Question

# The oxidation state of sulphur in the anions S2O2−4, S2O2−4 and S2O2−6 follows the order:

A
S2O26-<S2O24<SO23
B
S2O24<SO23<S2O26
C
SO23<S2O24<S2O26
D
S2O24<S2O26-<SO23
Solution
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#### Oxidation state of S2O2−42(x)+4(−2)=−2 2x=8−2 2x=6x=3Oxidation state of SO2−3x+3(−2)=−2 x=6−2 x=4Oxidation state of S2O2−62(x)+6(−2)=−2 2x=12−2 2x=10x=5So the oxidation state of sulphur in the anions S2O2−4, S2O2−4 and S2O2−6 follows the order.S2O2−4<SO2−3<S2O2−6.Hence option B is correct.

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