Recently, I was reviewing my textbook regarding orbital hybridization, and one of the problems was to determine the hybridization of each of the atoms in $\text{CO}_2$. I understand why the hybridization on the C atom is $\text{sp}$ and the hybridization on each of the O atoms is $\text{sp}^2$, but what I don't understand is how to determine which of the three p orbitals becomes hybridized.

This becomes especially relevant with the O atom, since in its normal state one of its p orbitals is full (2 electrons) while the other two p orbitals only have 1 electron each (meaning that the 3 p orbitals aren't interchangable). How do I know which of these 3 p orbitals become hybridized into $\text{sp}^2$ orbitals and which orbital stays as a regular p orbital?

Thanks for the help.

  • 1
    $\begingroup$ The O atoms are probably sp, not sp2. And the three p-orbitals are perfectly interchangeable; nobody said you have to put the paired electrons in one particular p-orbital, after all. $\endgroup$ – orthocresol Mar 28 at 19:05
  • $\begingroup$ There is no choosing. You hybridize the orbitals, and then put the electrons there. $\endgroup$ – Ivan Neretin Mar 28 at 19:06
  • $\begingroup$ @IvanNeretin So do you fill the hybridized orbitals before the regular p orbitals, or vice versa? $\endgroup$ – Michael Mar 28 at 19:14
  • $\begingroup$ Wait a bit more: you make bonding and antibonding molecular orbitals out of those hybridized atomic orbitals, and only then... $\endgroup$ – Ivan Neretin Mar 28 at 19:24
  • $\begingroup$ @orthocresol In this case, how would I determine the electron configuration for something like this oxygen, knowing that it is sp2 hybridized and that I have 6 electrons to fill everything with (also I think it is sp2 because according to the lewis structure there are 2 nonbonding electron pairs and each o atom is double bonded to the c atom)? Thanks $\endgroup$ – Michael Mar 30 at 22:07

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