HW problem 3.45

[Rosamari Orduna 1D]

3.45 asks: Draw the Lewis structures that contribute to the resonance hybrid of nitryl chloride, ClNO2 (N is the central atom).

Could someone please explain to me why N cant have a double bond with each O and a single bond with Cl?
Does it have something to do with the lone pairs? Or taking away/adding electrons ?

[Bree Perkins 1E]

N only has 5 valence electrons, and once you single bond to both O's and the Cl, you only have 1 lp (2 e-s) left. That being said, it can only form one more bond, and once it does, you can count all the bonds Nitrogen makes (4, each representing 2 e-s) which will equal 8 e-s (Octet). If it formed 2 bonds with each O and one with Cl, than that would be 5 bonds (10 e-s) which N is not able to do.

Hope this helps!

[Yadira Flores 1G]

Why when doing the resonance structure can we not give Cl-N a double bond? Or can we?

[Chem_Mod]

5 bonds = 10 electrons and N cannot have "expanded octet"

Also, having double bond with Cl and N would affect formal charges and you probably are not minimizing the number of formal charges for the most stable structure.

[Joshua Yang 1H]

Why when doing the resonance structure can we not give Cl-N a double bond? Or can we?

It is technically possible, but we're looking for the lewis structure with the lowest formal charge

If you calculate the formal charge of Cl, it comes out as 0, meaning that there is no need to change the electron configuration around Cl