Asked by hiwot
the oxidation number
N in NH4F
N in NH4F
Answers
Answered by
DrBob222
Oxidation number of compounds = 0
so N + 4*H + F = 0
N + 4*1 + -1 = 0
N + 3 = 0
N = -3 so that -3 + 3 = 0
so N + 4*H + F = 0
N + 4*1 + -1 = 0
N + 3 = 0
N = -3 so that -3 + 3 = 0
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