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632 g of sodium thiosulphate (Na2S2O3) r...

632 g of sodium thiosulphate `(Na_2S_2O_3)` reacts with copper sulphate to form cupric thiosulphate which is reduced sodium thiosulphate to give cupros compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothiosulphate `(Na_4[Cu_6(S_2O_3)_5])`.
`CuSO_4+Na_2S_2O_3toCuS_2O_3+Na_2SO_4`
`2CuS_2O_3+Na_2S_2O_3toCu_2S_2O_3+Na_2S_4O_6`
`3Cu_2S_2O_3+2Na_2S_2O_3to underset("Sodium cuprothiosulphate")(Na_4[Cu_6(S_2O_3)_5])`
In this process, 0.2 moles of sodium cuprothiosulphate is formed. `(O=16, Na=23, S=32)`
The average oxidation states of sulphur in `Na_2S_2O_3 and Na_2S_4O_6` are respectively.

A

`+5 & +2`

B

`+2 & +2.5`

C

`+5 & 2.5`

D

`+2 & +4`

Text Solution

Verified by Experts

The correct Answer is:
B

Suppose oxidation number of sulphur in `Na_2S_2O_3` is x
`2+2x+3(-2)=0`
or 2x=6-2
or 2x=4
x=+2
Suppose oxidation number of sulphur in `Na_2S_4O_6` is y
2+4y+3(-2)=0
or 4y =12-2
or 4y=10
y=+2.5
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