From here and here I found that the bond angle between the equatorial $\ce{F}$'s in $\ce{SF4}$ is less than that in $\ce{SOF4}$. I understand that is because of greater repulsion by lone pair than by double bond.

But the angle between the axial $\ce{F}$'s in $\ce{SF4}$ is more than that in $\ce{SOF4}$. I am not able to understand why is this so. Shouldn't the lone pair repel more than double bond here too?

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    $\begingroup$ I'd say the effects are too subtle to be explained that easily. $\endgroup$ – Ivan Neretin Feb 6 '19 at 5:37
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    $\begingroup$ You are trying to compare 2 different geometries: one with 4 vertices and the other with 5! Of course angles won't match $\endgroup$ – SteffX Feb 6 '19 at 14:44
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    $\begingroup$ @SteffX Try counting again and don't forget about lone pair. $\endgroup$ – andselisk Feb 6 '19 at 15:05
  • $\begingroup$ I suspect that some subtle molecular orbital effects are going on here. Oxygen has two pairs of pi-symmetry electrons to donate and is tending to pull the fluorines towards a symmetric structure to optimize that interaction. $\endgroup$ – Oscar Lanzi Feb 6 '19 at 15:25

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