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I am not able to understand the difference between structure of $\ce{SCl_{2}(OCH_{3})_{2}}$ and $\ce{SF_{2}(OCH_{3})_{2}}$.

I am already well versed with Molecular Orbital and VSEPR theories. I found the geometries, but don't understand why they are:

enter image description here

Why is the $\ce{F}$ at axial positions and $\ce{Cl}$ on the other hand in equatorial positions? (I am not able to understand Bent's Rule to be straight-forward, I have read Wikipedia already.)

A.K.
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Gaurav Agarwal
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    Related: Why does F replace the axial bond in PCl5? – orthocresol May 11 '17 at 18:57
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    As a general reminder: The involvement of d-orbitals into the bonding is one of the many shortcomings of VSEPR theory. The sp³d hybridisation simply is not evident (and it is not necessary) in the bonding of these molecules. I can't stress that enough. Textbooks which still rely on this explanation are outdated and desperately need revision. – Martin - マーチン May 12 '17 at 07:47
  • Ok, I think I understand it now..... Things became more clearer by Ron's answer to the question suggested above by Orthocresol (Thanks!) and I got to know that my textbook is outdated, cheers to Martin. – Gaurav Agarwal May 13 '17 at 11:17
  • @orthocresol I'm not sure that Bent's rule is straight forward to use here, and I am especially not sure, whether the given geometries are indeed correct. From my initial calculations, the lowest energy structure also has a co-linear arrangement of Cl-S-Cl. – Martin - マーチン May 19 '17 at 12:54

1 Answers1

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Check the bond angles. The larger the angles around a bond, the more S character. Electronegative elements prefer higher P character(Therefore less the bond angles). Axial position has lesser bond angles.Hence higher electronegative element takes axial position.( F>O>Cl)

This concept however may not be applicable to every compound.

user91439
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