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Gen chem question (expanded octets)
- Thread starter alanan84
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very good question 🙂
I don't know how we know it.
But with my experience, I just know how to draw XeF4. I know that this is not a good answer.
I think there are list of 'unusual' atom that forms more than octet.
I think Xe is one of them and then I think S or Se is another unusual exception.
http://www.saskschools.ca/curr_content/chem20/covmolec/movies/expand.html
This link teaches about expanded octets.
how about this.
Just write XeF4. total valance electrons are 36
Draw Xe surroudned by 4F with single bond = 36-8 = 28
Then make F octet = 28-24 = 4
put these 4 around Xe.
Usually, if we don't have enough electrons to satisfy every thing octet, we move 2 electrons and make it into double bond thus two atoms can share them and overcome the problem. But in this case, we made both Xe and F octet and still have extras, so I think the only thing we can do is putting 4 electrons to the Xe.
Try this one, PCl5.
total valence is 40
P surrounded by 5 Cl and linked with 1 bond each = 30 left
make Cl octet = used all~ 🙂
now look at P, it's more than octet~
I don't know how we know it.
But with my experience, I just know how to draw XeF4. I know that this is not a good answer.
I think there are list of 'unusual' atom that forms more than octet.
I think Xe is one of them and then I think S or Se is another unusual exception.
http://www.saskschools.ca/curr_content/chem20/covmolec/movies/expand.html
This link teaches about expanded octets.
how about this.
Just write XeF4. total valance electrons are 36
Draw Xe surroudned by 4F with single bond = 36-8 = 28
Then make F octet = 28-24 = 4
put these 4 around Xe.
Usually, if we don't have enough electrons to satisfy every thing octet, we move 2 electrons and make it into double bond thus two atoms can share them and overcome the problem. But in this case, we made both Xe and F octet and still have extras, so I think the only thing we can do is putting 4 electrons to the Xe.
Try this one, PCl5.
total valence is 40
P surrounded by 5 Cl and linked with 1 bond each = 30 left
make Cl octet = used all~ 🙂
now look at P, it's more than octet~
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Joonkimdds is right. Just draw the lewis structure, and from there it's pretty basic. Just know what shapes are formed with 3, 4, 5, or 6 electron groups, and realize that lone pairs take more space than bonding pairs.
Joonkimdds, I think u know it well😉. We r waiting for ur DAT score😀
So as joonkimdds said it:
In general:
1. Sum the valence electrons from all atoms. For an anion, add an electron. For a cation, subtract an electron for each +. (here is 36)
2. Connect all surrounded atoms to the central atom with a single bond.(Central atom is generally less electronegative than the atoms surrounding it). Then subtract all bonds from the total valence electrons (36-8=28)
3. Complete the octets of the atoms bonded to the central atom.(28-24=4)
4. Place any leftover electrons on the central atom. ( third period and beyond), therefore 4 on Xe
5. If there are not enough electrons to give the central atom an octet, make them double or triple.(here not needed), But sometimes even octet are satisfied for all atoms, but you still need to try multiple bonds to reduce the separated formal charges on atoms.
The electron-domain geometry of XeF4 is octadehral but here asked for the MOLECULAR geometry, therefore square planar.
So as joonkimdds said it:
In general:
1. Sum the valence electrons from all atoms. For an anion, add an electron. For a cation, subtract an electron for each +. (here is 36)
2. Connect all surrounded atoms to the central atom with a single bond.(Central atom is generally less electronegative than the atoms surrounding it). Then subtract all bonds from the total valence electrons (36-8=28)
3. Complete the octets of the atoms bonded to the central atom.(28-24=4)
4. Place any leftover electrons on the central atom. ( third period and beyond), therefore 4 on Xe
5. If there are not enough electrons to give the central atom an octet, make them double or triple.(here not needed), But sometimes even octet are satisfied for all atoms, but you still need to try multiple bonds to reduce the separated formal charges on atoms.
The electron-domain geometry of XeF4 is octadehral but here asked for the MOLECULAR geometry, therefore square planar.
Number of electron groups / Geometric Family(no lone pair) / if 1 lone pair/ 2 lone pair.
2 / Linear
3 / Trigonal planar / Bent
4 /tetrahedral/ Trigonal pyramid / Bent.
5 /Trigonal bipyramid/ See-saw / T-shaped.
6 / Octahedral/ Square pyramid/ Square planar.
I got it from princeton book. so once you determine the bond you ll figure out the geometric family then see how many lone pairs and you ll have overall shape.
2 / Linear
3 / Trigonal planar / Bent
4 /tetrahedral/ Trigonal pyramid / Bent.
5 /Trigonal bipyramid/ See-saw / T-shaped.
6 / Octahedral/ Square pyramid/ Square planar.
I got it from princeton book. so once you determine the bond you ll figure out the geometric family then see how many lone pairs and you ll have overall shape.
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Thank you. This should be very helpful because I struggled with VSEPR. 👍
I see wikipedia also could be beneficial for some visualization http://en.wikipedia.org/wiki/VSEPR_theory#AXE_Method
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ahahahahahahahahaha.
Just wait a little longer~ it will be over in about 2 months.
