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According to Bent's rule, in $\mathrm{sp^3d}$ more electronegative element is placed on the axial position, so between $\ce{Cl}$ and $\ce{OCH3}$ in $\ce{SCl2(OCH3)2}$ which should be placed on axial as oxygen is more electronegative, but wouldn't $\ce{CH3}$ reduce its overall electronegativity due to its positive inductive effect?The electronegativity of Oxygen is more than that of chlorine but the negative inductive effect of OCH3 is less than that of Chlorine

But my teacher said that $\ce{OCH3}$ should still be placed on axial position considering electronegativity of oxygen.

Can someone please explain this? Because we are comparing a group's electronegativity (which cannot be found on the Pauling scale) to that of an atom.

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    $\begingroup$ What can be done to improve the quality of the question? $\endgroup$ – Schwarz Kugelblitz Aug 25 at 15:49
  • $\begingroup$ Related: Geometries of SCl2(OCH3)2 and SF2(OCH3)2 $\endgroup$ – andselisk Aug 25 at 16:19
  • $\begingroup$ This post does not explain why OCH3 is placed at axial...I have seen this post before and found it to be giving a conceptual understanding of Bent's rule at a basic level but my question is slightly different $\endgroup$ – Schwarz Kugelblitz Aug 25 at 16:22
  • $\begingroup$ EN of $\ce{O > Cl}$, but -I effect of $\ce{OCH_3^- < Cl^-}$. May I add this point in question? $\endgroup$ – rv7 Aug 25 at 17:23
  • $\begingroup$ Thanks for the suggestion have added that in question...am hoping for a proper answer to this now $\endgroup$ – Schwarz Kugelblitz Aug 27 at 7:11
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I had the same doubt as yours. Therefore, I talked with my teacher over this and they gave the following explanation:

Here, two factors are involved in determining the shape of molecule:

  • Electronegativity: As the −I effect of $\ce{OCH_3^-}$ is less than (but comparable to) $\ce{Cl^-}$. Hence, electronegativity becomes a minor factor here. From Wikipedia:

    $$\ce{-NH3+} > \ce{-NO2} > \ce{-SO2R} > \ce{-CN} > \ce{-SO3H} > \ce{-CHO} > \ce{-CO} > \ce{-COOH} > \ce{-F} > \ce{-COCl} > \ce{-CONH2} > \ce{-Cl} > \ce{-Br} > \ce{-I} > \ce{-OR} > \ce{-OH} > \ce{-NR2} > \ce{-NH2} > \ce{-C6H5} > \ce{-CH=CH2} > \ce{-H}$$

  • Steric: As we know, $\ce{Cl^-}$ is a bulky substituent than $\ce{OCH_3^-}$. Because, the three lone pairs (one 3s and two 3d) of $\ce{Cl^-}$ are more diffused in relative to two lone pairs (one 2s and one 2p) of $\ce{OCH_3^-}$.

Ultimately, our need is to minimize bp-lp repulsions, as stability drives geometry. Hence, we would place $\ce{OCH_3^-}$ at axial position.

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