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So basically according to university textbooks, it is said that the second lone pair in a trigonal bypyramidal arrangement would also reside on the trigonal plane, causing the two lone pairs to be beside each other and compressing the other 3 bonded pairs (resulting in T-shape).

However, my question is: Why doesn't the two lone pairs move to the two axial groups so that the the three bonded pairs reside on the trigonal plane? Surely this would result in the less repulsion within the arrangement? Wouldn't the repulsion between a lone pair and 3 bonded pairs at 90 degrees be less than the repulsion between a lone pair and another lone pair & 3 bonded pairs?

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marked as duplicate by Geoff Hutchison, Jan, Todd Minehardt, Curt F., orthocresol Dec 13 '15 at 5:49

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  • $\begingroup$ chemistry.stackexchange.com/questions/9623/… $\endgroup$ – Mithoron Oct 12 '15 at 22:29
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    $\begingroup$ The 90° interactions are the most significant (highest energy). How many lone pair - lone pair, lone pair - bonding pair and bonding pair - bonding pair 90° interactions does each of the two possible structures have? $\endgroup$ – ron Oct 12 '15 at 22:42
  • $\begingroup$ For the one accepted in the text book there is one lone pair - lone pair and 3 lone pair - bonded pair. And for the I propose there are just 3 lone pair - bonded pair. However, the difference is that mine is three 90 degree interactions whereas the ones in textbook are unstated $\endgroup$ – Jack Tan Oct 12 '15 at 23:31
  • $\begingroup$ If both lone pairs are in the equatorial plane, then there are no 90° lone pair - lone pair 90° interactions; your count of the 90° lone pair - bonded pair interactions for this structure is also incorrect (should be 4). $\endgroup$ – ron Oct 13 '15 at 0:01
  • $\begingroup$ Yes, the 90° interactions are the most important. It's also worth pointing out that the differences in energy are small, and a Berry pseudorotation interconverts axial and equatorial in these species. $\endgroup$ – Geoff Hutchison Dec 12 '15 at 21:02

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