# What is the Lewis electron dot formula (Lewis structure) of Nitrous Oxide (#N_2O#)?

##### 1 Answer

I learned this by the counting-electron method, and then assigning **formal charges** to determine the most likely distribution of **valence electrons**.

The number of *valence* electrons available in the structure are:

#(N# :#5 e^(-))xx 2 = 10 e^(-)# #O# :#6 e^(-)#

#10 + 6 = 16# total available valence electrons.

*We have two nitrogens and one oxygen, which suggests that either we have oxygen in the middle or two nitrogens in a row.*

Notice how if you had oxygen in the ** middle**, the formal charges of both nitrogens have

**no way of being distributed well without exceeding 8 electrons for oxygen**:

One way of determining formal charges is:

#"Formal Charge" = "Expected Electrons" - "Owned Electrons"#

Since the *left nitrogen* "owns" **five** valence electrons (two lone pairs and one from the **five**, its formal charge is

Since the *right nitrogen* "owns" **four** valence electrons (one lone pair gives two electrons; then, two electrons from the **five**, its formal charge is

Since the *oxygen* "owns" **seven** valence electrons (two lone pairs gives four electrons; then, one electron from the left **six**, its formal charge is

*Oxygen has TEN, but it cannot have more than EIGHT valence electrons, so that structure is ruled out.*

Thus, **one of the nitrogens must be in the middle**.

Now we can get two plausible possibilities, which are both **linear** molecular geometries (NOT bent!!! Two electron groups!):

*CHALLENGE: Can you determine why the formal charges are how they are? Based on electronegativity differences between oxygen* (*and nitrogen* (*which is the more favorable resonance structure?*