What are the respective oxidation states of s | vedclass

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(B) In the tetrathionate ion $(S_4O_6^{2-})$,the structure consists of a chain of four sulfur atoms. For the terminal sulfur atoms (numbered $1$ and $

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$+5, 0, 0, +5$

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(B) In the tetrathionate ion $(S_4O_6^{2-})$,the structure consists of a chain of four sulfur atoms. For the terminal sulfur atoms (numbered $1$ and $4$): Each is bonded to three oxygen atoms (two double-bonded and one single-bonded) and one sulfur atom. The oxidation state is calculated as: $x + 2(-2) + 1(-2) + 1(-1) = -1$ (considering the charge contribution). More simply,using electronegativity: $S$ bonded to $O$ gains $+2$ for each double bond and $+1$ for the single bond,totaling $+5$. For the central sulfur atoms (numbered $2$ and $3$): Each is bonded to two other sulfur atoms. Since the atoms are identical,the bond between them contributes $0$ to the oxidation state. Thus,the oxidation state of the central sulfur atoms is $0$. Therefore,the oxidation states for sulfur atoms $1, 2, 3, 4$ are $+5, 0, 0, +5$ respectively.

What are the respective oxidation states of sulphur atoms numbered $1$ to $4$ in the tetrathionate ion $(S_4O_6^{2-})$ shown below?

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