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Is this the right way of working out bond pairs and lone pairs?
I'm doing some questions working out the shapes of molecules and angles and i'm wondering if this is the right method.
Say for XeF4, you get the electron shell structure (just normal, without the spdf thingy), which would be 2:8:18:26 (cos of the 2n^2 for each level)
Then for the outermost shell, you split it into spdf, so it'd be:
4s^2 4p^6 4d^10 4f^8
Then you get the last one, so 4f^8, and that means there's 8 electrons in the outermost shell? Sooooo when you draw the atom's outer shell, there's 8 electrons.
But then for F, how come if you work out the outermost electrons, it's like 2:7:
2s^2 2p^5, which is wrong, because there's not 5 in the outermost shell and that would be wrong when you drew the atom?
There must be 7 in F, cos I know when you draw XeF4, each F atom only needs one more to be complete, and you end up with 4 bond pairs and 2 lone pairs.
So for F, you just take the number at the end of the 2:7, but when I worked out Xe, I had to write out the whole spdf thing. So my method's obviously wrong, but I don't get it.
Please help.
Say for XeF4, you get the electron shell structure (just normal, without the spdf thingy), which would be 2:8:18:26 (cos of the 2n^2 for each level)
Then for the outermost shell, you split it into spdf, so it'd be:
4s^2 4p^6 4d^10 4f^8
Then you get the last one, so 4f^8, and that means there's 8 electrons in the outermost shell? Sooooo when you draw the atom's outer shell, there's 8 electrons.
But then for F, how come if you work out the outermost electrons, it's like 2:7:
2s^2 2p^5, which is wrong, because there's not 5 in the outermost shell and that would be wrong when you drew the atom?
There must be 7 in F, cos I know when you draw XeF4, each F atom only needs one more to be complete, and you end up with 4 bond pairs and 2 lone pairs.
So for F, you just take the number at the end of the 2:7, but when I worked out Xe, I had to write out the whole spdf thing. So my method's obviously wrong, but I don't get it.
Please help.
Scroll to see replies
http://en.wikipedia.org/wiki/Aufbau_principle
i hope this helps
i hope this helps
DaveJ
Why are you doing s p d f subshells? Surely you just need to know the number of electrons in the outermost shell, ie, the group number? Or have I just not covered the proper way of doing these yet?
But then that means there's 26 outer electrons in Xe, in the 4th shell, which is ridiculous cos it would make a million lone pairs, which is untrue cos there's only 2 in XeF4.
DaveJ
Ignore this, I was wrong. 
Oh lol okay, but how do you work it out?
We got told to do it like this:
Xe is a noble gas which is effectively group 8, so it's outermost level has 8 electrons.
There are 4 Fs so we do (1 x 4)= 4
The total number of electrons= 8+4= 12
The number of electron pairs= 12/2=6
The number of bonding pairs would be 4 because it is XeF4 and there would be 2 lone pairs.
Therefore the shape is based on octahedral but it would be a square planer I think?
Xe is a noble gas which is effectively group 8, so it's outermost level has 8 electrons.
There are 4 Fs so we do (1 x 4)= 4
The total number of electrons= 8+4= 12
The number of electron pairs= 12/2=6
The number of bonding pairs would be 4 because it is XeF4 and there would be 2 lone pairs.
Therefore the shape is based on octahedral but it would be a square planer I think?
when n=4, 2n^2 = 32 btw
piece_by_piece
We got told to do it like this:
Xe is a noble gas which is effectively group 8, so it's outermost level has 8 electrons.
There are 4 Fs so we do (1 x 4)= 4
The total number of electrons= 8+4= 12
The number of electron pairs= 12/2=6
The number of bonding pairs would be 4 because it is XeF4 and there would be 2 lone pairs.
Therefore the shape is based on octahedral but it would be a square planer I think?
Xe is a noble gas which is effectively group 8, so it's outermost level has 8 electrons.
There are 4 Fs so we do (1 x 4)= 4
The total number of electrons= 8+4= 12
The number of electron pairs= 12/2=6
The number of bonding pairs would be 4 because it is XeF4 and there would be 2 lone pairs.
Therefore the shape is based on octahedral but it would be a square planer I think?
it's square planarpiece_by_piece
We got told to do it like this:
Xe is a noble gas which is effectively group 8, so it's outermost level has 8 electrons.
There are 4 Fs so we do (1 x 4)= 4
The total number of electrons= 8+4= 12
The number of electron pairs= 12/2=6
The number of bonding pairs would be 4 because it is XeF4 and there would be 2 lone pairs.
Therefore the shape is based on octahedral but it would be a square planer I think?
