Question

The oxidation states of $\mathrm{S}$ atoms in ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$from left to right respectively, are

• A. $+6,0,0,+6$
• B. $+3,1,+1,+3$
• C. $+5,0,0,+5$
• D. None of the above

Answer 1

The correct option is C

$+5,0,0,+5$

Explanation for correct option

(C)$+5,0,0,+5$

Oxidation number

• "Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to a more electronegative element."

Structure of ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$

• Two sulfur(2 and 3) that is linked to each other has oxidation numbers of zero.
• The oxidation number of the other two sulfur atoms (1 and 4)can be calculated as follows:-
• $$\begin{gathered} 2x+2\times 0+6\times \left(-2\right)=-2 \\ 2x-12=-2 \\ 2x=10 \\ x=+5 \end{gathered}$$
• So, the oxidation states of $\mathrm{S}$ atoms in ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$from left to right respectively, are $+5,0,0,+5$.

Explanation for incorrect option

The oxidation states of $\mathrm{S}$ atoms in ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$from left to right respectively, are $+5,0,0,+5$. Hence, options (A), (B), (D) are incorrect.

Hence, option (C) is correct. The oxidation states of $\mathrm{S}$ atoms in ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$from left to right respectively, are $+5,0,0,+5$.