N2o3 hybridization?

1 Answer
Apr 14, 2018

See below:
Warning: Somewhat long answer!

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: sp^3
3 Charge centres: sp^2
2 Charge centres: sp

Now the Lewis structure for N_2O_3 shows resonance, as it is possible to draw two different Lewis structures:
Quora

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 sp^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 sp^2 hybridized.

The "upper" oxygen atoms on the leftmost structure are both sp^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 sp^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 sp^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- sp^2)
Now the top oxygen has 4 charge centres and is sp^3 hybridized and the bottom oxygen has 3 charge centres- sp^2
( Note , I believe the diagram is faulty- oxygen cannot have an expanded octet so I assume that it is supposed to be sp^2, although the diagram shows sp^3)

Hopefully it helped!