Xe is a noble gas which is effectively group 8, so it's outermost level has 8 electrons.
There are 4 Fs so we do (1 x 4)= 4
The total number of electrons= 8+4= 12
The number of electron pairs= 12/2=6
The number of bonding pairs would be 4 because it is XeF4 and there would be 2 lone pairs.
Therefore the shape is based on octahedral but it would be a square planer I think?
Ahhh fantastic, so for ANY atom, you can just look in the periodic table, and whichever group its in is however many electrons in the outer shell? And that works for anything?
I don't get the rest of your method, but i've got the one I was taught involving drawings so that's ok.
BrightGirl
Oh lol okay, but how do you work it out?
Well I didn't bother with all the subshell stuff as I thought it was irrelevant in this question.
I did each F contributes one electon, and Xe adds one electon to each of those to form 4 bonds. This leaves Xe with 4 electrons left, which form 2 lone pairs.
BrightGirl
Ahhh fantastic, so for ANY atom, you can just look in the periodic table, and whichever group its in is however many electrons in the outer shell? And that works for anything?
I don't get the rest of your method, but i've got the one I was taught involving drawings so that's ok.
I don't get the rest of your method, but i've got the one I was taught involving drawings so that's ok.
Yeah, I think so!
BrightGirl
Ahhh fantastic, so for ANY atom, you can just look in the periodic table, and whichever group its in is however many electrons in the outer shell? And that works for anything?
I don't get the rest of your method, but i've got the one I was taught involving drawings so that's ok.
I don't get the rest of your method, but i've got the one I was taught involving drawings so that's ok.
Yeah, that's what I was saying, but then assumed I was wrong cos I thought I just hadn't covered your method in class.
piece_by_piece
Yeah, I think so!
DaveJ
Yeah, that's what I was saying, but then assumed I was wrong cos I thought I just hadn't covered your method in class.
Wooo thank you! I totally get it now! So when you're drawing atoms and only their outer shells, if it's in group 4, it has 4, and if it's in group 3, it has 3!
Wait hang on a minute.
For SnCl2, Sn has 4 in its outer shell, and Cl has 7. So there's 1 lone pair and 2 bond pairs, which can't be right cos in my table, we've only done 3 scenarios (3 bond pairs + 1 lone, 2 bond pairs + 2 lone, 4 bond pairs + 2 lone)
For SnCl2, Sn has 4 in its outer shell, and Cl has 7. So there's 1 lone pair and 2 bond pairs, which can't be right cos in my table, we've only done 3 scenarios (3 bond pairs + 1 lone, 2 bond pairs + 2 lone, 4 bond pairs + 2 lone)
DaveJ
Yeah, I'm confused about how you said Xe would have 26 outer electrons though.
Oh cos the first shell holds 2, the second 8, the third 18 and the fourth 32, and in the fourth, Xe only has 26.
BrightGirl
Wait hang on a minute.
For SnCl2, Sn has 4 in its outer shell, and Cl has 7. So there's 1 lone pair and 2 bond pairs, which can't be right cos in my table, we've only done 3 scenarios (3 bond pairs + 1 lone, 2 bond pairs + 2 lone, 4 bond pairs + 2 lone)
For SnCl2, Sn has 4 in its outer shell, and Cl has 7. So there's 1 lone pair and 2 bond pairs, which can't be right cos in my table, we've only done 3 scenarios (3 bond pairs + 1 lone, 2 bond pairs + 2 lone, 4 bond pairs + 2 lone)
I think that scenario can occur.
BrightGirl
Wait hang on a minute.
For SnCl2, Sn has 4 in its outer shell, and Cl has 7. So there's 1 lone pair and 2 bond pairs, which can't be right cos in my table, we've only done 3 scenarios (3 bond pairs + 1 lone, 2 bond pairs + 2 lone, 4 bond pairs + 2 lone)
For SnCl2, Sn has 4 in its outer shell, and Cl has 7. So there's 1 lone pair and 2 bond pairs, which can't be right cos in my table, we've only done 3 scenarios (3 bond pairs + 1 lone, 2 bond pairs + 2 lone, 4 bond pairs + 2 lone)
Sn has 4 outer electrons
Cl donates 1 x 2 electrons= 2
The total number of electrons= 4+2=6
The number of electron pairs= 6/2=3
The number of bonding pairs is 2, and there is 1 lone pair.
The shape is based on trigonal planer, but is bent linear I think?
Perhaps the best way to think about VESPR is that it only deals with s and p subshells (and so you won't be asked questions on transisiton metal compound shapes because of the d subshells). This is why looking at the (old) group number will give you the outer electrons, so for Xe it has 2 s and 6 p electrons - eight 'outer electrons'
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