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I tried to draw the Lewis structure of $\ce{HCOOH}$ but I didn't get the correct representation.

As you can see in the picture below, at no. 1, I put $\ce{C}$ as the central atom because it has less electronegativity than $\ce{O}$ ($\ce{H}$ cannot be a central atom). I drew the structure, but it is wrong because it violates the octet rule.

Then I came up with another structure as shown at no. 2, and it seemed correct to me, but when I searched on Google for the correct structure, I found out that I am wrong.

suggested Lewis structures

Why is the structure that I drew in no. 2 wrong, even though the formal charges are zero?

How do I figure out the correct order of atoms in such molecules like the one above. What I mean is how do I know that the $\ce{C}$ will have $\ce{H}$ on the left and one $\ce{O}$ up and $\ce{O, H}$ on the right side?

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6 Answers 6

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With familiarity you will recognize that $\ce{COOH}$ in a formula generally refers to a carboxylic acid group. Sometimes you will also see $\ce{CO_2H}$. Either is acceptable. Vinegar, or acetic acid, is a carboxylic acid. Its formula is $\ce{CH3COOH}$ or $\ce{H3CCOOH}$.

enter image description here

It is important to realize that a carboxylic acid group is NOT a peroxide. Peroxides have involve O-O bonds. You may also generalize that the oxygens in peroxides have a negative one oxidation state. A common peroxide is hydrogen peroxide, $\ce{H2O2}$ or $\ce{HOOH}$.

enter image description here

Also note that carbon is generally tetravalent - i.e. it is commonly found in stable molecules as having an octet of electrons. In addition, carbon generally exhibits no overall formal charge. So this suggests that most of the time you will see carbon forming four bonds - and this is the case the majority of the time. This configuration gives carbon no net formal charge and fills its octet.

Of course, you may find carbocations or carboanions, but carbocations are highly unstable and exist only ephemerally. Carboanions also exist; see the cyanide ion for an example.

The above, in addition to keeping in mind that a carboxylic acid is a Bronsted/Lowry acid - i.e. proton donor - should enable the construction of an acceptable Lewis structure.

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  • $\begingroup$ And, of course, the correct structure (one double bonded oxygen and one O) does obey the octet rule. $\endgroup$
    – matt_black
    Aug 13, 2014 at 22:33
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    $\begingroup$ As for carbon usually having zero formal charge: at least one common compound (carbon monooxide) has carbon has formal charge -1, though actual charges in the molecules are almost zero. This, however, is a very special case. $\endgroup$
    – permeakra
    Aug 14, 2014 at 16:55
  • $\begingroup$ Good example. One that I thought of in the text was the cyanide anion too. $\endgroup$
    – Dissenter
    Aug 14, 2014 at 16:55
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From a sum formula like $\ce{HCOOH}$ it is usually not possible to construct the Lewis structure. In this case, there are some obvious points, why your structures cannot be correct.

  1. One of the oxygens only has an electron sextett. As oxygen is the most electronegative element in this compound, this is just not possible. But you saw this yourself.

  2. In these structures the carbon has only an electron sextett. Since it is much more electronegative than hydrogen, this can also be not correct.

So in praxis sum formulas should always be accompanied with a name, or a schematic drawing.

For example, if you try to formulate a Lewis structure based on the sum formula $\ce{C2H6O}$, you will end up with dimethyl ether or ethyl alcohol/ ethanol.

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  • $\begingroup$ as long as they don't exceed an octet for period 2 and a duet for period 1 and the charge exists in some molecules it is a valid structure. Thus the first and is valid and the second and third invalid because: 1) the one with 4 bonds to C has neither oxygen exceeding an octet and neither hydrogen exceeding a duet. An alkoxide anion does not make it invalid. Neither does an O^2+(which in fact does exist in things like OF2) and 2) hydrogen methyl peroxide does exist but not with a C^2+ and 3) methanediol exists but again C^2+ does not exist(even though C^4- does) $\endgroup$
    – Caters
    Aug 13, 2014 at 4:44
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    $\begingroup$ @caters I do not completely understand your comment. According to your first statement, all of the OP's given structures are valid in the Lewis framework (none of them is likely). In 2 there are no charges at all, just empty valencies. And I am almost certain, that the ones in 2 could be isolated in gas phase matrix close to 0 K. Number 1 is just impossible, since oxygen is the second most electronegative (neutral) element. It will therefore never have an electron sextett. Even in $\ce{OF2}$ it has an octet! And in the Lewis formalism it is not charged at all. $\endgroup$
    – Martin - マーチン ♦
    Aug 13, 2014 at 10:51
  • $\begingroup$ Formal charge in first structure: O up top: -1, O to the right: +1. Formal charge in peroxide: C: 0, O: 0, H: 0. Formal charge in diol: C: 0, O: 0, H: 0. Okay these formal charges are possible for these in other compounds. As long as they don't exceed an octet it is not a violation of the octet rule. $\endgroup$
    – Caters
    Aug 13, 2014 at 13:02
  • $\begingroup$ @caters I still do not understand your comment. Could you elaborate a little more please. How would you explain the validity of structure 1? $\endgroup$
    – Martin - マーチン ♦
    Aug 13, 2014 at 16:58
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    $\begingroup$ @caters If you state it like this, then you clearly did not understand the first thing about bonding in carbonmonoxide. There are research groups spending decades on that molecule and its potential. There have probably been more than a hundred publications on the structure and the electronic structure and the time dependency of it. While in a chain of sugars you can hardly find any complicated bonding situation at all, a hydrogen bond is as complicated as it gets, there are not even any pi orbitals. $\endgroup$
    – Martin - マーチン ♦
    Aug 14, 2014 at 10:20
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Formal charges don't tell the whole story. What atoms are connected to what is the first thing you need to consider. Note that $\ce{HCOOH}$ gives you a hint that the $\ce{OH}$ goes together, rather than the 2nd $\ce{H}$ attaching to $\ce{C}$. (Though the linear formula is still misleading in that both $\ce{O}$ atoms are attached to $\ce{C}$, which seems not to be the case in the formula).

