How do you draw the lewis structure for NO2?

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How do you draw the lewis structure for NO2?

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Answer 1 · 2018-08-05T04:56:48

Refer to the video in the explanation.

Explanation:

Answer 2 · 2018-08-05T05:52:19

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...

$O = {. N}^{+} - O^{-}$ $O=stackrel(dot)N^(+)-O^(-)$

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 $0$ $0$ , $+ 1$ $+1$ , and $- 1$ $-1$ . Of course, we could write the other resonance structure...

$^{-} O - {. N}^{+} = O$ $""^(-)O-stackrel(dot)N^(+)=O$

And so $N O_{2}$ $NO_2$ is a NEUTRAL, BENT molecule...whose geometry is based on a trigonal plane... $∠ O - N - O = 120^{\circ}$ $/_O-N-O=120^@$ to a first approx., but $115^{\circ}$ $115^@$ in actuality....

Why don't you try it for nitrite ion, $N O_{2}^{-}$ $NO_2^(-)$ ...how does the Lewis structure evolve..?

And just to add, that the formal Lewis structure of $N O_{2}$ $NO_2$ is useful when we assess reactivity... $N O_{2}$ $NO_2$ can READILY dimerize (alternatively $N_{2} O_{4}$ $N_2O_4$ can readily undergo dissociation.

$(^{-} O) O = + . N + + . N = O (O^{-}) \to^{-} (O) O = + . N - + . N = O (O^{-})     $ $(""^(-)O)O=stackrel(+dot)N + stackrel(+dot)N=O(O^(-)) rarr underbrace(""^(-)(O)O=stackrel(+dot)N-stackrel(+dot)N=O(O^(-)))$

$N_{2} O_{4} \equiv dinitrogen tetroxide$ $N_2O_4-="dinitrogen tetroxide"$

The nitrogen atom in $N O_{2}$ $NO_2$ or $N_{2} O_{4}$ $N_2O_4$ is said to be quaternized, i.e. it bears a formal POSITIVE charge....

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How do you draw the lewis structure for NO2?

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