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# Oxidation state of $S$ in sodium tetrathionate isA. $+5,0,0,+5$B. $+6,+6,0$C. $+4,+2,+5$D. $+2,0,+2,+3$

Last updated date: 12th Aug 2024
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Hint: The structure of sodium tetrathionate contains two atoms of sodium and four atoms of sulphur along with six atoms of oxygen attached to one another.
The molecule has overall charge zero which indicates that all the oxidation states of individual atoms involved in the molecule balances the charges possessed by each other.

As we can see, there are four sulphur atoms in one molecule of sodium tetrathionate, and the overall charge in the molecule is zero. Now we know that general oxidation state of oxygen is $-2$ and that of sodium is $+1$, so if we calculate the total oxidation state, without the oxidation number of sulphur atoms we get
$(2\times 6)-2=10$
Where, the total number of oxygen atoms are $6$ and that of sodium atoms is $2$.
So the sulphur atoms which are present at the middle of the structure will have zero oxidation state, and the terminal sulphur atoms will have $+5$ oxidation state each.