AQA CHEM1: 15th May 2012

Could anyone help me out with a query?

When trying to determine the shape of SO2, I drew it out as covalent bonding. However it seems to me that S has 10 electrons in it's outer shell and not 8? (I thought 8 was standard). Having said that, why is SO2 then a bent shape instead of trigonal pyramidal?

Also does anyone have a good method for determining the shape of a molecule (VSEPR)?
Thanks

Edit: I just discovered the AXE method, which does give the correct names even for SO2, so I guess i'll use this from now to be safe

Doesn't S have 6 electrons in its outer shell? So two electrons are bonded with the two oxygens and then it has two lone pairs, just like H2O

Does anyone know why the markschemes vary so much when it comes to mole calculation answers? I've always been taught to write them to three significant figures (standard form for the ISA as that seemed the norm) but for some questions they'll only accept the answer in standard form (mainly the pre 2006 questions like the ones at the end of the chapter in the AQA textbook) without specifying that in the question. More recent papers seem to accept either and usually the mark scheme only specifies the answer to three significant figures but I just did the June 2009 paper and got all the mole calculations right, writing them to three significant figures as I've been taught to BUT the mark scheme wants all the answers to five significant figures without saying it in the question!

I'm so confused as some mark schemes penalise you for writing more than three significant figures but then I've just lost all those marks for not answering to five significant figures...how are we supposed to know what they want?

S does have 6 electrons in it's outer shell, but so does O so in SO2 the only way for the oxygens to end up with a full outer shell is to have double bonding - it bonds like this: hence 10 electrons in Sulphur's outer shell. I've read a bit more about it, I think it just seems to be an exception to the rule

looking forward to get this exam over and done with so i can focus on unit2 more

I did OK on it but it was really just one of them papers where you have to be ready for it! Don't think anyone expected it lol.

To some degree I think if you read the question really carefully you'd be able to pick marks

same here! Ive done a few chem 2 past papers but the memorising is alot!

same have done 2 past paper under exam conditions and got 73/100 75/100 hoping to improve that to 80/100

Great! i have been getting between 60s to 70s! Hopefully would improve after memorising! Ive got three exams this week Bio1,chem1,m1 then got 5 days gap inbetween so got alot of time to memorise! I think chem 2 is quite straight forward if you know your stuff!

nice i got HBIO 1 and Chem1 this week and HBIO2 Chem2 next week and finaly Psya2 the week after i think chem2 is also pretty straight foward only troubles i'm hving atm are answering IR spec questions and answering observation questions e.g colour changes and precipitates

Yh those are the group 7 halogens! Again so much to memorise..i personally find redox a bit annoying and ionic equations!

ye silly halogens :/ also not too fond of group 2 metals. i like redox questions pm me if you need any help with them and i'll gladly help

Thanks Mate!

Ah thanks for that
I take Biology, Chemistry, Maths and Economics How about you?

Draw out all the possible chain isomers c6h14, c5h12ch3, c4h10ch3ch3. And then look at where you can move the CH3 group(s) if anywhere. Such as in c5h12ch3, you can move the ch3 group into 2 places forming 2 isomers, and for c4h10ch3ch3, the 2 methyl groups can either be position on the second carbon in the c4 chain, or they can both be positioned on separate carbons (carbon #2 and 3) creating 2 more isomers. Therefore the total number of isomers of c6h14 is 5.

Hope that helps.

thanks but ii thought you dont include the compound already given as an isomer?

Where it says C6H14 that doesn't actually give any information about how the chain is arranged, so you have to count the straight chain molecule as an isomer because it's a possible arrangement of those atoms.
If they had given it as a a structural formula, eg CH3CH2CH2CH2CH2CH3, then you wouldn't count the straight chain because they've already given you the information that it exists.
I hope that made sense!