How do you draw the lewis structure for NO2?

Aug 5, 2018

Refer to the video in the explanation.

Explanation:

Aug 5, 2018

So how many valence electrons we got...?

Explanation:

Five from nitrogen, and 12 from the oxygen atoms...and so we got 8 and a HALF electron pairs to distribute...17 electrons in total

As is typical, the least electronegative atom is placed in the middle to give...

O=stackrel(dot)N^(+)-O^(-)

From left to right as we face the page, each atom is associated with SIX, FOUR, and SEVEN valence electrons, giving rise to formal charges of $0$, $+ 1$, and $- 1$. Of course, we could write the other resonance structure...

""^(-)O-stackrel(dot)N^(+)=O

And so $N {O}_{2}$ is a NEUTRAL, BENT molecule...whose geometry is based on a trigonal plane...$\angle O - N - O = {120}^{\circ}$ to a first approx., but ${115}^{\circ}$ in actuality....

Why don't you try it for nitrite ion, $N {O}_{2}^{-}$...how does the Lewis structure evolve..?

And just to add, that the formal Lewis structure of $N {O}_{2}$ is useful when we assess reactivity... $N {O}_{2}$ can READILY dimerize (alternatively ${N}_{2} {O}_{4}$ can readily undergo dissociation.

(""^(-)O)O=stackrel(+dot)N + stackrel(+dot)N=O(O^(-)) rarr underbrace(""^(-)(O)O=stackrel(+dot)N-stackrel(+dot)N=O(O^(-)))

${N}_{2} {O}_{4} \equiv \text{dinitrogen tetroxide}$

The nitrogen atom in $N {O}_{2}$ or ${N}_{2} {O}_{4}$ is said to be quaternized, i.e. it bears a formal POSITIVE charge....