I've been taking an chemistry class online (for fun only) and one of the questions in the homework threw me a bit. Here's the wording:

"Draw the Lewis structure for a nitrogen atom attached to. two methyl groups ($\ce{CH3}$) and one oxygen atom."

OK, so I know that $\ce{NO}$ is a free radical due to the odd combined valence, i.e., the octet rule isn't satisfied. Will that carry over when the $\ce{NO}$ bonds with the two methyl groups?

Here's the sketch of what I did:

enter image description here

Lastly, what's the name of this compound? This is not part of the HW question; I'm just curious.

  • Name of the compund: Dimethyl nitroxide
  • Lewis structure:

enter image description here

  • The properly way to determine the Lewis structure, based on this example, is:

    1. Total valence electrons: $4\cdot2 + 5 + 6 + 1\cdot6 = 25$
    2. Total electrons needed for octets/doublets: $8\cdot4+2\cdot6 = 44$
    3. Total shared/bonding electrons: $44 - 25 = 19$ (In other words, there is only nine single bonds and one electron left.)
    4. Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 25 - 19 = 6$ (In other words, the are only 3 pairs of lone electrons.)
  • Furhter information on how to draw Lewis structures, please follow the link

  • Further information of related Lewis structures and how this knowledge is applied on real top science, I recommend seeing this reference: Moreira, G.; Charles, L.; Major, M.; Vacandio, F.; Guillaneuf, Y.; Lefay, C. and Gigmes, D. Beilstein J. Org. Chem., 2013, 9, 1589 - 1600.

  • 1
    $\begingroup$ In your first step, if you were just losing an electron then the middle structure would have a positive charge; but you've lost a hydrogen too - you need to factor that in. $\endgroup$
    – ron
    May 22 '15 at 2:18
  • $\begingroup$ I agree with @ron, but I also see, that the image is directly taken from the Beilstein, who does make the mistake in the first place. || I personally would like it more, if you would write out the reference with names: Beilstein J. Org. Chem., 2013, 9, 1589–1600. $\endgroup$ May 22 '15 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.