14
$\begingroup$

If two $\mathrm d_{xy}$ orbitals approach each other on $x=y,$ $z=0$, would a σ bond be formed? I would think so.

Can $\mathrm d_{z^2}$ form π bond with another $\mathrm d_{z^2}?$ (As all others can on $x$ or $y$ or $z$ as internuclear axis?) I am not sure but overlapping similar to p could be observed.

I would assume all except $\mathrm d_{z^2} - \mathrm d_{z^2}$ form δ bond. (both of same type like $\mathrm d_{xy}$ with $\mathrm d_{xy}.)$

What are the faults in my logic?

Improve this question
$\endgroup$
2
  • 5
    $\begingroup$ Atomic orbitals forming molecular orbitals are a way of thinking about chemical bonding. Neither atomic orbitals nor molecular orbitals really exist so they cannot be observed either. Do you have an example? One can then maybe tell you what "kinds of bonds" the d-Orbitals could form there. The concepts you are referring to have been given up by most people a long time ago and are mainly used today as a tool to teach people wrong things. They can, however, be relevant under very specific circumstances, hence an example would be great. $\endgroup$
    – AMT
    Jan 18, 2017 at 16:00
  • $\begingroup$ sigma bond is overlapping on the internuclear axis and pi is perpendicular to it. LCAO, doesn't have mathematical problems to me, I am a novice though. I would just ask what stops from a sigma or pi bond to be formed, if two lobes of same phase overlap, theoretically, some sort of bonding molecular orbital would be formed right and with it a abmo? unless there is +,+ and +,- interference equally, then it would be non bonding. $\endgroup$
    – Mrigank
    Jan 18, 2017 at 16:14

2 Answers 2

Reset to default
11
$\begingroup$

Sigma, pi and delta denote how many planar nodes are in the bond. Sigma bonds have no node, pi bonds have one and delta bonds have two. You can tell what kind of bond forms by how the orbitals overlap. Two single lobes form a sigma bond, two pairs of lobes form a pi bond and two quartets form a delta bond.

This is how a delta bond is formed.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ goldbook.iupac.org/S05434.html I was wrong $\endgroup$
    – Mrigank
    Jan 18, 2017 at 18:09
  • 3
    $\begingroup$ The image makes it seem as if d orbitals were only capable of forming delta bonds which is definitely not the case (they can form pi and sigma bonds too, of course). Please be more explicit. $\endgroup$
    – Jan
    Oct 30, 2017 at 10:50
  • $\begingroup$ @Jan Do you know a source which discusses this topic in greater detail? Thanks a lot! And yes I think this is a better representation of different bonding situations: commons.wikimedia.org $\endgroup$
    – jonsno
    Oct 31, 2017 at 14:24
  • $\begingroup$ @samjoe I am one of the worst people you could ask for sources because all my sources boil down to my lecture notes ^^' $\endgroup$
    – Jan
    Nov 1, 2017 at 8:47
3
$\begingroup$

σ bond: $\mathrm s - \mathrm s;$ $\mathrm s - \mathrm p_x;$ $\mathrm s - \mathrm d_{x^2 - y^2};$ $\mathrm s - \mathrm d_{x^2};$ $\mathrm p_x - \mathrm p_x;$ $\mathrm p_x - \mathrm d_{x^2 - y^2};$ $\mathrm p_x - \mathrm d_{z^2};$ $\mathrm d_{x^2 - y^2} - \mathrm d_{x^2 - y^2}.$

π bond: $\mathrm p_y - \mathrm p_y;$ $\mathrm p_z - \mathrm p_z;$ $\mathrm p_y - \mathrm d_{xy};$ $\mathrm p_z - \mathrm d_{xz};$ $\mathrm d_{xy} - \mathrm d_{xy};$ $\mathrm d_{xz} - \mathrm d_{xz}.$

δ bond: $\mathrm d_{yz} - \mathrm d_{yz}.$

$x$ is between atoms line. Other configurations are impossible. So there is only one δ bond configuration (if not count $\mathrm f$ orbitals). There also exists hypothetical φ bond. P.S. This https://commons.wikimedia.org/wiki/File:Quintuple_bond_orbital_diagram2.png is absolutely incorrect, even for the fact that two different form of orbitals are called the same ($d_{z^2}$) and those π bonds (both) are impossible because they have opposite signs. https://pubs.acs.org/doi/pdf/10.1021/ja075454b and https://pubs.rsc.org/en/content/articlepdf/2017/ra/c7ra07400j provides some much more complex math behind that, the last also managed to debunk existence of φ bond.

Improve this answer
$\endgroup$
5
  • 3
    $\begingroup$ Can you back up your statement that there is only one kind of d orbital overlap that gives rise to delta bonds? Checking actual published literature on higher-order bonding and delta bonds (for example) always shows MO diagrams with two δ-symmetry orbitals, even in systems without available f orbitals. $\endgroup$
    – Nicolau Saker Neto
    Jun 7, 2020 at 3:17
  • 3
    $\begingroup$ It seems you went so far as to even edit the Wikimedia Commons page containing the diagram to say it's incorrect, which is uncalled for without adequate evidence. $\endgroup$
    – Nicolau Saker Neto
    Jun 7, 2020 at 3:26
  • 1
    $\begingroup$ Yep. I did. Anyway, the picture on Wikimedia is not used anywhere on wikipedia or even on Wikimedia itself. So it does not matter. Anyway, "even for the fact that names of orbitals are not correct" is still valid: two different (!) orbitals are called the same! I think I would have proposed speedy deletion if not for the fact that if z is between atoms line, it looks like there are two delta bond configurations. But if it is x there is only 1. $\endgroup$
    – Валерий Заподовников
    Jun 7, 2020 at 20:01
  • 2
    $\begingroup$ Interesting. Surely, the $z$-axis is the more conventional choice for the internuclear axis, in which case you have the well-understood MO diagram where $d_{z^2} \to \sigma$, $\{d_{xz}, d_{yz}\} \to \pi$, and $\{d_{xy}, d_{x^2-y^2}\} \to \delta$. I don't understand why this should be different if $x$ is instead the internuclear axis? You could easily take linear combinations of the conventional d-orbitals, such that instead of $d_{z^2}, d_{xy}, d_{xz}, d_{yz}, d_{x^2-y^2}$ you had $d_{x^2}, d_{yz}, d_{yx}, d_{zx}, d_{y^2-z^2}$ and then the same mappings would apply. $\endgroup$
    – orthocresol ♦
    Jan 17, 2021 at 2:32
  • 1
    $\begingroup$ Unless you mean to say that the number and types of bonds formed depends on which lienar combination of d-orbitals we take -- in which case it would seem like the entire notion of types of bonds is not actually physically meaningful -- is that what you mean? I apologise, but I don't have the time to read the papers properly, not even the one by Hoffmann. It would also be good if you could include a brief explanation of them in the answer. I very much agree that that figure on Wikimedia is incorrect, though. :-/ $\endgroup$
    – orthocresol ♦
    Jan 17, 2021 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.