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I was asked to " Use VSEPR theory to predict the most stable structure for $\ce{SF3Cl}$ and justify your answer. "

I managed to interpret the structure as seesaw and with a lone pair on the equatorial position. However, I had chosen that the most stable structure was that with Chlorine on the equatorial position because I thought that the Chlorine-Sulfur bond length would be greater than that of Chlorine-Fluorine.

However, the answer key stated the Fluorine was more electronegative, and therefore would be more stable on the axial (with chlorine as equatorial).

I was wondering if the answer key was accurate and if so, how would I compare the effects of electronegativity and bond length? I had considered electronegativity but I had thought it would simply make the bond even shorter, increasing repulsion.

Thanks!

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The bond lengths have little to do with the stability of the atom positioning around the central $\ce{S}$ atom. Rather, the relative sizes of the electron clouds are what should be considered when positioning these atoms. An atom of $\ce{Cl}$ has a 8 more electrons than does an atom of $\ce{F}$, so naturally its electron cloud will be larger and therefore more repulsive. The $\ce{Cl}$ should therefore be positioned in the equatorial position for the same reason you predicted that the lone pair would be present in the equatorial position--because these positions offer the least repulsion between the electron clouds of the atoms.

In the axial position, the 3 equatorial positions, present at 90 degree angles to the axial position, repel the electron cloud.

In an equatorial position, however, only the 2 axial positions are present at 90 degree angles to repel the electron cloud. The other 2 equatorial positions are positioned away from the equatorial bond in question, and repel its electron cloud only slightly. For this reason unbonded electron pairs and atoms with large electron clouds occupy these spaces first-- they offer the least repulsion and the most stable positioning.

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