K3[Fe(NO3)6]
N is present in nitrate ion
nitrate ion = NO3-
assume N = X
O = -2 oxidation state = 3 O atoms so 3 (-2) = -6
we have to equilize -1 as -1 charge present on ion.
now
X + 2(-3) = -1
X - 6 = -1
X = -1 + 6
X = +5
oxidation state on N = +5
oxidation number oxidation number for N in K3Fe(NO3)6 please explain how to find it
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check my work. if it's wrong please explain.
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