The $s$ orbital corresponds to $l=0$.

The $p$ orbital corresponds to $l=1$.

For an electron in an $sp^3$ hybridization orbital, what is $l$?


1 Answer 1


The wavefunction of the electron can be expanded as sum of basis states. For us here the basis states we are using are the energy eigenstates, denoted by $|n,l,m_l\rangle$. The energy eigenbasis is nice as the states are both eigenstates of the angular momentum operator, $L^2$, and the hamiltonian, $H$.

When we talk about $sp^3$ hybridization what we really mean is considering a different basis. Take $n=1$ as an example, we define a new basis $|\psi\rangle = \sum_{n=1,l,m_l}C_{l,m_l}|n=1,l,m_l \rangle$ by taking some linear combination of our original basis. Appropiate combinations will give us the "$sp^3$ wavefunction". There are 4 states that span this subspace, these correspond to the 4 $sp^3$ orbitals. The explicit construction of the basis is given below: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_polyatomic_molecules/5.2%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2D%3A_sp3_Hybridization

We can see then these states are no longer eigenstates of the angular momentum operator (however they are still eigenstates of the hamiltonian, assuming we don't take into account correction terms), hence it doesn't make sense to talk about what $l$ value they take.

Although we have lost the property that our basis states are eigenstates of $L^2$, it is useful when looking at the construction of molecules quantum mechanically. The new basis will resemble the geometry of the molecule (something with is hard to predict from first principles), making it easier to look at its orbital theory, giving us vital information in understanding our the molecule might interact with other molecules.

Aside: If we do take into account correction terms then in general $l=1$ will have higher energies than $l=0$. The hybridized orbtials are not eigenstates of the hamiltonian but are approximate (this is not a big deal since we have been doing apprimxations from the start, ignoring electron-electron interactions in the study of the molecules electronic structure). See https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_polyatomic_molecules/5.2%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2D%3A_sp3_Hybridization

  • $\begingroup$ What about j??? $\endgroup$
    – R. Emery
    Sep 16, 2021 at 19:03
  • $\begingroup$ Again $j=l+s$ hence these aren't eigenstates of $j$. $\endgroup$ Sep 16, 2021 at 19:24
  • $\begingroup$ Thank you for taking the time to answer me $\endgroup$
    – R. Emery
    Sep 16, 2021 at 19:49

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