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Reply 200
Original post by Freddy-Francis
How are Van Der Waals formed?


1) random movement of electrons in shells unbalances distribution of charge across the molecule
2)at any one moment there is an instantaneous dipole across molecule
3)this induces dipoles in neighbouring molecules
4)which induces further dipoles in other molecules
5)small induced dipoles attract eachothother this is van der waals forces

down a group van der waals forces therefore get stronger due to increasing number of electrons and larger induced dipoles
Original post by panda001
1) random movement of electrons in shells unbalances distribution of charge across the molecule
2)at any one moment there is an instantaneous dipole across molecule
3)this induces dipoles in neighbouring molecules
4)which induces further dipoles in other molecules
5)small induced dipoles attract eachothother this is van der waals forces

down a group van der waals forces therefore get stronger due to increasing number of electrons and larger induced dipoles


Jeez. U guys are well prepared. :'( I only knew that it is caused by the movement of electrons within an atom :frown:
Tell me the important bits which is more likely to come up please.
I am a late reviser.
Reply 202
Original post by Freddy-Francis
Do we have to know the 5 bonding regions. Triagnol Bipyramid. There are two angles for it.. 90deg at equator and axis, 120deg at equator

I haven't seen it in any of my text books or past papers, and doubt it will come up as it's not one of the typical ones. Under 'Bonding and Structure' the specification just says to know "Theshapes of simple molecules and ions". But if you feel you'd rather be on the safe side go for it :wink:
Original post by sladyy96
I haven't seen it in any of my text books or past papers, and doubt it will come up as it's not one of the typical ones. Under 'Bonding and Structure' the specification just says to know "Theshapes of simple molecules and ions". But if you feel you'd rather be on the safe side go for it :wink:


My brain will explode, :frown:
I learned so many things by heart in 1 day..
Reply 204
Original post by Freddy-Francis
My brain will explode, :frown:
I learned so many things by heart in 1 day..

well I'm not going to go through any trouble to learn it as I think it's unlikely to be in the paper, and I haven't actually been taught it if that makes you feel any better :smile:
any tips on which bits to learn the most?
Do we get given the avagardos constant?
I hope that tomorrows chem paper isn't as awful as the bio one yesterday.:mad:
Original post by Freddy-Francis
Do we get given the avagardos constant?


I'm literally just looking at a question to do with that, it says 'using Avogadro's constant blah blah' and it doesn't give the actual value. It's 6.02x10^23
Original post by moonziggy
I'm literally just looking at a question to do with that, it says 'using Avogadro's constant blah blah' and it doesn't give the actual value. It's 6.02x10^23


oh ok. thanks.
Cruel people :frown:
does anyone know where to find jan13 chem paper?
Original post by sladyy96
Not sure if this answers your question directly, but the way I look at it is according to the number of regions of electron densities...
6 regions: 6bonds(octahedral)
4 regions: 4bonds(tetrahedral); 3bonds and 1lonepair(pyramidal); 2bonds and 2lonepairs(non-linear)
3 regions: 3bonds (trigonal planar); 2bonds and 1lonepair (non-linear)
2 regions: 2bonds (linear)

if you just remember the angles for octahedral, tetrahedral, trigonal planar, and linear, you can work out the others. Everytime there's a lone pair instead of a bond, the angle decreases by 2.5deg.
For example, for the shapes with 4 electron density regions:
4 bonds is 109.5deg; 3bonds and 1lonepair is 107deg; 2bonds and 2lonepairs is 104.5deg.


Wow, that was amazingly helpful! Never really got how the angles worked, just guessed most of the time, but I'll remember this, thank you :biggrin:
Reply 212
Original post by Freddy-Francis
Do we get given the avagardos constant?


Original post by moonziggy
I'm literally just looking at a question to do with that, it says 'using Avogadro's constant blah blah' and it doesn't give the actual value. It's 6.02x10^23


Yeah, it's on the sheet with the periodic table.
Original post by Freddy-Francis
Do we get given the avagardos constant?


In the data sheet we do, back page
Original post by moonziggy
I hope that tomorrows chem paper isn't as awful as the bio one yesterday.:mad:


We are in 2013. The paper is most definitely going to be cruel. Just do your best.
Reply 215
Original post by HelenPaddock
Wow, that was amazingly helpful! Never really got how the angles worked, just guessed most of the time, but I'll remember this, thank you :biggrin:

No problem, glad it helped! :biggrin:
Original post by HelenPaddock
In the data sheet we do, back page


oh ok then. good people :colondollar:
Reply 217
Anyone know the bond angle and shape of H2O2?
Reply 218
Original post by Rida94
Anyone know the bond angle and shape of H2O2?


If your referring to the bond angle of H-O-O then it Should be 104.5, as it consists of 2bp and 2lp in both cases. Try sketching the molecule out, it should help.
(edited 10 years ago)
Reply 219
Original post by Freddy-Francis
oh ok. thanks.
Cruel people :frown:


we get it on the data sheet they give us!

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