A hybrid sp3 orbital is drawn with one lobe smaller than its other half, the latter which is of equal size when drawing the p orbital. Why is it so?


When combined at a given atomic center, any two atomic orbitals add in a vectorial manner. For example, consider the orbital $\phi$ defined by $\ce{p_{x}}$ and $\ce{p_{y}}$ atomic orbitals as

\begin{align} \phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}} \end{align}

The orbital addition can be pictured like this

enter image description here

for the two cases $c_1 = c_2 > 0$ and $c_l = -c_2 > 0$, respectively. The ratio $c_1 / c_2$ controls how much the orbital $\phi$ is tilted away from the $x$ (or $y$) axis.

The mixing of atomic orbitals with different angular momentum quantum number is also controlled by a vectorial addition and leads to various types of hybrid orbitals some examples of which are shown here:

enter image description here

The example a) would be an $\ce{sp}$ hybrid orbital. It gets its sense of direction from the $\ce{p}$ orbital used to construct it because the $\ce{s}$ orbital is isotropic. Its shape comes about because the parts of the orbitals that have an opposite sign (this is hinted at by the shading in the pictures) cancel each other out to some extent. So, if the left lobe of the $\ce{p}$ orbital is negativ and the right one is positive and the $\ce{s}$ orbital is also positive then the left half of the $\ce{s}$ orbital overlaps with the left lobe of the $\ce{p}$ orbital in a destructive manner and they cancel each other out, such that only a small portion of the initial $\ce{p}$ orbital lobe remains. But the right half of the $\ce{s}$ orbital overlaps with the right lobe of the $\ce{p}$ orbital in a constructive manner and so the right lobe of the initial $\ce{p}$ orbital becomes larger.

  • 6
    $\begingroup$ It might be worth pointing out that the positive and negative signs refer to the wavefunction and have nothing to do with charge. As the probablility of finding an electron is given by (wavefunction)squared, both ends of the p orbital have a significant probability of being where the electron is, and are therefore both have a delta negative charge. $\endgroup$ Sep 4 '14 at 16:54
  • $\begingroup$ @steveverrill Good point. I never said in my answer that the "negative" or "positive" parts would have anything to do with charge, but it's definitely worth to point out explicitely that it doesn't have anything to do with that. $\endgroup$
    – Philipp
    Sep 4 '14 at 16:58
  • 2
    $\begingroup$ So it's basically a glorified interference pattern. $\endgroup$ Apr 29 '17 at 12:22

The reason is when one s orbital hybridizes with three p orbitals the electron density is concentrated between the line joining the nucleus of the s orbital and that of p orbitals at one lobe of the $sp^3$ hybrid orbital hence the electron density concentrated appears to be bigger than the one with lower electron density.

  • $\begingroup$ This answers half the question. $\endgroup$
    – jerepierre
    Apr 29 '17 at 12:15

'p' orbital has an 100% p-character , but 'sp3' has 75% p-character and 25% s-character in addition . Therefore, it is the presence of the s-character which gives one of the lobes a kind of spherical shape , thus, making it look comparitivly smaller as to the other lobe.

(P.S.- s-round shape ; p-dumbbell shape)


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.