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This question already has an answer here:

An element have many oxidation numbers, then which one should we consider in the given question?

e.g. In the compound $\ce{HCN}$ as $\ce{C}$ {carbon} has +4,+2,-4 oxidation numbers and $\ce{N}$ {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers

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marked as duplicate by user15489, M.A.R. ಠ_ಠ, ron, bon, Martin - マーチン Jul 13 '15 at 1:30

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e.g. In the compound HCN as C {carbon} has +4,+2,-4 oxidation numbers and N {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers

That is not correct. Each atom only has one oxidation number in a compound, although one element can have different oxidative states (i.e. sodium thiosulfate). Your statement should be:

C {carbon} has +4,+2,-4 oxidation numbers, and N {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers. In the compound HCN, carbon has +2, hydrogen has +1, and nitrogen has -3.

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