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Chemistry Question on Oxidation Number

Oxidation state of sulphur in anions SO32−.S2O42−SO^{2-}_{3}. S_{2}O^{2-}_{4} and S2O62−S_{2}O^{2-}_{6} increases in the orders:

A

S2O62−<S2O42−<SO32−S_{2}O^{2-}_{6}< S_{2}O^{2-}_{4}< SO^{2-}_{3}

B

SO62−<S2O42−<SO62−SO^{2-}_{6}< S_{2}O^{2-}_{4}< SO^{2-}_{6}

C

S2O42−<SO32−<S2O62−S_{2}O^{2-}_{4}< SO^{2-}_{3}< S_{2}O^{2-}_{6}

D

S2O42−<S2O62−<SO32−S_{2}O^{2-}_{4}< S_{2}O^{2-}_{6}< SO^{2-}_{3}

Answer

S2O42−<SO32−<S2O62−S_{2}O^{2-}_{4}< SO^{2-}_{3}< S_{2}O^{2-}_{6}

Explanation

Solution

In SO3−−SO_{3}^{--} x+3(−2)=−2;x=+4x+3\left(-2\right)=-2; x=+4 In S2O4S_{2}O_{4} 2x+4(−2)=−22x+4\left(-2\right) =-2 2x−8=−22x-8 = -2 2x=6;x=+32x=6; x=+3 In S2O62−S_{2}O^{2-}_{6} 2x+6(−2)−22x+6\left(-2\right)-2 2x=10;x=+52x= 10; x=+5 hence the correct order is S2O4−−<SO3−−<S2O6−−S_{2}O_{4}^{--} < SO_{3}^{--} < S_{2}O_{6}^{--}