What does it mean when it is said that an sp³ orbital has 25% s character?

Question

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?

Answer 1

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.

Answer 2

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.