[tex]\large \bold {SOLUTION}[/tex]
Looking at our picture above indicates the reaction between the two compounds. As much as possible we should remember that in making Lewis structures, there should be a minimal amount of lone pairs to minimize the atoms and make it more stable.
A lone pair on each oxygen atom can be transfered in the nitrogen atom to be a bond, in which a bond covers 2 valence electrons. We covered it with double bonds in order for them to be stable. A stable atom should have 8 valence electrons.
To find the formal charge of every atom use, the formula below.
FC = Ve - (b + d)
FC = formal charge
Ve = number of valence electrons
b = number of bonds
d = number of dots
There are 5 valence electron for Nitrogen and there are 6 valence electrons in Oxygen.
We will solve it in order (from left to right)
For the first oxygen atom (with the double bond)
FC = Ve - (b + d)
FC = 6 - (2 + 4)
FC = 6 - 6
FC = 0
For the Nitrogen atom (without dots)
FC = Ve - (b + d)
FC = 5 - (4 + 0)
FC = 5 - 4
FC = 1
For the Oxygen Atom (with the single bond)
FC = Ve - (b + d)
FC = 6 - (1 + 6)
FC = 6 - 7
FC = -1
For the Nitrogen Atom (to the right)
FC = Ve - (b + d)
FC = 5 - (3 + 2)
FC = 5 - 5
FC = 0
For the Oxygen Atom (to the right)
FC = Ve - (b + d)
FC = 6 - (2 + 4)
FC = 6 - 6
FC = 0
❤
