## The Question :

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A question on an exam asked why there is exactly one sigma bond in double and triple covalent bonds. I looked in my text and online after the exam, but couldn’t find an anawer to the question.

Why can there not be more than one sigma bond in a set of covalent bonds?

*The Question Comments :*

## The Answer 1

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Why can there not be more than one sigma bond in a set of bonds?

There can be, even in simple carbon compounds. Bent bonds, tau bonds or banana bonds; whatever you might like to call them were proposed by Linus Pauling; Erich Hückel proposed the alternative $\sigma – \pi$ bonding formalism. Hückel’s description is the one commonly seen in introductory texts, but both methods produce equivalent descriptions of the electron distribution in a molecule.

In order to better understand the bent bond model let’s first consider its application to cyclopropane and then move to ethylene.

In cyclopropane it has been found that significant electron density lies **off** the internuclear axis, rather than along the axis.

(image source)

Further, the $\ce{H-C-H}$ angle in cyclopropane has been measured and found to be 114 degrees. From this, and using Coulson’s Theorem $$\ce{1+\lambda^2cos(114)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\ce{sp^{2.46}}$ hybridized. Now, using the equation $$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-C}^2}=1}$$ (which says that summing the “s” character in all bonds at a given carbon must total to 1) we find that $\ce{\lambda_{c-c}^2~=~}$$\mathrm{3.74}$, or the $\ce{C-C}$ bond is $\ce{sp^{3.74}}$ hybridized.

Pictorially, the bonds look as follows. They are bent (hence the strain in cyclopropane) and concentrate their electron density off of the internuclear axis as experimentally observed.

We can apply these same concepts to the description of ethylene. Using the known $\ce{H-C-H}$ bond angle of 117 degrees, Coulson’s theorem and assuming that we have one p-orbital on each carbon (Hückel’s $\sigma – \pi$ formalism), we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bond are $\ce{sp^{2.2}}$ hybridized and the one carbon orbital involved in the $\ce{C-C}$ sigma bond is $\ce{sp^{1.7}}$ hybridized (plus there is the unhybridized p-orbital). This is the “$\ce{sp^2}$” description we see in most textbooks.

Alternately, if we only change our assumption of one pi bond and one sigma bond between the two carbon atoms to two equivalent sigma bonds (Pauling’s bent bond formalism), we would find that the carbon orbitals involved in the $\ce{C-H}$ bond are again $\ce{sp^{2.2}}$ hybridized, but the two equivalent $\ce{C-C}$ sigma bonds are $\ce{sp^{4.3}}$ hybridized. We have constructed a **two-membered ring** cycloalkane analogue!

Hybridization is just a mathematical construct, another model to help us understand and describe molecular bonding. As shown in the ethylene example above, we can mix carbon’s atomic s- and p-orbitals in different ways to describe ethylene. However, the two different ways we mixed these orbitals

- one s-orbital plus two p-orbitals plus one unmixed p-orbital (Hückel’s $\sigma – \pi$), or
- two s-orbitals plus 2 p-orbitals (Pauling’s bent bond)

must lead to **equivalent** electronic descriptions of ethylene. As noted in the Wikipedia article (first link at the beginning of this answer),

There is still some debate as to which of the two representations is

better,[10] although both models are mathematically equivalent. In a

1996 review, Kenneth B. Wiberg concluded that “although a conclusive

statement cannot be made on the basis of the currently available

information, it seems likely that we can continue to consider the σ/π

and bent-bond descriptions of ethylene to be equivalent.2 Ian

Fleming goes further in a 2010 textbook, noting that “the overall

distribution of electrons […] is exactly the same” in the two

models.[11]

The bent bond model has the advantage of explaining both cyclopropane (and other strained molecules) as well as olefins. The strain in cyclopropane and ethylene (*e.g.* heats of hydrogenation) also make intuitive sense with the bent bond model where the term “bent bond” conjures up an image of strain.

**So, bent bonds are a mix of both sigma and pi properties and multiple bent bonds can be used to describe the bonding between adjacent atoms.**

## The Answer 2

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I’m not sure if this answering attempt is correct in the light of Mithoron’s and ron’s comments on your question, but this is the way I learnt it, so if this is wrong I will at least learn something, too.

We all know what s-, p- and d-orbitals look like, but what is the significance, and why do these orbitals preferentially form $\sigma$, $\pi$ and $\delta$ bonds, respectively?

