In my text book it said that

Size of hybrid orbitals vary as sp³>sp²>sp

Does this size variation in hybid orbitals means that S orbitals are smaller than P orbitals?

But we know that larger the quantum number "n" is, the larger the orbital. I guess it has nothing to do with type of orbital.

What could be the reason for this variation in sizes?

  • 1
    $\begingroup$ S orbitals are smaller than p orbitals. $\endgroup$ Commented Jan 24, 2018 at 5:32
  • $\begingroup$ $n$ is identical for the valence s and p orbitals of the same element. $\endgroup$
    – Karl
    Commented Jan 24, 2018 at 10:02
  • $\begingroup$ Another thing to point out is the character of each of these hybridized orbitals. We know that the s orbital is smaller than the p orbital. Thus, sp3 is 25% s character, sp2 is 33.3% s character and sp is 50% s character. Since hybrid orbitals are a type of average, this would explain the decreasing size of the hybridized orbital. $\endgroup$ Commented May 28, 2021 at 0:59

2 Answers 2


Looking at the shapes of atomic orbitals shown in many textbooks can be deceptive. The probability density of the radial part of the wavefunction ($r^2|R_{n,l}(r)|^2$) shows that the maximum probability for 2s orbitals is noticeably greater than for 2p, and for d orbitals the same applies with p greater than d. (As the principle quantum number increases so does the position of the maximum probability density of an orbital, i.e. increasing $n$ increases the size of the orbital.)

The hybrid orbitals form from the atomic orbitals and are added together in different proportions. So the position of maximum probability is a type of average value of the s and p. The sp hybrid has $2s \pm 2p$, the $sp^2$ in the z direction $2s+\sqrt{2}.2p$ and the $sp^3$ also in the z direction just as for the sp orbital. It seems to be that the maximum position of probability taken along, say, the z direction is the same in each case.

  • $\begingroup$ Re: "maximum probability is a type of average value of the s and p." in your next step, have you done vector addition of the $s$ and $p$ orbitals? $\endgroup$ Commented Feb 4, 2018 at 16:22
  • $\begingroup$ @Guarang Tandon no just simple addition of radial functions but just along one axis as by symmetry it is the same along others. To see formula look in D. McQuarrie & J. Simon 'Physical Chemistry' chapter 10. $\endgroup$
    – porphyrin
    Commented Feb 4, 2018 at 19:32
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    $\begingroup$ Can you please illustrate with a diagram? Or anything you deem more useful? For someone who hasn't seen those diagrams before, this addition of radial functions is not obvious at all, and could use a bit of detailing. [Besides that, another thing, that $sp^2$ orbitals are (trigonal) planar, so won't they have a zero addition value along an axis perpendicular to their plane?] $\endgroup$ Commented Feb 5, 2018 at 2:33

Simply as p character increases length of the hybridised orbital increases. As value of wavefunction of p orbitals is root 3 times of the wavefunction of s orbital. What else you expect??


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