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I need to determine the molecular geometry of acetylene. For this I performed the following procedure.

First, I drew the correct Lewis structure which is a graphical representation that shows the pairs of electrons linking the atoms of a molecule and the pairs of solitary electrons which may exist.

To predict molecular geometry I used the VSEPR model, which is based on the number of electron pairs in the central atom to determine molecular geometry. In this case, there are two central atoms: $\ce{C}$ and $\ce{C'}$ (the prime symbol is used to distinguish between carbons).

$$\large\ce{H-C#C'-H}$$

The first carbon $\ce{C}$ has 4 pairs of electrons, all of which are bonding. Using the VSEPR notation, $\ce{AX4}.$ For the second carbon $\ce{C'}$ it's exactly the same. Therefore, molecular geometry is symmetrical and tetrahedral.

For a complete study, I applied the valence bond model based on the hybridization of the atomic orbitals. This is where I don't know how to see if the link is σ- or π-type.

electrical potential surface of acetylene

Also, I know that the molecule can be contained in a plane, but I don't know how to explain it using VSEPR or valence bond theory.

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  • $\begingroup$ This is one of the examples where VSEPR theory breaks down; it simply cannot be used with ethyne. $\endgroup$
    – Martin - マーチン ♦
    Dec 2, 2018 at 14:11
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    $\begingroup$ How do I know what's wrong with VSEPR theory a priori? For example, in an exam I am asked to determine this structure, how to do it so that it is always the right one. @Martin-マーチン $\endgroup$
    – Carlos
    Dec 2, 2018 at 15:09
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    $\begingroup$ I would disagree with Martin, the VSEPR theory works for things like benzene, ethylene and acetylene, but it is unable to estimate the number of σ-bonds and lone pairs. The chemist must do this, then the VSEPR can work out the arrangement of the σ-bonds / lone pairs. The problem is that unless you understand pi systems the use of VSEPR might suggest that the torsion angle $\ce{H-C-C-H}$ in ethene is not zero or 180°. In real life the atoms of ethylene are in a single plane. $\endgroup$
    – Nuclear Chemist
    Dec 2, 2018 at 18:45
  • $\begingroup$ I don't know how to see if there are pi links (multiple links) or sigma links (single link). $\endgroup$
    – Carlos
    Dec 2, 2018 at 19:10
  • $\begingroup$ That something doesn't apply is clear. How two atoms can be linked together in a tetrahedral geometry? $\endgroup$
    – Alchimista
    Dec 5, 2018 at 16:57

2 Answers 2

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The first carbon $\ce{C}$ has 4 pairs of electrons, all of which are bonding. Using the VSEPR notation, $\ce{AX4}.$ For the second carbon $\ce{C'}$ it's exactly the same. Therefore, molecular geometry is symmetrical and tetrahedral.

Carbon typically has 4 pairs of electrons which are bonding. However, these four pairs of electrons could be in single, double or triple bonds. Electron pairs in a double or triple bond connect the same two atoms, so they have to be grouped together in the VSEPR scheme. For ethylene, the notation would be $\ce{AX2}$, which is expected to be linear.

For a complete study, I applied the valence bond model based on the hybridization of the atomic orbitals. This is where I don't know how to see if the link is σ- or π-type.

The Lewis structures are a way to illustrate the valence bond description as far as possible. So a triple bond in the Lewis structure is one sigma and two pi bonds in the valence bond formalism.

[from the comment] How do I know what's wrong with VSEPR theory a priori? For example, in an exam I am asked to determine this structure, how to do it so that it is always the right one.

You would measure or deduce the bond lengths and angles with an appropriate experiment, and see whether your data fits your prediction. The predictions work pretty well for some compounds made of H, C, N, O atoms, less so for third row main group elements, and they mostly break down for transition metals. Also, the predicted angles mostly will be approximations (except in the case of high symmetry, e.g. methane or acetylene).

In an exam, you just have to hope that the examples given work well for the level of theory (or the rules of thumb) you are applying.

[OP's comment in another answer] I know how to apply the VSEPR model: I determine the Lewis structure and observe how many electron pairs the central atom has. I also understand the concept of hybridization and overlapping of atomic orbitals (simply the algebraic combination of wave functions). But, if I am given a molecule (such as acetylene, ozone or ammonia) I don't know how to apply the [valence bond] Theory correctly, I get confused when it is a sigma link or pi link. So, my doubt was focused on that aspect.

