Using my BedokFunland
JC formula for Oxidation State (OS) aka Oxidation Number (ON),
we have :
Oxidation State = Formal Charge + Electronegativity
Consideration.
There are 2 resonance contributors and 1 resonance hybrid.
Considering the N=N=O resonance contributor, the terminal N atom
has
Oxidation State = Formal Charge + Electronegativity Consideration =
(-1) + (0) = -1
Considering the N=N=O resonance contributor, the central N atom
has
Oxidation State = Formal Charge + Electronegativity Consideration =
(+1) + (+2) = +3
Hence average OS of N in this resonance contributor = ((-1) +
(+3)) / 2 = +1
Considering the N(triple bond)N-O resonance contributor, the
terminal N atom has
Oxidation State = Formal Charge + Electronegativity Consideration =
(0) + (0) = 0
Considering the N(triple bond)N-O resonance contributor, the
central N atom has
Oxidation State = Formal Charge + Electronegativity Consideration =
(+1) + (+1) = +2
Hence average OS of N in this resonance contributor = ((0) +
(+2)) / 2 = +1
Since O is more electronegative than N, the N(triple bond)N-O
resonance contributor is more significant compared to
the N=N=O resonance contributor, thus in the resonance
hybrid (ie. the actual molecule), the OS of the terminal N atom is
in between, but slightly closer to 0 than -1; and the OS of the
central N atom is in between, but slightly closer to +2 than
+3.
Hi wondering if anyone can help with this question:
using this structure of N2O,
looking at just the one on the left, (more likely one i
suppose), whats the proper way to calculate the oxidation number
for each of the atoms?
and will the average O.N. of the 2 Nitrogen atoms give +1 (e.g.
first N is 0 while second N is +2) which is nitrogen's oxidation
state in N2O or do both nitrogen atoms(in this case) have to have
an oxidation number of +1?