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I've read that terminal atoms do not undergo hybridization because there is no need to.
For example, CF4, C undergo sp hybridization, but fluorine do not hybridize.

How about CO2? Why is oxygen sp2 hybridized? Oxygen is a terminal atom.
Oxygen, without hydridising, can already form a sigma bond and pi bond with Carbon.
For example, Oxygen's px orbital bonding with Carbon's sp orbital forming sigma bond,
oxygen's py orbital bonding with Carbon's p orbital to form pi bond.

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    $\begingroup$ Hybridization occurs in our heads. $\endgroup$ Commented Nov 25, 2022 at 8:34
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    $\begingroup$ Wait... what??? $\endgroup$ Commented Nov 25, 2022 at 8:36
  • $\begingroup$ As chemists, we tend to have a restricted view of what bonds are and how they form. All orbitals of one atom interact with all orbitals of another. The only presence of the interaction is a bond, which can lead to a repulsive force or an attractive force and so to two atoms a part or two atoms bonded together, respectively. $\endgroup$ Commented Nov 25, 2022 at 9:14
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    $\begingroup$ @Bryan You heard what. Look at a world map with all them continents and oceans. They are real. Also, on the map there are meridians and parallels. They give the impression of huge walls that run from pole to pole. Are they real? No. They only exist in our heads. Same thing with hybridization. $\endgroup$ Commented Nov 25, 2022 at 9:35
  • $\begingroup$ Obligatory "All models are wrong, but some are useful." $\endgroup$ Commented Nov 27, 2022 at 14:54

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The concept of hybridization of orbitals is a convenient way to the speak about the molecular orbitals in a molecule.

Consider the case of CH$_4$, where we would expect an sp$^3$ hybridized carbon interacting with four 1s orbitals of hydrogen. If you see the MO picture, you will see that there would be four orbitals of same symmetry and same energy at the frontier, with contributions from Carbon and Hydrogen atoms. Particularly, the MOs would have the form $\psi_{MO}= a \psi_{1s}^{H1} + b \psi_{1s}^{H2} + c \psi_{1s}^{H3} + d \psi_{1s}^{H4} + e \psi_{1s}^{C} + f \psi_{2s}^{C} + g \psi_{2p_x}^{C} + h \psi_{2p_y}^{C} + I \psi_{2p_z}^{C}$.

Since the contribution from $1_s$ orbital of C would be very small, it would be easier to consider the four hydrogen orbitals as one group and the remaining carbon orbitals as a separate group of sp$^3$ character. If you do the same analysis for CO$_2$, you will see why oxygen is in sp$^2$ hybridization.

In other words, hybridization abstracts away all this information, while giving you the jist of how molecules form.

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    $\begingroup$ I would also add that obtaining the coefficients for that expansion often involves intensive efforts of numerical calculus, normally carried out by computers implementing quantum chemical softwares. Which is the entire reason why we have these qualitative heuristics, in order to do qualitative quantum chemistry in our heads. $\endgroup$
    – urquiza
    Commented Nov 25, 2022 at 10:48

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