AQA CHEM1: 15th May 2012

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They've done it before. From Gallium to Selenium. However it's just the same as it would be for the elements above them so no biggie

Reply 281

SHAH :)

Original post by Jack_Smith

1s2.2s2.sp6.3s2.3p6.4s2.3d10.4p6.5s2.4d10.5p1

but i dont get why you dont fill the 4f sub shell before the 5s one :s

Reply 282

564767

Original post by TaraBelle

They've done it before. From Gallium to Selenium. However it's just the same as it would be for the elements above them so no biggie

ahh okay, so just your typical increase due to increased nuclear charge?
what dicks

Reply 283

TaraBelle

Original post by joker12345

What on earth are T shapes?!

It's the same trigonal bipyramidal, but imagine two of the bonds being lone pairs instead

Reply 284

Voglie

Original post by wenger16

how do you work out how many electrons in each shell?

The number of e-'s an element has is equal to the atomic number of that element.

The s sub-shell stores 2 electrons,

the p sub-shell stores 6 electrons,

and the d sub-shell stores 10 electrons.

The sub-shell order is: 1s 2s 2p 3s 3p 4s 3d 4p, and depending on which type of sub-shell it is, that's how many e-'s you're allowed to put into it.

Just keep on filling the sub-shells in that order until you have filled in all the e-'s the element has.

Reply 285

dobe

Original post by wenger16

how do you work out how many electrons in each shell?

the 1s orbital fills first, then the 2s, 2p, 3s, 3p, 4s, 3d, etc
s orbitals have 2 electrons, p orbitals have 6 electrons, d orbitals have 10 and f orbitals have 14.
you then just count along until you have used all the electrons.
for example, phosphorus will be 1s2, 2s2, 2p6, 3s2, 3p3 (this equals to 15 electrons)
sorry if this made no sense whatsoever

Reply 286

564767

What do you guys reckon is the hardest part of the bonding chapter?

Reply 287

wenger16

Can someone explain all the shapes business? like trignal bypyramid etc?

Reply 288

Madara

The hardest part for me is working out some of the wierd bond angles like in NH3 its a triangular pyramid right but why is its bond angle 107? thats the only one which confuses me. You'd expect is to be 120-2.5 so 117.5 so could someone explain this? :s-smilie:

Reply 289

Madara

Original post by W4L1Y

Hi
how do you turn dm^3 into m^3
and cm^3 into m^3

This is for Pv = nRT

To convert dm3 to m3 you *1000

to convert cm3 into m3 you *100^3 so (100*100*100)= so your *1,000,000

Reply 290

bad8oy

Original post by Madara

its tetrahedral so 109.5-2.5=107
it has 4 pairs of electrons therefore tetrahdral but has the shape of a triangular pyramid because of the 3 arms.

Reply 291

564767

My teacher printed this for me.

http://www.google.co.uk/imgres?q=VSEPR+geometries&um=1&hl=en&client=firefox-a&rls=org.mozilla:en-GB :redface:

fficial&channel=np&biw=1366&bih=665&tbm=isch&tbnid=RXro2BboPKeG7M:&imgrefurl=http://chemistry.ncssm.edu/labs/molgeom/index.html&docid=bpa8jp_qq7HnQM&imgurl=http://chemistry.ncssm.edu/labs/molgeom/vseprgeo.jpg&w=692&h=549&ei=ZWKxT6nnGoGw0QXa45CqCQ&zoom=1&iact=rc&dur=369&sig=112453562921027911141&page=1&tbnh=131&tbnw=165&start=0&ndsp=24&ved=1t:429,r:20,s:0,i:115&tx=117&ty=55

Reply 292

Madara

Original post by bad8oy

its tetrahedral so 109.5-2.5=107
it has 4 pairs of electrons therefore tetrahdral but has the shape of a triangular pyramid because of the 3 arms.

Ohh I see... thanks you helped clear some stuff up

Reply 293

AspiringDoctor

That's really useful - thank you! :biggrin:

Reply 294

564767

Original post by AspiringDoctor

That's really useful - thank you! :biggrin:

yeah, cram that table like hell.
try to remember it as, 4 bonding pairs 3 lone pairs/3 bonding pairs 1 lone pair etc.

Reply 295

W4L1Y

to convert dm^3 into m^ 3 isnt it X 10^-2

Reply 296

Voglie

Original post by Madara

To convert dm3 to m3 you *1000

to convert cm3 into m3 you *100^3 so (100*100*100)= so your *1,000,000

Stop confusing people :P.

I've replied to this question earlier in the thread.

cm^3 -> m^3 = DIVIDE BY 10^6 (or multiply by 10^-6)
dm^3 -> m^3 = DIVIDE BY 10^3 (or multiply by 10^-3)