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 ?
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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 SizonCommented Dec 12, 2015 at 17:33
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1$\begingroup$ Dupe of this question. Also, Bent's rule...again. $\endgroup$– bonCommented Dec 12, 2015 at 18:15
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1 Answer
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Let's take a look at the electronic configuration of $\ce{ClF3}$.
For $\ce{ClF3}$ actually three isomers are possible.
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.sstatic.net/YpQef.gif i.sstatic.net/quW3e.gif $\endgroup$– user23446Commented Dec 12, 2015 at 18:31
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$\begingroup$ Good answer, though poor picture clarity, btw why aren't you adding the other two structure to your answer post? $\endgroup$ Commented Dec 12, 2015 at 18:34
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$\begingroup$ @SujithSizon i'm unable to add more than two images on my answer. $\endgroup$– user23446Commented Dec 12, 2015 at 18:35
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2$\begingroup$ Invoking d-orbitals in main group elements gives you $-1$ from me. $\endgroup$– JanCommented Dec 12, 2015 at 21:52
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2$\begingroup$ Re editing: I think you can edit answers to closed questions. At least I can. Re d-orbitals: They are too far removed energetically from the others (higher than the next-level s-orbital in ground state!) to reasonably take part in bonding for main-group elements. $\endgroup$– JanCommented Dec 13, 2015 at 19:48