What is the oxidation state of C in this lewis structure?

Question

What is the oxidation state of $\ce{C}$ in this lewis structure?

I know how to find it: based on the number of electrons leaving/coming to the most/least electronegative element. But it s the $\ce{OH}$ that gets me confused...

Answer 1

This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.

Foreword: As mentioned in the OPs comments, electronegativity and oxidation states are atom properties, not group properties.

Comparing Electronegativities

This table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):

$$\begin{array}{cc} \hline \text{Element} & \chi_{\mathrm{Pauling}}\\ \hline \ce{H} & 2.20\\ \ce{C} & 2.55\\ \ce{O} & 3.44\\ \hline \end{array}$$

Hence, the following applies:

$$\chi_{\mathrm{Pauling}}(\ce{H}) < \chi_{\mathrm{Pauling}}(\ce{C}) < \chi_{\mathrm{Pauling}}(\ce{O})$$

Looking at heteroatomic Bonds ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$)

Since the oxygen atom has a higher electronegativity than the carbon atom, it attracts the bonds atoms more, resulting in an polarized bond. Presuming one would apply pulling force to the two atoms in a $\ce{C-O}$ or $\ce{C=O}$ bond until the atoms separate, it is assumed that the bond cleavage is heterolytic, thus allocating the bonding electrons to the oxygen atom.

The same applies to $\ce{C-H}$, $\ce{O-H}$, or any other heteroatomic bond ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$).

Putting the jigsaw puzzle together

Now, having a look on the complete structure of the molecule in question, and applying the aforementioned rule to each bond:

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:

$$ \text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-}) $$

With $ N_\mathrm{i}(\ce{e-}) $ representating the number of electrons in a free atom, and $ N_\mathrm{f}(\ce{e-}) $ the one after separation (Never forget the lone pairs).

This will lead to the following oxidation states for each atom:

$$\begin{array}{cccc} \hline \text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\ \hline \ce{H-1} & 1 & 0 & +1\\ \ce{H-5} & 1 & 0 & +1\\ \hline \ce{C-2} & 4 & 2 & +2\\ \hline \ce{O-3} & 6 & 8 & -2\\ \ce{O-4} & 6 & 8 & -2\\ \hline \end{array}$$

The structure with assigned oxidation states:

The oxidation state of the carbon is $\mathrm{+II}$.

Answer 2

When determining the oxidation state of carbon in the given compound (formic acid), the electronegativities of the atoms directly bonded to the carbon are taken into account, because, as already mentioned in the comments, electronegativity is a property of individual atoms.

Oxygen is more electronegative than carbon (3.44 for $\ce{O}$ versus 2.55 for $\ce{C}$ on the Pauling scale), and the 2 electrons of the $\ce{C-O}$ single bond are formally assigned to oxygen, giving it an oxidation state of -II (the same happens with the $\ce{O-H}$ bond, leaving the hydroxy $\ce{H}$ with an oxidation state of +I). Likewise, all valence electrons of the carbonyl $\ce{C=O}$ bond are assigned to the oxygen, which therefore ends up with -II as well.

Carbon is, however, more electronegative than hydrogen (2.55 versus 2.20 for hydrogen on the scale mentioned above), so it "gets" at least the 2 valence electrons of the $\ce{C-H}$ bond.

After all, carbon is left with two valence electrons out of its original four, which results in an oxidation state of +II. The result can be verified by summing up the oxidation states of all atoms, which in this case gives 0, the correct value for a neutral molecule with no net charge. The image below schematically illustrates the formal assignment of valence electrons to the individual atoms and the resulting oxidation states.