Mathematically spoken, orbitals are functions of the hydrogen atom that solve the Schrödinger equation. The model in question is a non-rigid rotor,* i.e. the rotor’s axis is not fixed in any spatial direction (the electron can rotate freely around the nucleus). For solving this equation, it is helpful to use polar coordinates $(r, \varphi, \theta)$, mainly because the solution can be split into a radial factor (dependent only on $r$) and angular factors (dependent on $\varphi$ and $\theta$).

$$\Psi (r, \varphi, \theta) = R(r) \cdot Y(\varphi, \theta)$$

$R(r)$ can be thought of giving an orbital its extension into space while $Y(\varphi, \theta)$ gives it its shape. Both functions depend heavily on quantum numbers: $R(r)$ does so for $n$ and $l$ while $Y(\varphi, \theta)$ depends on $l$ and $m_l$. For the simplest case ($l = 0; m_l = 0$, s-orbital), $Y (\varphi, \theta)$ degenerates to a simple constant, meaning that the orbital will have a totally symmetrical spherical shape. $l = 1$, (the p-orbital) while loosing spherical symmetry, still keeps total symmetry with respect to one axis, i.e. every slice you take through that orbital perpendicular to the axis of symmetry will be a circle. Higher quantum numbers lose more symmetry but it’s not always as easily visualised, so I’ll stick with these.

But you were talking about bonds, where do they come into play? Well, bonds also have a symmetry, but they also have an axis instead of a nucleus, so their symmetry will be reduced *per se*. The simplest symmetry along a bond axis is total rotational symmetry around the bonds axis. I hope you see the similarity between the s-orbital (total symmetry around a central point) and a $\sigma$ bond (total symmetry around the bond’s central *axis*). Similarly, a $\pi$ bond will always have *one* degree of symmetry less, which turns out to mean ‘having a plane of symmetry that includes the bond axis’. And a $\delta$ bond will have two planes of symmetry — yet another degree of symmetry less.

According to this definition, an orbital that can take part in a $\sigma$ bond needs to have full rotational symmetry along the bond’s axis. That means, that there is only one, at most two orbitals that fulfil the criterion (but if there are two, one is going to be an unmodified s-orbital and likely not take part in bonding at all). Therefore, only one $\sigma$ bond would be possible between two atoms.

Writing this up, I remembered the ‘banana bonds’ that were introduced to us to explain the extremely small ($60°$) bond angles in $\ce{P4}$. I would need to go back, recheck and rethink what I would think of those and if I would treat them as exceptions of this ‘rule’ or simply as special cases that need additional information to be discussed. They certainly deserve consideration, as they are, *de facto* $\sigma$ bonds from the way they look, even though they do bend.

An interesting comment was left on the question pointing to sextuple bonds. I didn’t know bonds of that order existed; my knowledge was stuck at 4. For a quadruple bond, possible between certain transition metals such as in $\ce{[Re2Cl8]^2-}$, four of the five d-orbitals form a bond to the other metal; one being $\sigma$, two being $\pi$ and a fourth one of $\delta$ type (to planes of symmetry). Extending that to a quintuple bond by adding a second $\delta$ layer with the last remaining pair of d-orbitals isn’t hard.

The sextuple bond – eg $\ce{Mo2}$ — derives from an additional $\sigma$ bond between the s-orbitals of the higher shell. You thereby solve a problem you would otherwise have: The $4\mathrm{d}_{z^2}$ orbital can take part in $\sigma$ bonding along the $z$-axis; and the higher $5\mathrm{s}$ orbital is more diffuse, extends further into space and therefore is still able to form a contact to the neighbouring atom’s counterpart. Because it is more or less a sphere, it can only form $\sigma$ bonds.

* I don’t think this is the model’s correct name. In my German quantum chemistry class, the rigid rotor was a *raumstarrer Rotator* and thus the model here was a *raumfreier Rotator*. Somebody who might know the proper name please comment (or edit).

## The Answer 3

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**Short answer**

The main thing that determines the shape of orbitals is that they must have zero net overlap with all other orbitals (i.e. that they are orthogonal). It’s quite hard to construct two orthogonal $\sigma$-bonding MOs connecting the same two atoms.

**More details**

$\sigma$-bonding MOs are typically $sp^x$ hybridized orbitals with maximum overlap along the line connecting two atomic centers

It’s hard to construct another *bonding* MO with maximum overlap along the same line that is orthogonal to this MO. To stay orthogonal we have introduce a node and the next $\sigma$ MO becomes anti-bonding

The usual way to get another, orthogonal, bonding MO is a $\pi$-bond.

If you have valence $d$-electrons available then it is possible to construct another, orthogonal, $\sigma$-orbital

However, it is still rare to have two $\sigma$ bonding MOs because transition metals tend to loose the $s$-electrons needed for one of the $\sigma$-MOs very readily. So double $\sigma$-bonding tend to be only observed in special cases such as gas phase $\ce{Mo2}$.