|Oct11-07, 11:58 PM||#1|
1. The problem statement, all variables and given/known data
Give the formula of a fluoride molecule that:
is octahedral in shape
is trigonal bypyramidal in shape
3. The attempt at a solution
Ok, I'm not really actually looking for an answer, because I already know the answers are SF6, PF5, and ClF3. And I have no trouble understanding VSEPR structures. What I'm wondering is if there is an systematical/mathematical way to approach this problem rather than trial and error like I did?
Thank you very much
|Oct12-07, 04:15 AM||#2|
I don't think there's any simple yet mathematically rigorous way of finding which geometry yields the more stable molecule , at least not that I know of.
|Oct12-07, 09:16 AM||#3|
This isn't a mathematical treatment, but it is systematic. Write out the lewis structures and account for the lone pairs and the ligands. Assume the lone pairs are ligands and assign the structure.
|Oct13-07, 07:12 AM||#4|
an octahedral molecule will have 6 electron pairs around the central atom. And all of them will be bond pairs. Now, there will have to be 6 Fluorine atoms. You should look for a central atom which can exhibit valency 6. Group 6 elements, such as sulphur. The group 6 element should also have vacant and accessible d orbitals. Oxygen will not fit.
Trigonal bipyramidal molecules have 5 electron pairs. All of them are bond pairs. If one of them were to be lone pairs, the shape would not be trigonal bipyramidal. Following the same logic as above, there should be 5 fluorine atoms. You should search for a Group 5 element, having vacant and accessible d orbitals(not nitrogen). Phosphorous is ok.
T shaped molecules have 5 electron pairs, since this is the only combination, where 3 are bond pairs and 2 are lone pairs. It resembles the trigonal bipyramidal molecule. The two lone pairs will minimise their repulsions by being at 120 degrees mutually, and be on the flat trigonal part. There will be 3 fluorine atoms across the three bond pairs.
Hence the central atom should have 7 outermost electrons (3 for bond pairs and 4 for the 2 lone pairs). A group 7 element, will be ok, such as iodine or chlorine.
this is the best way i found to draw structures. First determine the number of B.P. and L.P., and then find the central atom.
hope it helps
|Oct15-07, 07:54 PM||#5|
If the central atom of the molecule has 2 bond pairs and 0 lone pairs, then it will be linear. If it has 2 bond pairs and 2 lone pairs, then it will be v-shaped. If it has 3 bond pairs and 1 lone pair then it will be pyramidal. If it has 3 bond pairs and 0 lone pairs then it will be trigonal planar. If it has 3 lone pairs and 1 bond pair then it will be linear. I just memorize these.
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