[rruro]

So I've attached the molecule below, I usually use the formula where you subtract valence electrons on bonded atoms -total number of valence electrons... But I'm sure I keep getting the wrong answer? is 40 even possible?

I would really appreciate anyone's insight or advice on how to work out the lone pairs of electrons on this molecule



Many thanks!

[rruro]

I know there are 6 Pi bonds, but I'm not sure how to count lone pairs..

[Babcock_Hall]

If one assumes that all formal charges are shown (a good assumption IMO), then the number of lone pairs on an atom can be found from the equation to calculate formal charge.  FC = (# of valence electrons) - (# of bonds + # of electrons in lone pairs).  The number of valence electrons and the number of bonds are both known.  The number of electrons is lone pairs is the only unknown.

[AWK]

Draw electron pairs to make up the oxygen and nitrogen octets and count them.

[Babcock_Hall]

I had overlooked the nitro group in the condensed structure; therefore, I am grateful to AWK for fleshing the drawing out and including the formal charges on N and O.  In general formal charges should be made explicit.  With time they will become second nature to you.