# N2o3 hybridization?

Apr 14, 2018

See below:

#### Explanation:

The first step in determining hybridization is to determine how many "charge centres" surrounds the atoms in question, by looking at the Lewis structure.

1 charge centre is the equivalent of either:
A single covalent bond.
A double covalent bond.
A triple covalent bond.
A lone electron pair.

And then Hybridization is divided into the following:
4 Charge centres: $s {p}^{3}$
3 Charge centres: $s {p}^{2}$
2 Charge centres: $s p$

Now the Lewis structure for ${N}_{2} {O}_{3}$ shows resonance, as it is possible to draw two different Lewis structures:

Lets begin inspecting the hybridization, starting with the leftmost Lewis structure.

One nitrogen is bonded by 1 double bond and 2 single bonds, so it must have 3 "charge centres"- it is $s {p}^{2}$ hybridized.
The other nitrogen atom is bonded by 1 single bond, 1 double bond and has a lone pair. It has 3 "charge centres" so it is therefore also $s {p}^{2}$ hybridized.

The "upper" oxygen atoms on the leftmost structure are both $s {p}^{2}$ hybridized as they have 2 lone pairs and 1 double bond- 3 charge centres.
However, the bottom oxygen has 3 lone pairs and 1 double bond, that makes 4- so it is $s {p}^{3}$ hybridized.

Now moving to the right structure we can see that the Nitrogen atoms are unchanged in the amount of charge centres(3), so they retain their $s {p}^{2}$ hybridization.

However, now the top-left and bottom left oxygen have a different amount of charge centres than on the left. (The rightmost oxygen is unchaged- $s {p}^{2}$)
Now the top oxygen has 4 charge centres and is $s {p}^{3}$ hybridized and the bottom oxygen has 3 charge centres- $s {p}^{2}$
( Note , I believe the diagram is faulty- oxygen cannot have an expanded octet so I assume that it is supposed to be $s {p}^{2}$, although the diagram shows $s {p}^{3}$)

Hopefully it helped!