Aqueous solution of `Na_(2)S_(2)O_(3)` on reaction with `CI_(2)`, gives

Options

A. `Na_(2)S_(4)O_(6)`
B. `NaHSO_(4)`
C. `NaCI`
D. `NaOH`

Answer

B

Explanation

In acidic medium, `S_(2)O_(3)^(2-)` undergoes oxidation to give `S_(4)O_(6)^(2-)` ("tetrathionate").
`2S_(2)O_(3)^(2-)rarr S_(4)O_(6)^(2-)+2e^(-)` , `("n factor" =(2)/(2)=1)`
In basic medium, `S_(2)O_(3)^(2-)` undergoes oxidation to give `SO_(4)^(2-)` ions.
`S_(2)O_(3)^(2-)+10overset(Θ)OHrarr S_(4)^(2-)+5H_(2)O+8e^(-)` , `("n factor" = 8)`
In aqueou or neutral medium, `SO_(2)O_(3)^(2-)` undergoes to give `HSO_(4)^(Θ)` ions.
`S_(2)O_(3)^(2-)+5H_(2)O rarr 2HSO_(4)^(Θ)8H^(o+)+8e^(-)` , `("n factor" = 8)`
So in aqeous medium, `Na_(2)S_(2)O_(3)` ions react with `CI_(2)` to give `NaHSO_(4)`.
`Na_(2)S_(2)+4CI_(2)+5H_(2)O rarr 2NaHSO_(4)+8HCI`
`S_(2)O_(3)^(2-)rarr 2HSO_(4)^(Θ), CI_(2)rarr 2CI^(Θ)`