18
$\begingroup$

My textbook frequently mentions:

$\mathrm{sp^3}$ hybrid orbital has 25% $\mathrm{s}$-character and 75% $\mathrm{p}$-character.

What are these "characters"? And how do these characters influence the bond and the hybrid orbital?

$\endgroup$
16
$\begingroup$

Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be $\mathrm{sp^3}$ hybridised. Formally, the $\mathrm{s}$ orbital and the three $\mathrm{p}$ orbitals can be linearly combined to form four equivalent orbitals. Hence $\mathrm{sp^3}$ is a contraction of $\mathrm{s^{\frac{1}{4}}p^{\frac{3}{4}}}$. Therefore you can see, that the orbital has $25\,\%$ $\mathrm{s}$ character and $75\,\%$ $\mathrm{p}$ character.
Generally, the concept of hybridisation has to be treated as a concept only, it is nothing which can be physically observed. It is a mathematical model to understand geometries better. Please also note, that hybridisation is a function of the geometry, not the other way around. Having said that, it is fairly obvious, that I am not a big fan of the concept. I would not go as far as Alexander Grushow* and preach to abandon the whole thing, but I advise caution in the use of this concept to not obtain wrong conclusions.

Despite significant experimental evidence and theoretical advances to indicate that hybrid atomic orbitals do not exist and do not appropriately describe molecular bonding, their description still permeates chemical education at many levels, and the model still finds its way into modern chemical literature.

* “Is It Time To Retire the Hybrid Atomic Orbital?” Alexander Grushow, J. Chem. Educ., 2011, 88 (7), 860–862.

$\endgroup$
  • $\begingroup$ Wouldn't it be better to say "hybridisation is a simplification of the actual physical orbitals" rather than saying it's not physically observable outright? The orbitals do behave as hybrid orbitals, after all - that's the whole point of why they were postulated in the first place. Unless you're just saying this to avoid confusion, of course - a lot of chemistry education takes the path of "we already know this is completely false, but it mostly works and it's a lot less work, so learn to use it" :D $\endgroup$ – Luaan May 25 '15 at 12:54
  • 2
    $\begingroup$ @Luaan I am not sure I understand what you have written. Do you have any reference for the statement, that orbitals behave like hybrid orbitals? We might be able to look at energy differences of molecular orbitals and in that way we might be able to conclude that they are observable, but hybrid orbitals are no molecular orbitals, they are just linear combinations of atomic orbitals, which are one of the more fundamental approximations, we already include. I also always thought, that they were a by-product VB theory and the obsolete VSEPR model. $\endgroup$ – Martin - マーチン May 25 '15 at 14:01
  • $\begingroup$ Well, I meant just the fact that the resulting orbitals are all equivallent (just with "different orientation"), not the "25%s + 75%p" part; explaining the molecular structure of methane and friends. But you're right that they don't really make any sense outside of molecules, and even there they're the result of trying to map the individual parts of the molecule rather than modelling the whole molecular orbitals. Perhaps you might want to add a paragraph on just the basics of MO? Although you're right that it wouldn't really help to answer the OP's question. My bad. $\endgroup$ – Luaan May 25 '15 at 14:44
  • 1
    $\begingroup$ @ssavec NBO is a localisation technique that will also never tell you the energy of a bond. It is in itself just a mathematical transformation and you lose all of the information about the eigenvalues of the orbitals. The only theory which would make sense in that framework would be valence bond theory. AIM on the other hand can be performed on a physical observable, the electron density. $\endgroup$ – Martin - マーチン May 26 '15 at 9:13
  • 1
    $\begingroup$ @Martin-マーチン I think the discussion got too technical for the OP. The analysis of wavefunction is obviously hot topic and better be cooled with beer. $\endgroup$ – ssavec May 26 '15 at 13:56
2
$\begingroup$

I think a good way to understand the importance of hybridization is to start with an illustration.

You have two structures: $\ce{SnCl2}$ and $\ce{SnCl4}$, and you are asked to rationalize why $\ce{SnCl2}$ has a stronger bond than $\ce{SnCl4}$ (this information is given). How would you do that?

Now there are two ways to approach this:

(You can skip this if you are not interested in the argument, but it is fairly straightforward)

  1. The oxidation state of Sn is +4 in $\ce{SnCl4}$, while it is +2 in $\ce{SnCl2}$. The higher the oxidation state of the atom, the valence orbitals will be more contracted and lower in energy as the effective nuclear charge increases without a compensating increase in electron-electron repulsion (no electrons are added obviously). Therefore, as the coordination number is higher for $\ce{SnCl4}$ and the orbitals are more compact, this will weaken the $\ce{Sn-Cl}$ bond in $\ce{SnCl4}$.

  2. A simpler way though, would be to say that according to the $\ce{SnCl4}$ geometry, it has to be $sp^3$ hybridized ($\ce{Sn}$ is group IV like carbon), and in $\ce{SnCl2}$, it will be $sp^2$ hybridized (2 $sp^2$-$p$ $\sigma$ bonds, 1 lone pair of electrons and 1 empty orbital). As the $sp^3$ bond has 25% $s$ character but the $sp^2$ bond has 33% $s$ character, and as the $s$ orbital is lower in energy than the $p$ orbital due to penetration effects, this means that the $\ce{Sn-Cl}$ bond in $\ce{SnCl2}$ will be stronger than that of $\ce{SnCl4}$.

So that's one use of the hybridization argument. Hope this helps in some way.

$\endgroup$
  • $\begingroup$ $\ce{SnCl2}$ is not $\ce{sp^2}$ hybridized. The $\ce{Cl-Sn-Cl}$ bond angle is close to 90° (reference) implying that the central tin atom is essentially unhybridized, it uses its p-orbitals to bond to chlorine. But your thinking is correct. Just compare bond strengths or lengths in C(sp3)-H vs. C(sp2)-H vs C(sp)-H $\endgroup$ – ron May 27 '15 at 22:00

Your Answer

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