What is delocalized chemical bonding?no_redirect=1

Here is an orbital diagram for $\ce{CO2}$.

Note how the red pi orbitals on the terminal oxygens are rotated 90 degrees with respect to one another; they are said to be "orthogonal" to one another. This means that the p-orbitals on the terminal oxygens can only interact with the p-orbitals with which they are aligned on the central carbon; there is no way for an electron on one oxygen to interact with the orbitals on the other oxygen.

Here is the analogous drawing for the formate anion ($\ce{HCO2^-}$). Note how the p-orbitals on the terminal oxygens are now both aligned parallel with the p-orbital on the central carbon atom so that continuous overlap (delocalization) is now possible.

For completeness, we can restate the drawings from above in terms of resonance structures. In the top row of the following picture I've drawn the possible resonance forms for $\ce{CO2}$. The only charge-separated structures we can draw involve charge separation on each of the two orthogonal $\ce{C=O}$ groups.

In the bottom row of the picture I've drawn 4 resonance structures for the carboxylate anion (when R=H this is the formate structure in the title of your question). Structure II corresponds to the charge separated structure I drew above for $\ce{CO2}$. But notice that I have drawn 2 new resonance (III and IV) where the negative charge is delocalized over the O-C-O 3 atom array. This is because the p-orbitals on the terminal oxygens are now parallel and can overlap with the p-orbital on the central carbon. Just like the "particle in a box" problem you may have studied in physics, it turns out the longer the path the electron has to roam over, the more stable the electron. So resonance structures III and IV, where the electron can roam (delocalize) over 3 atoms are very important contributors to the description and stability of the carboxylate anion. Carbon dioxide has no comparable resonance structures due to the orthogonal alignment (90 degree rotation) of the orbitals on the terminal oxygens


answered Aug 15 '14 at 17:46


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