When a molecule has resonance structure, does it mean there are delocalized electrons in the molecule?

When will delocalization happen?

Can the resonance strcuture be replaced by a delocalized valence bond picture?

May you use N2O (nitrous oxide) as a example?

I've worked out the delocalized picture myself, but in my picture, the central nitrogen has a lone pair, the molecule is not linear. it is wrong.

My picture:

_______  <- delocalization of unpaired pi electrons from 2 atom
:N=N:-O::

  

DISCLAIMER: I make no attempt to address molecular-orbital arguments in the body of this answer; I am deliberately staying within a valence-bond or even a Lewis model of bonding. However, I have given some MO-related information in sidebars like this one.

Part of your problem (that is, in getting a bent structure rather than the correct linear structure) is that you are ignoring the canonical (or grammatical) rules for generating resonance structures. The two canonical Kekulé structures (which means that all electrons are shown paired and all atoms have octets) of N₂O are

Molecular orbital calculations bear this out. MP2/6-31G* geometry optimizations predict three salient things about the structure of N₂O:

1. The singlet (all electrons paired) state is lower in energy than the triplet (diradical) state.
2. The singlet state is linear, and thus does not have a lone pair on the central nitrogen.
3. The triplet state is bent, and thus does have a lone pair on the central nitrogen.

We could have predicted this, of course, from consideration of our various resonance structures: the two "best" structures, in which all electrons are paired, also predict a linear geometry from VSEPR considerations. The other three structures all predict a bent geometry.

Contrast the situati on with NO₂.

The sort of representation you are after is the sort you see sometimes in drawings of, say, benzene,

An attempt to use the same conventions for N₂O would look like this

rather than like yours. Notice that I have simply taken the electron-pair that shifts from a N-N to a N-O bond in the resonance structures, and shown it as a dotted line connecting all three atoms. Notice also that the placement of the other bonds/lone pairs has not changed, so there is no lone pair on the central nitrogen!

However, there is a serious problem with my attempt to represent the two important resonance structures this way: a dotted line conventionally represents ˝ a bond (two electrons spread over three atoms), but what we want to represent is four electrons spread over three atoms, which is just impossible within Lewis structure rules! My attempt has dropped two electrons; yours adds them back in as a lone pair on the central nitrogen, at the cost of (a) giving the central nitrogen more than an octet and (b) predicting that the geometry of the molecule is bent when it isn't.

There is a way to save the day, but at the cost of violating Lewis octet rules. We can have a nitrogen-nitrogen triple bond and a nitrogen-oxygen double bond in the same structure,

but that puts ten electrons on the central nitrogen. You're better off following the rules, which include both octet-based Lewis structures and resonance. After all, this is a highly-empirical way of representing molecules and, as with anything empirical, trying to find a different set of rules often breaks the system!

MP2-calculated molecular orbitals for singlet N₂O show several π-type orbitals, but there is not a very excellent way to depict them. There are two kinds of lone-pair orbitals: two σ-type lone pairs coaxial with the molecule, either on the end nitrogen or on the oxygen; and two perpendicular π-type lone-pair orbitals, each with π "lone-pair" electron density on oxgyen and π-bonds between the two nitrogen atoms.

Relating these molecular orbitals to the resonance structures, which are after all valence-bond formalisms, is chancy at best.