If we try to calculate the oxidation state of nitrogen in $\ceN2O$ using the acquainted algebraic method, we get oxidation state $+1$ because that both nitrogen atoms and that"s what I uncovered when i looked it increase on the internet.

You are watching: Oxidation state of nitrogen in n2o

I make the efforts to perform it with the structure and also that"s where I gained confused:

$$\ce\overset-N=\overset+N=O$$

The $\ceN=N$ link is a name: coordinates bond, so the should provide a $-1$ oxidation state because that the left $\ceN$ and $+1$ because that the center one. Since the center one has actually a double bond through oxygen, it gets extr $+2$ for a total of $+3.$

But if I consider this resonance structure:

$$\ceN#\overset+N-\overset-O$$

The left $\ceN$ is acquiring an oxidation state that $0$ and the middle one it s okay $+2.$ So ns am confused regarding what is in reality correct.

Is it feasible for atom to exist in superpositions the oxidation states? Or maybe I to be overlooking some straightforward concept?


inorganic-gimpppa.org resonance oxidation-state lewis-structure pnictogen
share
enhance this question
monitor
edited january 29 in ~ 8:48
*

andselisk♦
33.5k1414 gold badges112112 silver- badges197197 bronze badges
asked january 29 at 6:33
*

PhysicsaPhysicsa
5888 bronze badges
$\endgroup$
3
include a comment |

1 prize 1


energetic earliest Votes
5
$\begingroup$
If we shot to calculation the oxidation state the nitrogen in $\ceN2O$ using the acquainted algebraic method, we gain oxidation state $+1$ because that both nitrogen atoms and also that"s what I discovered when ns looked it up on the internet.

Well … you get an average oxidation state. This calculation may be implicitly assumes that all nitrogen atom be equivalent. In some cases (e.g. Hydrazine) they are and the result you acquire algebraicly is as you would intend from a Lewis depiction. In other instances (e.g. Here) this is not the case as the nitrogens room not identical (only one is bound come oxygen). Thus, the is obvious that a simple algebraic method should fail.

But what about the ‘real’ result? What around the resonance structures? Well, this is wherein things get really difficult. You basically have two π equipment orthogonal to each other and also each occupied by 4 electrons which have the right to manifest as a lone pair on either end and a multiple bond to the various other atom. If girlfriend really want to pat the game, you could include an additional resonance framework as shown below, whereby all π lone pairs are centred on the terminal nitrogen:

$$\ceN#\overset+N-\overset-O \overset-N=\overset+N=O \overset2-N-\overset+N#\overset+O$$

(It is clear the this 3rd resonance structure contributes least to the overall picture as it has actually a higher charge separation and charges space separated opposite come what electronegativity would predict.)

The speculative structure mirrors that the $\ceN-N$ street is slightly much shorter than the $\ceN-O$ distance which one can use to assume the the $\ceN-N$ bond have actually a slightly larger bond order 보다 the $\ceN-O$ bond. But ultimately, they are still very comparable (the distinction is just $\pu4pm$) so equal bond orders might also be an option. Long story short: quick of calculating the electronic distribution (i.e. Solving the Schrödinger equation) girlfriend won’t have the ability to arrive in ~ a identify answer for the ‘real’ oxidation states.

See more: How Many Gallons Is 140 Quarts To Gallons, Convert 140 Quarts To Gallons

So what can you perform on paper? compare the resonance structures, the leftmost as I have actually ordered them is slightly much better than the main one as the officially charges are spread according come the various electronegativities. Thus, I would be lean to offer it a slightly higher weight and also – in a classroom setup – usage it to identify oxidation states. The said, it is too lot of one ambiguous example to be seriously used in any kind of examination unless the score of stated examination is to construct the chain of disagreements that formed this answer.