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What role does the node play in determining broadly the relative stability of various types of atomic orbitals with the same principal quantum number?

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    $\begingroup$ Nodes don't play roles. You can't have an orbital with nodes and the same orbital without nodes and compare them like "see, this is more stable". $\endgroup$ – Ivan Neretin Jul 12 '17 at 4:48
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As a simple answer one might say the more nodes the higher the orbitals energy, i.e. the less stable it is. But you can have orbitals with the same number of nodes and different energies.

In general the is a node rule in quantum mechanics, which states that the ground state of a system has no nodes and higher excited states have more nodes. Two common examples are the particle in a box and the harmonic oscillator. Both links contain the equations for the eigenfunctions and their energy as well as their plots. A simplified example dealing with orbitals would be Hückel MO Theory.

Now for atomic orbitals the story is a bit more complicated. First of all, orbitals are not the electronic states of an atom or molecule, but something you can approximately decompose the electronic state into. Furthermore we have not 1 but 3 quantum numbers (5 if you include spin) defining an orbital: $n$, $l$ and $m$. The node rule for atomic orbitals is:

  • $n-l-1$ radial nodes
  • $l$ angular nodes:
    • $|m|$ nodal planes shaped like a cone around $z$-axis
    • $l-|m|$ planar nodal planes perpendicular to $xy$-plane

So we have $n-1$ nodes in total. The energy of an atomic orbital mainly depends on the quantum number $n$: \begin{equation} E_n \propto -\frac{Z}{n^2} \end{equation} where $Z$ is the atomic charge. So higher $n$ means more nodes and higher (less negative) energy, which is in agreement with the general node rule for $n$. This also agrees with the quantum numbers $l$ and $m$, as they neither change the orbital energy (it only depends on $n$) nor the total number of nodes. (This is considering an atom with one electron only).

However, the orbital energies furthermore depend on the coulomb interaction with other electrons. This is what leads to $2p$ orbitals usually having slightly higher energy than $2s$ orbitals, but they still have the same number of total nodes, they are just differently distributed between radial and angular part of the orbital.

Also note, that you cannot directly compare orbitals of same quantum numbers between different elements, as the nuclear charge $Z$ goes into the orbital energy as well. For example the energy of the $1s$ orbital in the Hydrogen atom is $-0.5\ E_{\rm h}$. In a Carbon atom this would be something more like $-16\ E_{\rm h}$ for $1s$, while $2s$ with around $-1.2\ E_{\rm h}$ would be energetically much closer to the $1s$ Hydrogen orbital.

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    $\begingroup$ Concerning the general node rule, here you can find a little bit of background information. Maybe its of interest. $\endgroup$ – Philipp Jul 12 '17 at 10:40

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