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I am not sure of the explanation for the following statement:

The reducing property of dioxides of group 16 elements decreases from $\ce{SO2}$ to $\ce{TeO2}$.

  • Is the statement valid? If yes, how do I explain it theoretically?
  • If not, what would be the actual trend in that case?
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The statement is actually valid.If you go from S to Te, the compounds of Se and Te are stable at lower oxidation states due to inert pair effect.
So, the tendency of these group 16 elements to be in higher oxidation states decreases down the group.That means, tendency of the elements getting oxidised decreases down the group and thus their reducing property(reducing others, getting oxidised itself) decreases frm S to Te.

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  • $\begingroup$ So this is the trend for all types of compounds formed by these elements? I mean, whether it be hydrides, oxides, halides etc. The reducing property always decreases from S to Te ? $\endgroup$ – Mayur Prasanna Feb 9 '18 at 10:31
  • $\begingroup$ You don't have chalcogen(IV) hydrides. Otherwise yes, the +4 oxidation state becomes increasingly stable vis a vis +6 as you go from sulfur to heavier congeners. And sulfur itself tends to favor +4 at higher temperatures; sulfate salts tend to decompose forming sulfur dioxide upon heating. $\endgroup$ – Oscar Lanzi Feb 9 '18 at 12:12

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