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when we make a Valence shell electron pair repulsion theory based structure for an molecule after calculating its hybridisation ; when we have to decide the positions of the ions we put them in such positions so that repulsion is minimized .It follows the basic rule of lone pair-lone pair repulsion> lone pair-bond pair repulsion >bond pair -bond pair repulsion . So when it comes to deciding the position of lone pairs of electrons in ClF3 why dont we put them on the axial positions so that they have an angle of 180 deg between them as compared to 120 deg that they currently have ?enter image description here

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marked as duplicate by bon, ron, Jan, Todd Minehardt, Loong Dec 12 '15 at 22:34

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  • $\begingroup$ I think your diagram is a bit wrong, isn't it supposed to be a Bent-T shape? Also if the lone pairs occupy axial positions then there will be very high 6*(LP-BP) repulsion's as compared to earlier very high 2*(LP-BP) repulsion's. $\endgroup$ – Sujith Sizon Dec 12 '15 at 17:33
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    $\begingroup$ Dupe of this question. Also, Bent's rule...again. $\endgroup$ – bon Dec 12 '15 at 18:15
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Let's take a look at the electronic configuration of $\ce{ClF3}$.

enter image description here

For $\ce{ClF3}$ actually three isomers are possible.

enter image description here
enter image description here
enter image description here

It is highly unfavorable for the lone pairs to be 90° apart which rules one of the three. In order to rationalize the obervation that the isomer with both lone pairs in equatorial positions is the observed form, it is necessary to count the number of 90° bond pair - lone pair interactions. The observed isomer has four such angles and the isomer with the lone pairs in axial positions has six which is (presumably) less favorable. The molecule might be called T-shaped based on the atomic positions only.

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  • $\begingroup$ Here are the links to the other two possible structures i.stack.imgur.com/YpQef.gif i.stack.imgur.com/quW3e.gif $\endgroup$ – user23446 Dec 12 '15 at 18:31
  • $\begingroup$ Good answer, though poor picture clarity, btw why aren't you adding the other two structure to your answer post? $\endgroup$ – Sujith Sizon Dec 12 '15 at 18:34
  • $\begingroup$ @SujithSizon i'm unable to add more than two images on my answer. $\endgroup$ – user23446 Dec 12 '15 at 18:35
  • $\begingroup$ @VMiranda It's also important to consider the bonding that the various geometries imply. In the top molecule, is that a p orbital holding 4 electrons? If so, that's not possible - you can put at most 2 electrons in a single orbital. In your middle drawing, how many $\ce{sp^2}$ orbitals are there in that molecule? You can't have more than 3. In your bottom isomer, what kind of bonds are those involving the axial fluorines? The correct answer is as much about bonding (specifically hypercoordinate bonding, see my comment above for a relevant link) as electron repulsion. $\endgroup$ – ron Dec 12 '15 at 21:42
  • $\begingroup$ Invoking d-orbitals in main group elements gives you $-1$ from me. $\endgroup$ – Jan Dec 12 '15 at 21:52

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