## O having higher FC than N

$FC=V-(L+\frac{S}{2})$

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### O having higher FC than N

Why is it that in [NO]+, there is a triple bond where N has an FC of 0 and O has an FC of +1 when O is more electronegative? Why doesn't it form a double bond so that N has the higher FC and O is at 0? For reference, this is 3.48.a, thanks.

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### Re: O having higher FC than N

Hey Isa,

If you draw both Lewis structures out, you will see that having a double bond between each atom results in a full octet for O, but not N. With a triple bond, they both have full octets.