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miscellaneous
Posted: Sun Nov 14, 2004 11:52 am
 
What is the structure of ICl4- ??

I drew it out and I got I in the middle with 4 Cl, one on top, bottom, left, right. I only come up with 32 electrons and I know i need 36. Any help is appreciated.
 
Bob
Posted: Sun Nov 14, 2004 2:51 pm
 
On 14 Nov 2004 11:29:35 -0600, chomebole@yahoo-dot-com.no-spam.invalid
(miscellaneous) wrote:

[quote:74cea34429]What is the structure of ICl4- ??

I drew it out and I got I in the middle with 4 Cl, one on top, bottom,
left, right. I only come up with 32 electrons and I know i need 36.
[/quote:74cea34429]
So add 4 more electrons, just as you calculated. That would be two
lone pairs on the central I.

Any time a halogen atom forms more than one bond, you will have an
expanded octet. You have encountered this (maybe for S, P)???

So there are six sets of electrons around the central I, which means
the electron geometry is octahedral. I'll leave the ion geometry to
you.

bob
 
Farooq
Posted: Sun Nov 14, 2004 3:07 pm
 
miscellaneous wrote:
[quote:3d561ca538]What is the structure of ICl4- ??

I drew it out and I got I in the middle with 4 Cl, one on top,
bottom,
left, right. I only come up with 32 electrons and I know i need 36.
Any help is appreciated.

[/quote:3d561ca538]
In ICl4-, there are four bond pairs in the valence shell of iodine as
you did correclty accounting for the 32 electrons, so what about the
two lone pairs (= 4 electrons) on iodine, since they have to be
accomodated into the expanded octet. The lone pairs occupy the axial
position, the four bond pairs in the same plane. What does this
indicate square planar, tetrahedral or octahedral stucture of ICl4(-)?
 
Killinchy
Posted: Sun Nov 14, 2004 4:47 pm
 
"miscellaneous" <chomebole@yahoo-dot-com.no-spam.invalid> wrote in message
news:419795ff$1_2@Usenet.com...
[quote:9ed37005df]What is the structure of ICl4- ??

I drew it out and I got I in the middle with 4 Cl, one on top, bottom,
left, right. I only come up with 32 electrons and I know i need 36.
Any help is appreciated.

*-----------------------*
Posted at:
www.GroupSrv.com
*-----------------------*
[/quote:9ed37005df]

Count the valence electrons (pretend that H, if present, has seven of them)

I 7 electrons
4Cl 28 electrons
minus 1 electron total 36 (as you say)

Here's the fiendishly cunning bit divide by 8 this gives 4
remainder 4 divide by 2 this gives 2
the
first four is the number of bonding pairs the two is the number of lone
pairs

Total of six pairs. Parent shape is the octahedron - actual shape (4
bonding and 2 lone) is the the square plane.

Let me give another example

26 valence electrons divide by 8 this
give 3 bonding pairs, remainder 2. Halve the remainder and get 1 lone
pair.

total of four pairs - parent shape is the tetrahedron - 3 bonding and 1 lone
pair gives a trig pyramid
 
 
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