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Why is it that for a molecule with trigonal bipyramidal geometry, the lone pair is located in the equatorial region whereas for a molecular with an octahedral geometry, the lone pair is located in the axial region?

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marked as duplicate by Mithoron, A.K., Mathew Mahindaratne, a-cyclohexane-molecule, Tyberius Aug 13 '18 at 2:24

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    $\begingroup$ Rotate your octahedral structure 90 degrees and you'll find that an "axial" region becomes "equatorial". All six vertices of the octahedron are, in fact, equal. So, there is no difference between the "axial" or "equatorial" positions. $\endgroup$ – orthocresol Aug 11 '18 at 10:38
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In trigonal biyramidal geometry (tbp), the lone pair of electron is located in equitorial position due to two reasons:- 1. The equitorial position has more s character than axial position. So in equitorial position, lone pair is much more attracted by nucleus (as s orbital is closer to nucleus), hence additional stabilization is gained. 2. You may use bent rule which states that lone pair prefers to stay in orbital where s character is more. As for octahedral geometry, all bonds are identical. So no equitorial or axial position.

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