Assign formal charges to each atom in the resonance forms of n2o.

Here's what I get  

Explanation:

The three resonance forms of N₂O are shown in the first diagram below (you can also use horizontal dashes to represent the bonding pairs).

To get the formal charge (FC) on the atoms, cut each bond in half, as in the second diagram.

Each atom gets the electrons on its side of the cut.

Formal charge = valence electrons in isolated atom - electrons on bonded atom

FC = VE - BE  

(a) In Structure A

Left-hand N:

VE = 5

BE = 1 lone pair (2)+ 3 bonding electrons = 2 + 3 = 5

FC = 5 - 5 = 0.

Central N:

VE = 5

BE = 4

FC = 5 - 4 = +1

On O:

VE = 6

BE = 3 lone pairs(6) + 1 bonding electron = 7

FC = 5 - 6 = -1

(b) In Structure B

Left-hand N:

VE = 5

BE = 2 lone pairs (4)+ 2 bonding electrons = 4 + 2 = 6

FC = 5 - 6 = -1.

Central N:

VE = 5

BE = 4

FC = 5 - 4 = +1

On O:

VE = 6

BE = 2 lone pairs(4) + 2 bonding electrons = 6

FC = 6 - 6 = 0

(c) In Structure C

Left-hand N:

VE = 5

BE = 3 lone pairs (6)+ 1 bonding electrons = 6 + 1 = 7

FC = 5 - 7 = -2.

Central N:

VE = 5

BE = 4

FC = 5 - 4 = +1

On O:

VE = 6

BE = 1 lone pair(2) + 3 bonding electrons = 5

FC = 6 - 5 = +1

it means there are 3 atoms of oxygen