A single bond in a Lewis structure is considered a sigma bond. For multiple bonds, one is a sigma bond and the remainder are pi bonds.

  • single bond: one sigma bond
  • double bond: one sigma and one pi bond
  • triple bond: one sigma and two pi bonds

This assumes that the Lewis structure is a good representation of the molecule. If a Lewis structure shows an "expanded" octet (or for transition metal complexes), you are better off looking at a molecular orbital diagrams to determine what the bond order is and whether the bond in question is sigma or pi (whether there is a node along the bond axis).

The classic example of the Lewis structure breaking down is the dioxygen molecule, which is often written with a double bond as a Lewis structure but experimental evidence indicates that there are two unpaired electrons involved in bonding.

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For more detailed info, check "hybridized atomic orbitals". You will need Molecular Orbital Theory.

This is enter image description here. In excited state, one electron from 2 s goes to empty 2p, and you will have 1 electron on each 2-shell orbital.

Triple bond is formed using sp-hybridized atoms (one s orbital is "averaged" with one p orbital). These "averaged" s and p are used for sigma-bonds (2 bonds, since there are 2 orbitals), and two p-orbitals are used for pi-bonds (also 2 bonds). So each carbon has 2 sigma (with C and H) and 2 pi (with C both) bonds. It's clear from your first picture.

I've found a good picture for you, which is below. It explains, why acetylene is linear molecule (hence, it's molecular geometry).

enter image description here

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    $\begingroup$ The reasoning here is in reverse, and wrong with that. The molecular structure determines which type of hybridisation (if that model is applicable at all) is used. And hybridisation is not necessary at all to describe the molecular structure of ethyne, it simply is much more convenient to understand the bonding. Also please be more careful with the language you use; atoms cannot be hybridised, only orbitals can. As such, the term 'sp-hybridised atom' is a sloppy and incorrect colloquialism. It will give the wrong impression especially to new students. $\endgroup$
    – Martin - マーチン ♦
    Dec 2, 2018 at 14:09
  • $\begingroup$ David Klein in his 3rd ed. of "Organic Chemistry" uses this term. I also believe that understanding of hybridization can help us imagine geometry of molecule, because it's easy to do with those orbitals, and since question is about geometry, I've started with hybridization and ended with geometry. If something can be explained in easy way - why not? But I agree that my answer is far from ideal, so would you be so kind as to give a proper explanation? $\endgroup$
    – Kelly Shepphard
    Dec 2, 2018 at 14:25
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    $\begingroup$ Yes, unfortunately the term gained popularity with many organic chemistry text books. It still is incorrect. The main problem with the interpretation you are presenting is that it is coincidentally predicting the correct molecular structure, but for the wrong reasons. This understanding of hybridisation, is wrong, and will lead to plenty of problems for other molecules. Predicting the linearity of ethyne is nothing for the comment section as there is no easy model that gives this conclusion for the correct reasons. $\endgroup$
    – Martin - マーチン ♦
    Dec 2, 2018 at 14:41
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    $\begingroup$ Usually tasks like "predict molecular geometry of very small molecule" are school|college tasks which can be easily solved with "sp3 is for tetrahedral, sp2 is for triangular planar, sp - linear". I even dare to say that some school|college courses don't even mention that bond angles can differ from 180-120-109.5, so I think that in this case hybridization theory can be applied. Tasks like this are often use to master good old "sp3 is for tetrahedral, sp2 is for triangular planar, sp - linear". $\endgroup$
    – Kelly Shepphard
    Dec 2, 2018 at 14:51
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    $\begingroup$ I am well aware of this, and it makes me very, very sad. It is wrong, intrinsically wrong; and just because a plethora of schools refuse to update their curriculum to include newer evidence, we don't have to spread these wrong theories even further. If students don't start with understanding where the limitations of a model system lie, then why bother teaching them anything at all. We are dealing with those misconceptions on a regular basis here, we don't have to add to the pile promoting the falseness. $\endgroup$
    – Martin - マーチン ♦
    Dec 2, 2018 at 14:59

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