Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: RobertWithers on April 11, 2022, 09:46:25 PM
-
Hello all. So I am supposed to write the resonance structures for N2O42-.
This is the logic I used:
N - 5 valence e- times 2 = 10
O - 6 valence e- times 4 = 24
Given Charge of -2: +2
= 36
I've tried arrangements in which there are two double bonds between N=O but those don't add up to 36 e-. The only way I could reach 36 was with a double bond between the nitrogens and single bonds between the oxygens and nitrogens.
However this would only yield one resonance structure of I am not mistaken as the double bond cannot move?
I've spent far too much time worrying that I have made a stupid mistake so I've come here for guidance. Any advice would be appreciated.
-Rob
-
I think it is not existing with 2 negative charge. The neutral molecule can be found.
https://upload.wikimedia.org/wikipedia/commons/thumb/4/4e/Distickstofftetroxid.svg/1200px-Distickstofftetroxid.svg.png