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Question: The oxidation state of Sulphur in the anions \({S_2}{O_4}^{2 - }\), \(S{O_3}^{2 - }\) and \({S_2}{O_...

The oxidation state of Sulphur in the anions S2O42−{S_2}{O_4}^{2 - }, SO32−S{O_3}^{2 - } and S2O62−{S_2}{O_6}^{2 - } follows the order:
A. S2O62−<S2O42−<SO32−{S_2}{O_6}^{2 - } < {S_2}{O_4}^{2 - } < S{O_3}^{2 - }
B. S2O42−<SO32−<S2O62−{S_2}{O_4}^{2 - } < S{O_3}^{2 - } < {S_2}{O_6}^{2 - }
C. SO32−<S2O42−<S2O62−S{O_3}^{2 - } < {S_2}{O_4}^{2 - } < {S_2}{O_6}^{2 - }
D. S2O42−<S2O62−<SO32−{S_2}{O_4}^{2 - } < {S_2}{O_6}^{2 - } < S{O_3}^{2 - }

Explanation

Solution

Hint: Just try to recall that oxidation state is the number of electrons that an atom either gains or losses in order to form a chemical bond with another atom. Now by knowing the basic rules of oxidation state you can easily find the oxidation state of Sulphur in anions given in the question.

Complete step by step solution:
There are some rules to assign an oxidation state to a given element or compound.
The rules are as follows:
Rule 1: The oxidation state of any free element is equal to 0. For example: the oxidation number of NaNa is 0.
Rule 2: If any monatomic ion is given, then the oxidation state of that ion is its net charge.
Rule 3: In most compounds, oxidation state of oxygen atom is -2 except in case of peroxides where it is -1.
Rule 4: All alkali metals i.e. group 1 elements exhibit oxidation state of +1 in their compounds.
Rule 5: Oxidation state of all alkaline earth metals i.e. group 2 elements are taken +2 in their compounds.
Rule 6: Oxidation state of all halogens are taken as -1 except some cases. But the oxidation number of fluorine is always taken as -1 in their compounds
Rule 7: If polyatomic ions would have been given then in that case, the sum of all oxidation numbers of all atoms that constitute them equals the net charge of the polyatomic ion.

Coming to the question, now you can easily calculate the oxidation state of SS in anions S2O42−{S_2}{O_4}^{2 - },SO32−S{O_3}^{2 - } and S2O62−{S_2}{O_6}^{2 - } by using the above stated rules:
Calculation:
Oxidation state of SS in S2O42−{S_2}{O_4}^{2 - }: Oxidation state of oxygen atom, O= -2
Let the oxidation state of SS in S2O42−{S_2}{O_4}^{2 - } be x.
By Rule 7:
2×x+4×(−2)=−2⇒2x−8=−2⇒2x=6x=+3 2 \times x + 4 \times ( - 2) = - 2 \Rightarrow 2x - 8 = - 2 \Rightarrow 2x = 6 x = + 3
Oxidation state of SS in SO32−S{O_3}^{2 - }: Oxidation state of oxygen atom, O= -2
Let the oxidation state of SS in SO32−S{O_3}^{2 - } be x.
By Rule 7:
x+3×(−2)=−2⇒x−6=−2x=+4 x + 3 \times ( - 2) = - 2 \Rightarrow x - 6 = - 2 x = + 4
Oxidation state of SS in S2O62−{S_2}{O_6}^{2 - }: Oxidation state of oxygen atom, O=-2
Let the oxidation state of SS in S2O62−{S_2}{O_6}^{2 - } be x.
By Rule 7:
2×x+6×(−2)=−2⇒2x−12=−2⇒2x=10x=+5 2 \times x + 6 \times ( - 2) = - 2 \Rightarrow 2x - 12 = - 2 \Rightarrow 2x = 10 x = + 5

Hence, from above calculations we can easily say that option B is the correct option.

Note: It should be remembered to you that if the given species is neutral then, the sum of all the oxidation numbers of the constituent atoms equals to 0.
Also, it should be remembered that the maximum oxidation state of sulphur is +6 and the minimum oxidation state of sulphur is -2.