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jerepierre
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There are 5 pi bonds: three C-C in the ring and two S-O pi bonds. Add the ring, and the IHD is 6.

Because of it's ability to exist as higher oxidation states than the second row elements, I don't believe that there is any way to include sulfur in the IHD formula unambiguously. Consider the following two compounds, toluenesulfonyl chloride and an isomer.

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The formulas are the same, but the IHD of the left structure is 6 but the right structure is 4.

I think the best way to handle third row (and beyond) is to treat the large atom as if it were it's second row analog (P->N, S->O, etc.), and use the result as the lower limit on the IHD.

jerepierre
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