Solveeit Logo
Login

Question

Question: The oxidation state of sulphur in the anions \(SO_{3}^{2 -}\), \(SO_{4}^{2 -},S_{2}O_{4}^{2 -},S_{2}...

The oxidation state of sulphur in the anions SO32−SO_{3}^{2 -}, SO42−,S2O42−,S2O62−SO_{4}^{2 -},S_{2}O_{4}^{2 -},S_{2}O_{6}^{2 -} is in the order of

A

S2O42−>S2O62−>SO42−>SO32−S_{2}O_{4}^{2 -} > S_{2}O_{6}^{2 -} > SO_{4}^{2 -} > SO_{3}^{2 -}

B

S2O62−>SO32−>S2O42−>SO42−S_{2}O_{6}^{2 -} > SO_{3}^{2 -} > S_{2}O_{4}^{2 -} > SO_{4}^{2 -}

C

SO42−>S2O62−>SO32−>S2O42−SO_{4}^{2 -} > S_{2}O_{6}^{2 -} > SO_{3}^{2 -} > S_{2}O_{4}^{2 -}

D

SO32−>SO42−>S2O42−>S2O62−SO_{3}^{2 -} > SO_{4}^{2 -} > S_{2}O_{4}^{2 -} > S_{2}O_{6}^{2 -}

Answer

SO42−>S2O62−>SO32−>S2O42−SO_{4}^{2 -} > S_{2}O_{6}^{2 -} > SO_{3}^{2 -} > S_{2}O_{4}^{2 -}

Explanation

Solution

: SO32−→x+(−6)=−2⇒x=+4SO_{3}^{2 -} \rightarrow x + ( - 6) = - 2 \Rightarrow x = + 4

SO42−→x+(−8)=−2⇒x=+6SO_{4}^{2 -} \rightarrow x + ( - 8) = - 2 \Rightarrow x = + 6

S2O42−→2x+(−8)=−2⇒x=+3S_{2}O_{4}^{2 -} \rightarrow 2x + ( - 8) = - 2 \Rightarrow x = + 3

S2O62−→2x+(−12)=−2⇒x=+5S_{2}O_{6}^{2 -} \rightarrow 2x + ( - 12) = - 2 \Rightarrow x = + 5

Hence , SO42−>S2O62−>SO32−>S2O42−SO_{4}^{2 -} > S_{2}O_{6}^{2 -} > SO_{3}^{2 -} > S_{2}O_{4}^{2 -}