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I'm comparing sulfur tetrafluoride and thionyl tetrafluoride. The lone pair has more s character compared to a double bond. Therefore, sulfur in $\ce{SF4}$ should have a greater s character compared to sulfur in $\ce{SOF4}$.

Since $\ce{SF4}$ has a greater s character it's bond length would be comparativly smaller than that of $\ce{SOF4}$.

Therefore, $\ce{X}>\ce{Y}$. But my book says otherwise. Did I go wrong anywhere?

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The simplest way to compare these two compounds is to compare sulfur’s oxidation state. In $\ce{SF4}$, sulfur is in the $(\mathrm{+IV})$ state. In $\ce{SOF4}$, it is $(\mathrm{+VI})$. The higher a compound’s oxidation state, the higher its formal positive charge and thus the more stable (i.e. lower in energy) its orbitals are. Orbitals reduced in energy means contracted orbitals which in turn leads to shorter bonds.

Other than that, the two compounds are pretty identical. Both feature two two-electron-two-centre $\ce{S-F}$ single bonds and one four-electron-three-centre $\ce{F\bond{...}S\bond{...}F}$ bond; the latter can be understood with the following resonance depiction:

$$\ce{F^-\ \ S^+-F <-> F-S^+\ \ F-}$$

Even the structures are comparable although the angles are slightly different.

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  • $\begingroup$ explain this please: higher its formal positive charge and thus the more stable (i.e. lower in energy) its orbitals are $\endgroup$ – Prakhar Nov 26 '16 at 5:08
  • $\begingroup$ @PrakharSankrityayan Orbitals have an orbital energy. A lower energy is called ‘more stable’. Electrons, which populate orbitals, are negatively charged. If the atom is partially positively charged, electrons are better attracted to it. Hence, their orbitals are lower in energy. $\endgroup$ – Jan Nov 26 '16 at 22:16

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