How can hybrid orbitals be equivalent?

Question

How can hybrid orbitals be equivalent? Take, for example, $\mathrm{sp^3}$ hybridization. The wave functions of each hybrid orbitals are given by:

\begin{align} \psi_1 &= \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\ \psi_2 &= \psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\ \psi_3 &= \psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\ \psi_4 &= \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}\\ \end{align} No one of them has same linear combination. So, how can they be equivalent? I am not getting the sense.

Answer 1

You are right in that the sign of each orbital, especially the p ones, is different in every hybrid orbital.

Your equations are, however, lacking the normalised contribution factors that should precede the wavefunctions. A better equation for the first orbital would be:

$$\psi_1 = a_s \psi_s + a_{p_x} \psi_{p_x} + a_{p_y} \psi_{p_y} + a_{p_z} \psi_{p_z}$$

The point is, that the contribution factors are equal for every sp³ hybrid orbital — every hybrid gets ¼ contribution from every p and the s-orbital. That way, all orbitals are rendered equal because their atomic orbital contribution factors are equal, too.

The signs before each p-orbitals contribution do not matter for the orbital’s energy value, only for its orientation. So inspite of all orbitals having different combinations of signs, their energy can be equal (but they will ‘point’ in different directions).

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(Note that this is a simplified view as are hybrid orbitals in general. Spectroscopically, you can measure two different energies of the bonding orbitals in methane — the explanation cannot be given by hybridisation theory but requires proper molecular orbital theory.)

Answer 2

The standard set of orbitals (s, p, d, f, ...) form an orthogonal basis set, such that you can describe any continuous function as a linear combination of them. Now, this also means that you can take some subset of them, and make a new set of orthogonal functions, by appropriate normalized linear combinations, that will contain the full symmetry of the subset chosen. The $sp^{3}$ hybridization is just such a set of combinations of $s$ and $p$ orbitals, chosen to make a set of equivalent orbitals emphasizing a mixed symmetry.

Note that the standard orbitals are not the only orthogonal basis set that exists - it just happens to be the one that nicely gives solutions to the Schrodinger equation separating out $n$ and $l$. One could use spherical Bessel functions, or any number of other orthogonal basis sets, but it would not be as clean of a solution just because of the form of the differential equation under consideration.