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Why do lone pairs not reside in equatorial positions in species with octahedral geometry?

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    $\begingroup$ In octahedral geometry, all positions should be equivalent (before putting a bond or a lone pair there). It is by convention that the lone pairs are put on "axial" positions. However, they will definitely be across each other, consider XeF4. $\endgroup$ – TAR86 Feb 1 '17 at 15:26
  • $\begingroup$ @TAR86 I don't understand what you mean by "equivalent". Do you mean all positions have the same bond angle to each other? $\endgroup$ – b3nj4m1n Feb 1 '17 at 15:37
  • $\begingroup$ Can you give an example? $\endgroup$ – bon Feb 1 '17 at 15:41
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    $\begingroup$ @b3nj4m1n Consider a cube (such as dice). Octahedral geometry means that the central atom is in the center of the cube and the surrounding atoms are on the center of the faces of the cube. Thus, due to the high symmetry of the octahedral geometry, they are all equivalent. If this explanation does not help, you need to build a model. $\endgroup$ – TAR86 Feb 1 '17 at 17:41
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The main reason is repulsion with the electron pairs. Compare $\ce{SF6}$ & $\ce{SF4^2-}$ as an example. To get from $\ce{SF6}$ to $\ce{SF4^2-}$, we need to replace two fluorines with lone pairs. The first replacement is arbitrary; in an octahedral geometry, all the positions are equivalent. Once we have replaced that first fluorine however, the bond positions are no longer equivalent. We will have four equatorial bonds (90 degree separation from the lone pair) and one axial bond (180 degree separation from the lone pair). Lone pairs are very repulsive, especially with respect to other lone pairs, so we want the next lone pair to be as separated as possible from the first. The axial bond is further from the lone pair then any of the equatorial bonds, so it makes sense for the lone pair to go there.

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simply because in octahedral geometry both the equatorial position or base and the axial-equatorial position maintain same bond angle which is 90deg, so at first you can chose any position to place the first lone pair, but for the next electron pair the remaining positions are no more identical in the view of the bond angle.Cause if u chose eq. position for the second one, if it's a lonepair too then according to VSEPR theory you haven't made the right choice, cause you are still having an option to reduce the greatest replusion (LP-LP) between electron pairs, that is the axial position.So that angular distance between the LPs become 180. If the explanation above for octahedral is clear to you than you yourself can find why it is vice versa for a tetrahedron.:)

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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. This doesn't make much sense, first you state that all positions are equal, then they are not. You could put two lone pairs in equatorial positions and maintain 180° (If measuring angles between lone pairs would be possible.) It would also be better, if you avoid abbreviations and do not use emoticons. $\endgroup$ – Martin - マーチン Jun 7 '18 at 11:29
  • $\begingroup$ In case of two lone pairs in a tetrahedron it creates a dilemma for me too. But for 1 or more than 2 electron pairs it makes sense. And for the first electron pair in an octahedral shape all the positions are equal, but not for the next one. I don't see any problem here. $\endgroup$ – Deehan Jun 8 '18 at 11:53
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The VSEPR theory tells us that the order of repulsion is :
lone pair-lone pair >lone pair-bond pair >bond pair-bond pair.
It is clear that you need to place the lone pairs such that they have minimum neighboring bond pairs or lone pairs. An equatorial position in an octahedral geometry has 5 neighboring bond pairs/lone pairs ( 3 in the equatorial plane and two in the axial plane ) while an axial position has 4 neighboring bond pairs/lone pairs (4 in the equatorial plane, the other axial position is not a neighboring position).
That is the reason why lone pairs first occupy axial positions in an octahedral geometry. Hope this helps.

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  • $\begingroup$ Are we talking about the same thing? en.wikipedia.org/wiki/VSEPR_theory see SF6. All F's are equivalent. $\endgroup$ – TAR86 Feb 1 '17 at 18:13
  • $\begingroup$ wait @TAR86 so are all the positions equal or not? I'm confused now $\endgroup$ – b3nj4m1n Feb 2 '17 at 1:50
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    $\begingroup$ @b3nj4m1n I am utterly convinced that all positions of the surrounding atoms in an octahedral geometry are equal. This is due to the symmetry of the octahedron. $\endgroup$ – TAR86 Feb 2 '17 at 6:07
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    $\begingroup$ See, one lone pair isn't any issue, because all six sites are equivalent, as @TAR86 points out. OP is asking about why two lone pairs will go trans to each other. The use of "axial" or "equatorial" is highly misleading in my opinion. $\endgroup$ – orthocresol Feb 8 '17 at 1:47

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