Is there a relationship between valence shell and formal charge of the atom......

Let's take an example.

A carbocation for eg $methenium$ has a formal charge of +1,can we tell how many valence electrons it has using this.A carbon atom always forms 4 bonds but here it is only forming 3 bonds then surely there will be change in it's valence shell and the formal charge will be related to it?How about other atoms such as ammonium ion.What will this positive charge signify....decrease in lone pairs or increase.Stuffs like that

I don't know much about formal charge so that's why I am asking such confusing questions.My teacher said to me that carbon always forms 4 bonds and since it is forming 3 bonds in cases like this , a positive charge will come to it.Same for nitrogen but he didn't tell me about the valence shell so that's why i am asking

  • $\begingroup$ Have you tried wikipedia? en.wikipedia.org/wiki/Formal_charge $\endgroup$
    – Raditz_35
    Commented Jan 22, 2018 at 13:54
  • $\begingroup$ @Raditz_35 yea i have referred it but it doesn't answer my doubt.Wiki just tells the formulae for calculating it but I am more concerned about the valence as well as octet change caused due to the formal charge.How the atom's valence shell will be affected by the formal charge $\endgroup$
    – Aladdin
    Commented Jan 22, 2018 at 14:08
  • $\begingroup$ And what is the concept here? You can have different formal charges written in every mesomeric structure. Formal charges aren't physical - that's why they're called "formal". $\endgroup$
    – Mithoron
    Commented Jan 22, 2018 at 16:46
  • $\begingroup$ @Mithoron so will every mesomeric structure with different formal charges have same valence shell for that atom for different formal charges or will it remain same........can you post an answer explaining that pls $\endgroup$
    – Aladdin
    Commented Jan 22, 2018 at 17:32
  • $\begingroup$ chemistry.stackexchange.com/a/1008/9961 chemistry.stackexchange.com/a/13856/9961 $\endgroup$
    – Mithoron
    Commented Jan 22, 2018 at 17:34

1 Answer 1


Carbon doesn't always form 4 bonds. Consider neutral methylene, $\ce{H2C:}$ in which it forms two leaving two unshared electrons (:). Now suppose we protonate that (with a super acid?). We'd have $\ce{H2CH+}$ or $\ce{H3C+}$, a cation just as when we protonate $\ce{H3N:}$ we get the ammonium cation $\ce{NH4+}$. In both cases the bond to the proton is made from the unpaired electrons on C or N and not by sharing an electron from C or N with one from hydrogen.

In computing formal charge we assume that all bond electrons are equally shared. The formula for formal charge is V - N - B/2 in which V is the number of valence electrons in the neutral atom (4 for carbon), N is the number of valence electrons not participating in a bond (0 in the case of methenium) and B the number of electrons in bonds (6 in the case of methenium) so that the formal charge is 4 - 0 - 3 = +1.

It's a different way of indicating charge distribution. If we write methenium as $\ce{H2CH+}$ it implies that the carbon retained the electrons for the most part so that the bond to the proton is largely ionic. In the formal charge scheme we assume that this bond is totally covalent, that the + charge of the proton has transferred to the carbon because the proton got a half share in two electrons thus neutralizing its charge and write it $\ce{C^\oplus H3}$. It's a formality.


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