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The difference in the oxidation numbers of the two types of sulphur atoms in $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$ is:
A. $5$
B. $4$
C. $6$
D. $8$

Last updated date: 13th Jun 2024
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Hint: To find out the correct option, at first think about the oxidation number. The oxidation number is also referred to as the oxidation state and it is the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom.

Complete step by step solution:
First of all, let us see the structure of $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$. The structure is given as:
seo images

- From the above structure, we can see that there are two types of sulphur. One type of sulphur has double bonding with the oxygen atoms and is present to the peripheral side of the structure while the second type of sulphur has no bonding with other elements and this type is present in the middle of the structure.
- The oxidation number of the peripheral sulphur atoms is $+5$ each while the oxidation number of the middle sulphur atoms is $0$ each. And the difference in the oxidation number of the two types of the sulphur will be $5-0=5$.
So, the correct answer is “Option A”.

Note: Possibly you may be confused with the two types of sulphur. Remember that, there are two peripheral sulphur and two sulphur atoms present in the middle of the structure. Each peripheral sulphur atom has the same oxidation state and the middle sulphur has the same oxidation state. Sulphur is abundant, multivalent and non-metallic element and in normal conditions, it occurs in cyclic octa-atomic molecules.