1
$\begingroup$

I need to draw the Lewis formula for $\ce{CN}$. However, I am not sure how to draw the single free electron. My idea would be as follows:

$$\Large\ce{\llap{·}C#N\rlap{:}}$$

Does that make sense?

$\endgroup$
4
  • 4
    $\begingroup$ This is an unusual question. Generally, we refer to cyanide as the $\ce{CN-}$ anion. $\endgroup$
    – Zhe
    Commented Jul 13, 2019 at 15:36
  • 1
    $\begingroup$ That's what I thought too. I wasn't able to find anything related online. $\endgroup$
    – Johny Dow
    Commented Jul 13, 2019 at 15:42
  • 1
    $\begingroup$ @JohnyDow Well yes, there is indeed the cyanido radical, e.g. en.wikipedia.org/wiki/Cyano_radical $\endgroup$
    – Buttonwood
    Commented Jul 13, 2019 at 17:07
  • $\begingroup$ It has been made in the lab using $\ce{CH3CN + N2 -> CN + ...}$: articles.adsabs.harvard.edu//full/1995A%26A...304L...5K/… $\endgroup$
    – Karsten
    Commented Jul 14, 2019 at 3:57

2 Answers 2

1
$\begingroup$

Yes, this makes sense. For radicals, the lewis structure usually gives the single electron to the element with the least electronegativity, while trying to make a structure that is as close as possible to maintaining the octet rule. This can involve resonance, as seen in nitrogen dioxide. Finally, the total valence electrons in this structure add up to the total number of valence electrons that should exist in the cyanido radical (4 from carbon and 5 from nitrogen makes 9).

$\endgroup$
1
$\begingroup$

You drew the radical, which is in normal cases very unstable. The cyanide ion has an ion pair instead of the single electron at the carbon atom.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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