The difference in the oxidation numbers of th | vedclass

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(A) The structure of $Na_2S_4O_6$ is shown below:$Na^+O^--S(=O)_2-S-S-S(=O)_2-O^-Na^+$In $Na_2S_4O_6$ (sodium tetrathionate),there are two types of su

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$5$

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(A) The structure of $Na_2S_4O_6$ is shown below:$Na^+O^--S(=O)_2-S-S-S(=O)_2-O^-Na^+$In $Na_2S_4O_6$ (sodium tetrathionate),there are two types of sulphur atoms:$1$. The two terminal sulphur atoms are bonded to three oxygen atoms (two via double bonds and one via a single bond) and one sulphur atom. Their oxidation number is calculated as $+5$.$2$. The two central sulphur atoms are bonded only to two other sulphur atoms. Since the electronegativity of identical atoms is the same,their oxidation number is $0$.The difference in the oxidation numbers of the two types of sulphur atoms is $|5 - 0| = 5$.

The difference in the oxidation numbers of the two types of sulphur atoms in $Na_2S_4O_6$ is

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