
Reply 1044
Original post by krisshP
It does not say in the spec that we need to know the systematic nAvq,e of NO2- ion, yet they asked that. Isn't that illegal?

this is ocr we're talking about...
(edited 10 years ago)
Reply 1045
Original post by super121
Nitrate (III) is the systematic name


phew, thought i might have gone that wrong when people said it was nitrogen dioxide
Original post by juniorx
i thought it was n2o



Wouldn't that would mean that the oxygen forms two single bonds with the nitrogen and then has two lone pairs left over? :s-smilie:
Original post by juniorx
i said less co2 and the total enery was lower?


I said that :smile:
For the C-F bond... I just said "higher frequency UV required as more energy is required to break the C-F bond as C-F has a greater polarity than C-Cl and so greater bond enthalpy"
Original post by krisshP
It does not say in the spec that we need to know the systematic nAvq,e of NO2- ion, yet they asked that. Isn't that illegal?


It could have easily been worked out using the oxidation states (which is on the spec).
Original post by Tejmasta
phew, thought i might have gone that wrong when people said it was nitrogen dioxide


Nitrogen dioxide? :O I'm pretty sure it was an ion and N2O not NO2
(edited 10 years ago)
Original post by tigerz
Haha yup, when I did f331, someone sneakily copied it so its possible! Yeah that was standard though got 1.17 (3 sf) for the frequency and 7.something for the bit above!


We shall see! And ditto, got the same as well :biggrin:

Original post by MintCrepz
How on earth were we meant to know this at AS?


You worked out the oxidation state of nitrogen in that ion in the earlier part when it was talking about oxidation, and there's two oxygen atoms, so you know it's going to end in ate... Systematic means you need to put in the oxidation state of the central atom in brackets.

It's like the chlorate section in the AS revision guide :smile:
Reply 1052
Original post by tigerz
Haha yup, seems like i've gotten similar to quite a few people on here! :wink:



Yeah thats probably gonna be alright, defo the co2 bit

In fact for the C-F bond its more strong as its a smaller atom right :P

i said the c-f bond length is shorter and that flourine is more electronegative so will form a stronger bond (electrons occupy a shell closer to the positive nucleus-distance is less)
(edited 10 years ago)
Original post by abzy1234
It really doesn't matter tbh. You learn this in F335 in a bit more detail; the A2 revision guide also writes Nitrate (III) ion:

IMAG0965.jpg


If it's in F335 A2, why were we asked this question? It's not in the F332 spec, so it's an illegal question by evil OCR salters.
Original post by steviep14
For the C-F bond... I just said "higher frequency UV required as more energy is required to break the C-F bond as C-F has a greater polarity than C-Cl and so greater bond enthalpy"


Yeah that's correct.

I kinda said that the C-F bond was stronger than the C-Cl bonds as flourines bonds are closer to the nucleus or some crap like that :/


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Original post by abzy1234
We shall see! And ditto, got the same as well :biggrin:



You worked out the oxidation state of nitrogen in that ion in the earlier part when it was talking about oxidation, and there's two oxygen atoms, so you know it's going to end in ate... Systematic means you need to put in the oxidation state of the central atom in brackets.

It's like the chlorate section in the AS revision guide :smile:


Yeah yeah that's how I knew it was III but I thought only NO3 was nitrate nothing else, whoops.
Reply 1056
Original post by booooomblastruin
Wouldn't that would mean that the oxygen forms two single bonds with the nitrogen and then has two lone pairs left over? :s-smilie:


no, nitrogen is the central atom so there is a triple bond between it and the other nitrogen and a dative bond with the oxygen (both electrons from nitrogen) giving it a linear shape
Original post by anthonyrf
Yeah that's correct.

I kinda said that the C-F bond was stronger than the C-Cl bonds as flourines bonds are closer to the nucleus or some crap like that :/


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That is correct if you phrased it correctly... that's the reason C-F is more polar :smile:
Original post by Tejmasta
the answer was all in the prerelease text. all you had to do was basically copy out what it said


That's what I did. They said to throw in only one equation and I gave like 2/3, is this wrong?
Reply 1059
I can confirm that the shape of Nitrous oxide (N2O) was linear.

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