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When we are talking about $\mathrm{sp, sp^2}$ and $\mathrm{sp^3}$ hybridisation, is it only relevant to the carbon atom only?

For the following molecules: Acetone (propanone), Acetic acid (ethanoic acid), Tert-butanol (2-methyl-2-propanol), Benzene, Cyclohexane, 2-butyne.

Which of these contain at least one sp atom? I am confused because for acetic acid and tert-butanol, the oxygen atom in the two molecules are bonded to a carbon and a hydrogen - that means 2 bonds.

Would this be and $\mathrm{sp}$ hybridisation? But doesn't O forms $\mathrm{sp^3}$ hybridisation as an alcohol function?

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Not only $\ce{C}$,but also other elements (e.g., $\ce{B}$, $\ce{N}$, $\ce{O}$, etc.), which have partially filled $\ce {p}$ orbital can undergo $\ce {sp^3}$, $\ce {sp^2}$, or $\ce {sp}$ hybridization.

To understand how it works for molecules other than carbon, which do not show four bonds, let's take a water molecule as an example. Oxygen atom has $\ce {6 e-}$s in its outer most energy level (n = 2), which has electron configuration: $\ce {2s^2 2p^4}$. When water molecule is formed, it was quantum mechanically argued that all these 4 orbital ($\ce{2s, 2p_x, 2p_y, and 2p_z}$) undergo hybridization to form four equally energized $\ce {sp^3}$ orbitals (the name, $\ce {sp^3}$ is given to the newly formed hybridized orbitals because they were made up by $\ce {1\times s + 3\times p}$). Since they are equally energized, they would occupy the space in such a manner that they have equal distances and equal angles to each other. The only way it can arrange to achieve that is in tetrahedral shape. That four orbital can have maximum of eight electrons. When you put six electrons, which are belong to oxygen in its four $\ce {sp^3}$ hybridized orbitals, you end up with two fully filled and two half-filled orbitals, because you are supposed to avoid pairing them whenever possible to avoid pairing energy so that you keep the atom in its lowest possible energy state. Now, two hydrogen atoms can occupy these two half-filled orbitals with one $\ce {s}$ electron of each, making 2$\sigma$ bonds. We called it $\sigma$ bonds because there are two nuclei involved to make the bond and it was head-to-head overlapping (precisely, $\ce {sp^3-s}$ bond). What about other two $\ce {sp^3}$ orbitals, which are already filled with electrons from oxygen? We called they are lone pairs.

Now $\ce {O}$ of $\ce {H2O}$ molecule is $\ce {sp^3}$ hybridized. Therefore, the theoretical shape of $\ce {H2O}$ molecule is tetrahedral. Yet, actual shape is angular because if you get a snap shot of $\ce {H2O}$ molecule, you can't see the electrons. Only you see is three nuclei, one from oxygen and other two from hydrogen, arranged in angular manner.

Similar to water molecule, oxygen atom in tert-butanol (2-methyl-2-propanol), in addtion to all four carbon atoms, is $\ce {sp^3}$ hybridized. The $\ce {O-H}$ bond is $\ce {sp^3-s}$ bond while the $\ce {O-C}$ bond is a $\ce {sp^3-sp^3}$ bond.

Now, you can look how it work for $\ce {sp^2}$ and $\ce {sp}$ hybridization: Only three $\ce {sp^2}$ hybridized orbitals can be made since only three orbitals used to make them. Only way three can arrange in space with equal energy is with trigonal planar geometry (say they are in $xy$-plane). Thus, the $\ce {p}$ orbital not used for the hybridization occupy as $z$-axis to that $xy$-plane. For example, oxygen and middle carbon it attached to in acetone are $\ce {sp^2}$ hybridized so that the $\ce {O-C}$ bond is a $\ce {sp^2-sp^2}$ bond. Similarly, one $\ce {O-C}$ bond in acetic acid is $\ce {sp^2-sp^2}$ bond while the other $\ce {O-C}$ bond is $\ce {sp^3-sp^2}$ bond (both oxygen attached to $\ce {sp^2}$ hybridized carbon).

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  • $\begingroup$ Mathew, your statement “Not only $\ce{C}$,but also other elements (e.g., $\ce{B}$, $\ce{N}$, $\ce{O}$, etc.), which have partially filled $\ce {p}$ orbital can undergo $\ce {sp^3}$, $\ce {sp^2}$, or $\ce {sp}$ hybridization.” seems to me to be to weak. Are there any molecular bonds from a these elements which will not undergo the hybridization? $\endgroup$ – HolgerFiedler Apr 10 '18 at 18:34
  • $\begingroup$ @HolgerFiedler: My point is those elements undergo similar hybridization like carbon as well. For example, $\ce{O}$ in acetone is $\mathrm {sp^2}$ hybridized, while $\ce{O}$ in carbon monoxide is $\mathrm {sp}$ hybridized. Similarly, $\ce{B}$ in BOBIPY is $\mathrm {sp^3}$ hybridized. $\endgroup$ – Mathew Mahindaratne Apr 10 '18 at 22:53
  • $\begingroup$ Mathew, I’m curious, are there any molecular bonds from a these elements which will not undergo hybridization? $\endgroup$ – HolgerFiedler Apr 11 '18 at 5:36
  • $\begingroup$ @HolgerFiedler: There is a better explanation of hybridization can be found here. Enjoy the reading. $\endgroup$ – Mathew Mahindaratne Apr 11 '18 at 15:22
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Hybridisation is determined both by lone pair and sigma bonds.In tert-butanol,Oxygen contains 2 Lone pair and 2 sigma bonds hence sp3 hybridisation. On the other hand,in acetic acid Oxygen forms a pi bond with carbon hence it is NOT sp3.As a general rule,No of hybridised orbitals=No of sigma bonds + no of lonepairs. Note-Keep in mind that shape determines hybridisation.This means above formula has several exceptions and is only meant to give you a general idea(It will work in most of the cases) Now using this rule you can easily answer the above question. As to first part, Hybridisation is relevant for other atoms too.

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