Fractional values of oxidation numbers are possible as in Na2S4O6I,Fe3O4II,N3HIII
A
Only I
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B
Only II
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C
I and III
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D
I,II and III
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Solution
The correct option is DI,II and III Na2S4O6Fe3O42×1+4x+6(−2)=03×x+4(−2)=02+4x−12=03x−8=04x=+103x=8x=+104=+2.5x=+83or+223Average oxidation number ofAverage oxidation number of ironSis+2.5is+223 N3H3x+1=03x=−1x=−13Average oxidation number ofnitrogen is −13.