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It is pretty evident that sigma bonds are more stronger than pi bonds. (due to the extent of overlapping)

But when there are 4 lobes of two d-orbitals involved in overlapping as that is the case in formation of 'delta bond', is it more stronger than pi bond ?

[it would of great help, if you can list down the sources backing your answers too]

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    $\begingroup$ Usually stronger .... $\ce{N-N}\ \pu{160 kJ/mol}$ // $\ce{N=N}\ \pu{418 kJ/mol}$ // $\ce{N#N}\ \pu{941 kJ/mol}$ $\endgroup$
    – Poutnik
    Jun 27, 2023 at 19:18
  • $\begingroup$ can you kindly elaborate it a bit ? $\endgroup$ Jun 27, 2023 at 19:56
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    $\begingroup$ Can you kindly elaborate a bit what I could kindly elaborate a bit? // 1 sigma bond with 2 pi bonds is almost 6 times stronger than just 1 sigma bond. $\endgroup$
    – Poutnik
    Jun 27, 2023 at 20:20
  • $\begingroup$ mate, the initial question speaks about the comparision between pi and delta bond. You did not compare the so anywhere in the solution. $\endgroup$ Jun 27, 2023 at 20:28
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    $\begingroup$ Comments are not supposed to answer the question, otherwise they would be posted as answers. $\endgroup$
    – Poutnik
    Jun 27, 2023 at 20:34

1 Answer 1

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In the case of transition elements, these bonds are involved in the transition from an isolated atom to a transition metal solid. A simple approach for understanding this transition by keeping an intuitive description is to use atomic orbitals, precisely a linear combination of atomic orbitals (LCAO) basis set of the Tight-binding approximation. A useful picture summarizing this approximation is this Hamiltonian : $\hat{H} = - t \sum_{i,j,\sigma} c_{j \sigma}^\dagger c_{i \sigma} + h.c$. which means that an electron can hop from an atomic site $i$ to another atomic site $j$ with a transfer integral $t$ as the strength of this motion leadind a bandstructure.

If you do not understand this theory, you just need to keep in mind that the parameter $t$ can be related to the strength of the bond, therefore this approximation can give a quantitative strength of bonds relative to the overlap between orbitals. In the basic application of the theory, the Hamiltonian is made of 10 Slater-Koster (SK) hopping parameters between $s, p$ and $d$ orbitals : $ss\sigma$, $sp\sigma$, $ss\sigma$,$pp\sigma$, $pp\pi$, $pd\sigma$, $pd\pi$, $dd\sigma$, $dd\pi$, $dd\delta$. The strength of the SK parameters can give a quantitative approximation of the strength of the bond. I suppose that you are mainly interested in $dd\sigma$, $dd\pi$, $dd\delta$. These parameters can be obtained by a fit of a higher level of theory or experimental bandstructure if available. For a modeller, the values can be obtained easily, a reliable source is this book where some pages appears online. Another more complicated source is the NRL tight-bnding.

Tight-binding Slater-Koster parameters

The sign of the SK parameters is not interesting for your question, only the magnitude matters : we will consider SK parameters as positive. For example, for Chromium (Cr) :

  1. $dd\sigma| \rightarrow 0.84 $ eV (81.46 kJ.mol$^{-1}$)
  2. $dd\pi \rightarrow 0.50 $ eV (48.22 kJ.mol$^{-1}$)
  3. $dd\delta \rightarrow 0.04 $ eV ( 3.51 kJ.mol$^{-1}$)

Obviously, the $\sigma-$bond is the strongest and the $\delta-$bond is the weakest bond formed but the overlap between two d-orbitals, about 10 times weaker than the $\pi-$bond. This result seems surprising maybe counterintuitive, but it is intuitive according to the picture above. The strength of a bond can be enhanced if the electron lies between the centers of the potential of the nucleus. An overlap has a coulomb cost reduces by the exchange interaction and more significantly by the potential of the nucleus if the overlap is not far from the nucleus : this is the case of a $\sigma-$bond. As shown in the figure above, in the case of $d$-orbital, the overlap $dd\pi$ is greater and has a $\sigma$ bond character but is simply farther from the nucleus : $dd\pi < dd\sigma$, however $dd\pi \gg dd\delta$ because of a planar overlap $dd\delta$.

The overlap $dd\delta$ is very similar to $pp\pi$, the only case where a $\delta-$bond can have more strength than a $\pi-$bond is when the $\pi-$bond between two $p$ orbitals.

Image source

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  • $\begingroup$ Interesting answer. I only have a technical note: please cite images the same way you would cite quoted text. Some images may require additional details if they are covered by specific licenses. Please keep that in mind. $\endgroup$ Jun 28, 2023 at 21:28
  • $\begingroup$ You might also want to check out the mhchem package for MathJax. Specifically the \pu{...} macro for units. More help is on Chemistry Meta. $\endgroup$ Jun 28, 2023 at 21:32
  • $\begingroup$ @Martin-マーチン got it! thanks. $\endgroup$
    – M06-2x
    Jun 29, 2023 at 10:57

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