Solveeit Logo
Login

Question

Chemistry Question on Oxidation Number

The oxidation states of sulphur in the anions SO32−,S2O42−SO _{3}{ }^{2-}, S _{2} O _{4}{ }^{2-} and S2O62−S _{2} O _{6}{ }^{2-} follow the order -

A

S2O42−<S2O62−<SO32−S_2O_4^{ 2 - } < S_2O_6^{ 2 - } < SO_3^{ 2 - }

B

S2O62−<S2O42−<SO32−S_2O_6^{ 2 - } < S_2O_4^{2 - } < SO_3^{ 2 - }

C

S2O42−<SO32−<S2O62−S_2O_4^{ 2 - } < SO_3^{ 2 - } < S_2O_6^{ 2 - }

D

SO32−<S2O42−<S2O62−SO_3^{ 2 - } < S_2O_4^{ 2 - } < S_2O_6^{ 2 - }

Answer

S2O42−<SO32−<S2O62−S_2O_4^{ 2 - } < SO_3^{ 2 - } < S_2O_6^{ 2 - }

Explanation

Solution

Oxidation state of S in SO32−SO_3^{ 2 - }
\hspace19mm x + ( - 2 ×\times 3 ) = - 2
\hspace19mm x = + 6 - 2 = + 4
Oxidation state of S in S2O42−S_2 O_4^{ 2 - }
\hspace19mm 2 x 4- (- 2 ×\times 4) = - 2
\hspace19mm 2x = + 8 - 2 = + 6
\hspace19mm x=+62=+3x = \frac{ + 6 }{ 2 } = + 3
Oxidation state of S in S2O62−S_2 O_6^{ 2 - }
\hspace19mm 2 x 4 - (- 2 x×\times 6) = - 2
\hspace19mm 2x = 4 - 12 - 2 = 10
\hspace19mm x = 102=−5\frac{10}{ 2} = - 5
Hence, increasing order of oxidation states of S is
S2O42−<SO32−<S2O62−S_2O_4^{ 2 - } < SO_3^{ 2 - } < S_2O_6^{ 2 - }