Question

Oxidation state of sulphur in Sodium tetrathionate $\left( {N{a_2}{S_4}{O_6}} \right)$ is:
(a) $ + 2$
(b) $ + 4$
(c) $ + 1 \cdot 5$
(d) $ + 2 \cdot 5$

Answer 1

Hint: Oxidation number is the number of electrons an atom gains or loses. It can have values positive, negative or zero. A neutral atom in elemental state has zero oxidation state.

Complete step by step answer:
(1) For any compound, ${A_x}{B_y}$, we dissociate in generalize form $x{A^ + } + y{B^ - }$, assuming A to be the positive cation, and B to be the negative anion. Then, we multiply the oxidation state of each atom involved in the compound with their respective number in the compound, and equate it to the overall charge of the compound. For the taken example, it will be like:
($x{\text{ }} \times $ number of A atoms + $y{\text{ }} \times $ number of B atoms = 0), since ${A_x}{B_y}$ is neutral.
(2) For Sodium tetrathionate $\left( {N{a_2}{S_4}{O_6}} \right)$, oxidation of Sodium, ${\text{Na}}$ is $ + 1$ because losing one electron Sodium can attain noble gas configuration of Neon, and get stabilized. Oxidation number of Sulphur is to be found, so we assume that it has an oxidation state of ${\text{x}}$. Oxidation number of Oxygen is known to be $ - 2$, because by gaining two electrons, oxygen can attain the noble gas configuration of Neon to make itself stable. Lastly, the overall charge on the given compound is zero as $\left( {N{a_2}{S_4}{O_6}} \right)$ is neutral.
(3) So, the final equation will be:
$2 \times \left( {{\text{Oxidation number of Na}}} \right) + 4 \times \left( {{\text{Oxidation number of S}}} \right) + 6 \times \left( {{\text{Oxidation number of O}}} \right)$
$ \Rightarrow 2 \times \left( { + 1} \right) + 4 \times \left( {\text{x}} \right) + 6 \times \left( { - 2} \right)$
$ \Rightarrow 2 + 4{\text{x - 12 = 0}}$
$ \Rightarrow {\text{x}} = \dfrac{{10}}{4}$
$ \Rightarrow {\text{x}} = 2 \cdot 5$
Therefore, the oxidation number of each Sulphur atom in the given compound is $2 \cdot 5$.

Hence, option (d) $ + 2 \cdot 5$ is the correct answer.

Note:
In the above answer, a positive sign is given to indicate that it has a positive oxidation number as oxidation number can be positive integer, negative integer, fraction (decimal values) or zero.