VSEPR - ICl3

[HumanElement]

ICl3. Correct answer is T-shape - not trigonal planar. Why?

Can someone explain why ICl3 takes shape of T rather than planar? I know it is a T because of the 2 pair of non-bonding on the side of the plane. Wouldn't it be better or more favorable position if the lone pairs are on top and bottom and the 3 Cl on the horizontal plane?

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[TeachME]

Well since there is only one set of lone pairs for Iodine, it can only be at the top or the bottom or left or right. Either way it still comes out to a T-shape no matter where you put the lone pairs. I guess it is the norm to put it at the top and then the bottom and so on and so forth. Just mentally rotate the molecule. Hope this helps.

 

[HumanElement]

Well since there is only one set of lone pairs for Iodine, it can only be at the top or the bottom or left or right. Either way it still comes out to a T-shape no matter where you put the lone pairs. I guess it is the norm to put it at the top and then the bottom and so on and so forth. Just mentally rotate the molecule. Hope this helps.

There are 2 lone pairs and 3 shared around the I.

For example in this pic. The lone pairs would be at the big ball and tiny ball space while the Cl would be at the other medium ball. The result would end in T-shape.

I was thinking why can't the Cl position at Big ball, tiny ball, and the medium ball that are in same horizonal plane and the lone pairs stay on top and on bottom. In the latter configuration, it would result in a plane. My bad, I was thinking of plane earlier instead of linear since there are 3 Cl. Wouldn't it be better this way since they are all equal space apart rather than uneven distribution in T-shape configuration?

 

[Maygyver]

ICl3. Correct answer is T-shape not linear. Why?

Can someone explain why ICl3 takes shape of T rather than linears? I know it is a T because of the 2 pair of non-bonding on the side of the plane. Wouldn't it be better or more favorable position if the lone pairs are on top and bottom and the 3 Cl on the horizontal plane?

It helped me to look at the diagram here http://en.wikipedia.org/wiki/VSEPR.

I think no matter where you take the electrons from it will result in a T-shape. I find the easiest way for me to do VSEPR is just to memorize that diagram because some of it is slightly counter intuitive. I hope this helps.

 

[Maygyver]

There are 2 lone pairs and 3 shared around the I.

For example in this pic. The lone pairs would be at the big ball and tiny ball space while the Cl would be at the other medium ball. The result would end in T-shape.
http://cpa.hkcampus.net/%7Ecpa-kmt/tbp.gif

I was thinking why can't the Cl position at Big ball, tiny ball, and the medium ball that are in same horizonal plane and the lone pairs stay on top and on bottom. In the latter configuration, it would result in a plane. My bad, I was thinking of plane earlier instead of linear since there are 3 Cl. Wouldn't it be better this way since they are all equal space apart rather than uneven distribution in T-shape configuration?

I actually had the same problem when I first learned VSPER. I forget the reasoning behind the T-shape. :-\ I'll try to look into it some more.

 

[HoangDDS]

ICl3. Correct answer is T-shape - not trigonal planar. Why?

Can someone explain why ICl3 takes shape of T rather than planar? I know it is a T because of the 2 pair of non-bonding on the side of the plane. Wouldn't it be better or more favorable position if the lone pairs are on top and bottom and the 3 Cl on the horizontal plane?

It wouldn't be trigonal planar because because of exactly what you've said. The two lone pair is gonna cause replusion. Plus trigonal planar is AX3. While ICl3 is AX3E2 which is T shaped. Just basically following the VSPR rules.

(X = surrounding atoms, E = lone pair)

It can't be have 3 Cl on the same plane with one lone pair on top and bottom. There will just be too much electron repulsions between the unpaired electrons of Cl and the I 's lone pairs.

 

[AdaAda87]

Why is it trigonal pyramidal?

Shouldn't the 3 clorines be on the bottom, and then the lone pair be on the top?