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The difference in the oxidation numbers ...

The difference in the oxidation numbers of two types of sulphul atoms in `Na_(2)S_(4)O_(6)` is…..

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The correct Answer is:
5
Structure of `Na_(2)S_(4)O_(6)` is
`Na^(+).^(-)O ul(+5)underset(O)underset(||)overset(O)overset(||)(S)-overset(0)(S)-overset(0)(S)-underset(O)underset(||)overset(O)overset(||)(S)ul(+5)O^(-)Na^(+)`
Difference in O.N = 5 - 0 = 5
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The difference in the oxidation numbers of the two types of sulphur atoms in Na_(2)S_(4)O_(6) is: Oxidation number of sulphur atom involved in coordinate bond formation is (+5) and that of middle sulphur atom is zero. Hence the difference in oxidation number of two types of sulphur aton will be (+5).]

The difference in the oxidation states of the two types of sulphur atoms in Na_(2)S_(4)O_(6) is :

The two sulphur atoms in Na_(2)S_(2)O_(3) have