Oxidation state of 'S' in sodium tetrathionate is :

+ 5, 0, 0, + 5

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+ 6, + 6, 0

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+ 4, + 2, + 5

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+ 2, 0, + 2, + 3

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Solution

The correct option is A $+ 5, 0, 0, + 5$
In Sodium tetrathionate, sulphur is in $+ 5, 0, 0, + 5$ oxidation state.
The oxidation states of middle sulphur atoms are $0$ each and terminal sulphur is $+ 5$ each.

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Similar questions

The oxidation states of S in sodium tetrathionate $(N a_{2} S_{4} O_{6})$ and $N a_{2} S_{2} O_{3}$ is:

(N a_{2} S_{4} O_{6})

N a_{2} S_{4} O_{6}

Oxidation states of $P$ in $H_{4} P_{2} O_{6}, H_{4} P_{2} O_{6}, H_{4} P_{2} O_{7}$ are respectively:

P O C l_{3}, H_{2} P O_{3} a n d H_{4} P_{2} O_{6}

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