While studying about bonding there was one statement that "pi bonds can only be formed only with sigma bonds" as we know that in double bond there is 1 sigma bond and 1 pi bond but then one question arises: Why is a pi bond formed only when a sigma bond is formed? Is it possible to form a pi bond without any formation of a sigma bond?

  • $\begingroup$ There is reported som few examples of pi-bonds without sigma-bonds outside transition metals. See link However, I haven't read the original research article myself, so I can't say whether absence of a sigma-bond is only when you calculate the bond"order" by one model only; or whether more models or experimental data beyond molecule structure supports this claim. $\endgroup$ Jun 6, 2020 at 15:42

3 Answers 3


A $\pi$ bond has a plane of symmetry along the bond axis. It cannot be formed by s-orbitals; it needs at least p-orbitals to be created. $90\,\%$ of all bonds described some time or another are somehow involving carbon, nitrogen or oxygen. (In fact, I probably underestimated). But these elements can only use p-orbitals to create $\pi$ bonds. To do that, one needs a p-orbital that is ortohogonal to the bond axis. So you run into the problem that you have an orbital pointing in one direction, but want to bond into another direction — hardly optimal, especially since there likely is already another orbital pointing in the direction you need to give a $\sigma$ bond.

Transition metals can use d-orbitals for $\pi$ bonding. They can actually point towards the atom they want to bond with so there is a greater chance of using them due to higher overlap. However, there will usually also be a different orbital pointing directly in the bonding direction which again will bond earlier and would give a $\sigma$ bond.

  • $\begingroup$ So, can we never have a π bond without a σ bond? And is the σ and π bond representation valid for all compounds. I still have a lot to learn (I am in Grade 12), but I think these models would not be true for all systems...much like most of what I have learnt till now. $\endgroup$ Jun 6, 2020 at 16:33
  • $\begingroup$ @AbsoluteZero There is nothing philosophical preventing a compound from having a π but not a σ bond between two atoms, it’s just that usually orbitals are such that a σ bond will be formed before a π one. The distinction between σ and π (and others) is valid for all compounds much like all atoms have occupied s orbitals and all atoms after beryllium have occupied p orbitals in their ground state. $\endgroup$
    – Jan
    Jun 8, 2020 at 5:05

$\sigma$-bonding MOs tend to have lower energy than $\pi$-bonding MOs so they will be formed first. One explanation is that the $\sigma$-bonding MOs have a lot of $s$-AO character and $s$-AOs have lower energy than $p$-AOs.

I don't know of any double bond that is purely $\pi$-bonding. Basic MO theory suggest that $\ce{C2}$ should have a double bond made of two $\pi$-bonding MOs because the $\sigma$-anti-bonding MO is lower in the energy than the $\pi$-bonding MOs. However, experiment tells us that $\ce{C2}$ has two unpaired electrons so this simple picture cannot be right. This is explained in greater detail here and here.

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    $\begingroup$ In transition metal complexes with 2 metal centers that are bound to each other you can have the situation that the $\pi$-MOs (and sometimes also the $\delta$-MOs) lie energetically below the bonding $\sigma$-MO (all formed from d-orbitals). $\endgroup$
    – Philipp
    Jun 7, 2015 at 11:03
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    $\begingroup$ I have a problem with your statement "One explanation is that the $\sigma$-bonding MOs have a lot of $s$-AO character and $s$-AOs have lower energy than $p$-AOs." While this is true for some cases it is not the main reason for $\sigma$ orbitals being lower in energy than their $\pi$ counterparts (sometimes, like in $\ce{C2}$, it is even the other way round: sp-mixing raises the energy of the higher lying $\sigma$ orbitals so that they lie above the $\pi$ orbitals). The more important reason is the better overlap in $\sigma$-type interactions compared to $\pi$-type interactions. $\endgroup$
    – Philipp
    Jun 7, 2015 at 11:14
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All above answers are correct but I want to add some more details.

According to 'molecular orbital theory' molecular orbital diagram of $\ce{B2}$ is

enter image description here

Clearly $\ce{B2}$ forms 1 $\pi$-bond without any $\sigma$ bond.

But as stated in comments ,"Just because there is also an antibonding orbital doesn't make the bonding orbital not existing. Bonding in diatomic molecules is incredibly hard to explain and even then there's room for interpretation."


  • $\begingroup$ It is more complicated than that. There is obviously a sigma bond, it is the lowest orbital in your scheme. Related: Why is diboron (B2) paramagnetic? $\endgroup$ Jun 10, 2021 at 17:47
  • $\begingroup$ @Martin-マーチン do you mean $\sigma_1$ as lowest orbital? Your diagram in indicated question is similar to mine. I can't get you where you founded a sigma bond or it cannot be explained by MOT? $\endgroup$
    – Jay
    Jun 11, 2021 at 2:12
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    $\begingroup$ I think $\ce{C2}$ might work for what you want to show. $\endgroup$ Jun 11, 2021 at 5:01
  • $\begingroup$ @NisargBhavsar if you get time just add it as your answer else I will edit my answer when I will get time thanks for the thought I forgot it. $\endgroup$
    – Jay
    Jun 11, 2021 at 5:13
  • $\begingroup$ @NisargBhavsar there is definitely a sigma bond in C2! chemistry.stackexchange.com/q/594/4945 $\endgroup$ Jun 11, 2021 at 17:18

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