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QUESTION

# How can I draw the Lewis structure for NO?

You can find a procedure for at

http://socratic.org/questions/what-is-the-lewis-structure-of-nh4br

For NO, the skeleton structure is N-O.

The trial structure is

You have 14 in your trial structure.

The you have available are: 1 N + 1 O = 1×5 + 1×6 = 11. The trial structure has three extra electrons.

We have to insert one or more double bonds. With three electrons, we can make only one double bond with one electron left over: N=O.

With an odd number of electrons (11), we cannot give every atom an octet. We can write two possible structures.

The formal charge on each atom is:

Top structure: N = 5 - 3 - ½(4) = 0; O = 6 – 4 - ½(4) = 0 Bottom structure: N = 5 – 4 - ½(4) = -1; O = 6 - 3 + ½(4) = +1

The “best” Lewis structure is one that has the fewest formal charges — the top structure.

Thus, the Lewis structure of NO is