Oxidation state of S in $N a_{2} S_{2} O_{3}$ is

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-1

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Solution

The correct option is D 2
O.S. of Na = +1 (alkali metal)
O.S. of O = -2
$\Rightarrow N a_{2} S_{2} O_{3} \Rightarrow 2 (1) + 2 n + 3 (- 2) = 0$
$\Rightarrow X = \frac{4}{2} = 2$

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