Identifying Fluoride Molecules: Shapes & Formulas

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Discussion Overview

The discussion revolves around identifying fluoride molecules based on their geometric shapes, specifically octahedral, trigonal bipyramidal, and T-shaped configurations. Participants explore systematic approaches to determine molecular structures rather than relying solely on trial and error.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses a desire for a systematic or mathematical method to identify molecular shapes, noting their familiarity with VSEPR structures.
  • Another participant suggests that there may not be a simple mathematical approach to determine which geometry yields a more stable molecule.
  • A different participant proposes a systematic method involving Lewis structures, accounting for lone pairs and ligands to assign molecular shapes.
  • It is noted that an octahedral molecule requires six bond pairs, leading to the conclusion that six fluorine atoms are needed, with sulfur suggested as a suitable central atom.
  • For trigonal bipyramidal structures, it is mentioned that five bond pairs are necessary, and phosphorus is identified as a suitable central atom due to its valency.
  • In discussing T-shaped molecules, it is explained that they consist of five electron pairs, with three bond pairs and two lone pairs, suggesting chlorine or iodine as potential central atoms.
  • Another participant shares a memorization strategy for determining molecular shapes based on the number of bond pairs and lone pairs.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a singular systematic method for identifying molecular shapes, and multiple approaches are presented, indicating a lack of agreement on the best strategy.

Contextual Notes

Some assumptions regarding the availability of d orbitals for central atoms are mentioned, as well as the dependency on the specific elements chosen for central atoms, which may affect the applicability of the proposed methods.

neoking77
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Homework Statement


Give the formula of a fluoride molecule that:
is octahedral in shape
is trigonal bypyramidal in shape
is T-shaped
...etc

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
 
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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.
 
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.
 
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
 
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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