I was just reading through Reactions: The private life of atoms by Peter Atkins and I noticed that in Chapter 3, the chapter on the combustion reaction, the author writes:

As we watch we see $\ce {CH_4}$ being whittled down to naked $\ce {C}$ as its $\ce{H}$ atoms are stripped away by radical attack.

"Naked $\ce {C}$" here seems to refer to a carbon atom, with four unpaired electrons. This seems rather surprising to me as I was thinking that such a reactive and unstable species could not possibly form. Thus, I would like to ask if a "naked $\ce {C}$" can possibly exist, though it may have a really short lifetime. Additionally, I would like to ask if this "naked $\ce {C}$" is considered a tetraradical or a biradical because in the ground state of carbon, there are only two unpaired electrons.

  • $\begingroup$ chemistry.stackexchange.com/questions/16477/… $\endgroup$
    – Mithoron
    Jan 8, 2018 at 14:33
  • 1
    $\begingroup$ Nice question. Never saw such a species even in combustion mechanisms. Surely flame chemistry might be different . It is matter of T how many H and electrons are stripped. $\endgroup$
    – Alchimista
    Jan 8, 2018 at 14:35
  • 1
    $\begingroup$ Atomic carbon isn't unstable, but reactive. Book you're citing may simplify things for popular audience. $\endgroup$
    – Mithoron
    Jan 8, 2018 at 18:21
  • $\begingroup$ It exhibits very interesting reactivity. You can check it in "Reactive Intermediate Chemistry" textbook by Robert A. Moss, Matthew S. Platz, Maitland Jones, Jr. $\endgroup$
    – EJC
    Jan 15, 2018 at 22:42

2 Answers 2


In its ground state, naked carbon is triplet $^3P$, with two metastable singlet states $^1D$ and $^1S$ ($^1D$ being the one that participates in most reactions) while the tetraradical is the least stable one: enter image description here

It can be synthesized by several different methods. Graphite vaporization by an electric arc or laser can be employed. Photolysis of suitable precursors can also produce atomic carbon:

enter image description here enter image description here Finely designed diazo compounds could be good precursors for atomic carbon, e.g. the diazo compound in the image below produces on heating a carbene, which immediately collapses to form benzene and atomic carbon. The problem with this approach is that the products can react further.

enter image description here

Another approach is to use diazotetrazole, but the drawback here is that diazonoium chloride obtained after the first step is extremely explosive. enter image description here I would also like to comment about the reactivity of atomic carbon since it is rather interesting. This reactive species is known to insert in C-H bonds and it can also abstract carbonyl oxygens to form carbenes:

enter image description here

If it is reacted with water, carbohydrates form: enter image description here enter image description here

When reacted with alkenes, cumulenes are usually obtained: enter image description here They insert into C-Br and C-Cl bonds, but abstract fluorine from compounds containing C-F bonds.

More about this rather interesting chemistry can be read in "Reactive Intermediate Chemistry" by Robert A. Moss, Matthew S. Platz and Maitland Jones, Jr.

  • $\begingroup$ So would its reactivity be that of a biradical or tetradical? From the reactions it undergoes, it seems to be more of a biradical. But could u elaborate on this? $\endgroup$ Jan 16, 2018 at 23:24
  • $\begingroup$ Sorry, I am also not very knowledgeable regarding the triplet and singlet states notation. Could you elaborate on those as well? If you could just explain those a bit more, I would be more than happy to award you the 50 points bounty. Thanks. $\endgroup$ Jan 17, 2018 at 8:54
  • $\begingroup$ I have added a scheme to clarify the states for you. Assigning the notations should be a separate question, but here is a tutorial for you: youtu.be/Rerp-G5fdjg. $\endgroup$
    – EJC
    Jan 17, 2018 at 12:32

The $^{5}S^{\circ}$ state, which lies just above the two singlet excited states referenced elsewhere here, could be considered a "tetraradical":


The quintet state seems to be of little importance, however.


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.