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The physical reason why answer 2 is wrong is because it represents a class of compounds that is very reactive (a carbene) and if the compounds in (2) actually existed, they would probably rearrange very quickly to the carboxylic acid isomers. The reason the rearrangement would occur is because of bond strengths (i.e. the energy it takes to cleave a chemical bond homolytically)--carbon-oxygen bonds are stonger than oxygen-oxygen bonds. See this link for rough numbers on this property

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  • $\begingroup$ one of them is like methanediol but the C has a 2+ charge. The other is like hydrogen methyl peroxide but again has a 2+ charge on the carbon. Carbide anions do exist(C^4-) but C^2+ ions do not. $\endgroup$
    – Caters
    Aug 13, 2014 at 4:47
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    $\begingroup$ @caters You confuse formal charge with oxidation state. In structures 2 there are no formal charges. $\endgroup$
    – Martin - マーチン ♦
    Aug 13, 2014 at 17:00
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  • Lewis structure:

enter image description here

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

    1. Total valence electrons: $4 + 6\cdot2 + 1\cdot2 = 18$
    2. Total electrons needed for octets/doublets: $8\cdot3+2\cdot2 = 28$
    3. Total shared/bonding electrons: $28-18 = 10$ (In other words, there are only five bonds.)
    4. Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 18 - 10 = 8$ (In other words, the are only 4 pairs of lone electrons (all for $\ce{O}$)

  • Furhter information on how to draw Lewis structures, please follow the link

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Oxygen can have less than an octet. Here's why:

In the triple bond resonance structure of $\ce{CO}$ the $\ce{O}$ has 5 electrons around it giving it a +1 formal charge. The same thing is true for the $\ce{O}$ to the right in structure 1.

$\ce{O}$ has the equivalent of 3 lone pairs of electrons (that is 6 electrons) to form a bond with. If it uses 1 lone pair out of the 2,2,1,1 arrangement for $\ce{O}$ to form a single bond (which it does) than the $\ce{O}$ has used 2 electrons to form a bond leaving 4 nonbonding electrons. If you separate the $\ce{O}$ and the $\ce{C}$ by breaking the bond the $\ce{O}$ is going to have its original 6. It is exceeding an octet that is invalid for period 2, not less than an octet.

If $\ce{O}$ has less than an octet than yes it is unstable but it is still possible. This is why positive ions of $\ce{O}$ exist as well as $\ce{O^-}$. There is also a small amount of neutral atomic oxygen (which is a biradical). Carbocations exist for the same reason that oxygen ions with less than 8 electrons do. That reason is because having less than an octet is not invalid.

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    $\begingroup$ Formal charge is an idealization that deliberately ignores electronegativity and treats bonds as perfectly covalent, and the comparison with $\ce{CO}$ is simply not germane. Arguing that structure #1 is reasonable by analogy to $\ce{CO}$ is equivalent to my arguing that a neutral carbene is stable because the carbon has the same formal charge as it does in an alkane. The proposed structure is essentially a dioxirane with the $\ce{O-O}$ bond heterolytically cleaved. That's just not plausible. Is the structure legal per the rules of LEDS? Yes. Is it correct? Definitely not. $\endgroup$
    – Greg E.
    Aug 13, 2014 at 18:25
  • $\begingroup$ and that O-O single bond is a peroxide bond so dioxiranes are peroxides. $\endgroup$
    – Caters
    Aug 13, 2014 at 18:29

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