The correct order of S-S bond length in the f | vedclass

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(A) To determine the $S-S$ bond length,we analyze the hybridization of the sulfur atoms in each oxyanion.$(I)$ In $S_2O_4^{2-}$ (dithionite),each sulf

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$I > II > III$

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(A) To determine the $S-S$ bond length,we analyze the hybridization of the sulfur atoms in each oxyanion.$(I)$ In $S_2O_4^{2-}$ (dithionite),each sulfur atom is $sp^3$ hybridized with one lone pair. The $S-S$ bond is formed by the overlap of $sp^3$ orbitals.$(II)$ In $S_2O_5^{2-}$ (disulfite),one sulfur is $sp^3$ hybridized (with a lone pair) and the other is $sp^3$ hybridized (without a lone pair). The $S-S$ bond involves an $sp^3-sp^3$ overlap.$(III)$ In $S_2O_6^{2-}$ (dithionate),both sulfur atoms are $sp^3$ hybridized without lone pairs.According to Bent's rule,the $s$-character in a hybrid orbital increases as the electronegativity of the attached atom decreases or as the number of lone pairs on the central atom increases. Higher $s$-character leads to shorter bond lengths. Conversely,higher $p$-character leads to longer bond lengths.In $S_2O_4^{2-}$,the sulfur atoms have more $p$-character in the $S-S$ bond due to the presence of lone pairs,resulting in the longest $S-S$ bond.In $S_2O_6^{2-}$,the sulfur atoms have more $s$-character in the $S-S$ bond,resulting in the shortest $S-S$ bond.Therefore,the order of $S-S$ bond length is $I > II > III$.

The correct order of $S-S$ bond length in the following oxyanions is:
$(I)$ $S_2O_4^{2-}$ $(II)$ $S_2O_5^{2-}$ $(III)$ $S_2O_6^{2-}$

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The correct order of $S-S$ bond length in the following oxyanions is:$(I)$ $S_2O_4^{2-}$ $(II)$ $S_2O_5^{2-}$ $(III)$ $S_2O_6^{2-}$