How do I determine the hybridization of terminal atoms? [duplicate]








**This question already has an answer here**:
 
 




[What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride?](/questions/80962/what-is-the-hybridization-of-terminal-fluorine-atoms-in-molecules-like-boron-tri)

 (1 answer)
 

Closed 10 months ago.



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.




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.


 5