# What is the Lewis structure of N2O?

Jun 16, 2018

Well, we got 16 valence electrons....

#### Explanation:

16 valence electrons:

$2 \times {5}_{\text{nitrogen"+1xx6_"oxygen"="8 electron pairs}}$

...to distribute over 3 centres. And you must simply KNOW here that the oxygen is terminal. And given the example, there will be formal charge separation.

$N \equiv \stackrel{+}{N} - {O}^{-}$ versus ""^(-)N=stackrel(+)N=O

This is half the story inasmuch as we consider the data: i.e. the bond-length of dinitrogen, dioxygen, and nitrous oxide...

$N \equiv N$ $: \text{bond length} = 1.10 \times {10}^{-} 10 \cdot m$

$O = O$ $: \text{bond length} = 1.21 \times {10}^{-} 10 \cdot m$

$N \equiv \stackrel{+}{N} - {O}^{-}$ $: \text{N-O bond length} = 1.19 \times {10}^{-} 10 \cdot m$

$: \text{N-N bond length} = 1.13 \times {10}^{-} 10 \cdot m$