3.49


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Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

3.49

Postby Kelly Kiremidjian 1C » Sun Nov 05, 2017 7:28 pm

Can anyone explain 3.49? I had the correct lewis structure but am not getting the formal charge correct.
For part A, I had
Formal Charge of Oxygen= 6 valence e - (1 lone pair +6/2) which gives a +2 charge, and the book says Oxygen has a +1 charge.
For Nitrogen I had
Formal Charge of Nitrogen=5 valence e-(1lone pair+6/2) which gives a +1 charge, and the book says Nitrogen has no charge.

Any help would be appreciated. thanks!

Curtis Wong 2D
Posts: 62
Joined: Sat Jul 22, 2017 3:00 am

Re: 3.49

Postby Curtis Wong 2D » Sun Nov 05, 2017 7:46 pm

For 49, it's NO+, so in total there should be ten electrons. That means the formal charge for Nitrogen would be 5-(6/2 + 2 electrons [1 lone pair]) = 0. For oxygen, it would be 6- (6/2 +2 electrons [1 lone pairs]) = +1.

Therefore the charge for the ion is +1.

Julian Krzysiak 2K
Posts: 49
Joined: Fri Sep 29, 2017 7:07 am

Re: 3.49

Postby Julian Krzysiak 2K » Sun Nov 05, 2017 7:48 pm

When solving for formal charge, the L is the # of electrons in the bond pair, so it would be 2; you don't count the whole pair as 1.

So if we do that, then O would be 6 -(2+ 6/2)= +1, and N would be 5 -(2 + 6/2)= 0

Navleen Bajwa 3A
Posts: 20
Joined: Fri Sep 29, 2017 7:06 am

Re: 3.49

Postby Navleen Bajwa 3A » Tue Nov 14, 2017 11:25 pm

The rest of your calculations are right except for the L in the formal charge equation. The L represents the actual number of lone pair electrons. So one lone pair would have 2 lone pair electrons. Therefore, the formal charge of Nitrogen= 5-(2+6/2). Formal charge of Oxygen= 6-(2+6/2).


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