HW problem 3.45

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Rosamari Orduna 1D
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Joined: Fri Apr 06, 2018 11:04 am

HW problem 3.45

Postby Rosamari Orduna 1D » Fri May 18, 2018 1:13 am

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
Posts: 34
Joined: Fri Apr 06, 2018 11:04 am

Re: HW problem 3.45

Postby Bree Perkins 1E » Fri May 18, 2018 9:16 am

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
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Joined: Wed Nov 15, 2017 3:01 am

Re: HW problem 3.45

Postby Yadira Flores 1G » Sun May 20, 2018 11:28 am

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

Chem_Mod
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Re: HW problem 3.45

Postby Chem_Mod » Sun May 20, 2018 9:33 pm

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
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Joined: Mon Jan 08, 2018 8:19 am

Re: HW problem 3.45

Postby Joshua Yang 1H » Sun May 20, 2018 10:08 pm

Yadira Flores 1G wrote: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


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