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

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Solution

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Oxidation state of S2O2−4

2(x)+4(−2)=−2

2x=8−2

2x=6

x=3

Oxidation state of SO2−3

x+3(−2)=−2

x=6−2

x=4

Oxidation state of S2O2−6

2(x)+6(−2)=−2

2x=12−2

2x=10

x=5

So 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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