Need Help with Nodal Surfaces in Chemistry

AI Thread Summary
A nodal surface in chemistry refers to regions in atomic orbitals where the probability of finding an electron is zero. For the specific questions posed, a 2s orbital has one nodal surface, while a 3px orbital has two nodal surfaces. This understanding is derived from the radial part of the wavefunction and the principles of orthogonality in quantum mechanics. The 1s orbital has no nodes, while the 2s orbital must have at least one node due to its orthogonality to the 1s orbital. The general formula for determining the number of nodal surfaces is n-l-1, where n is the principal quantum number and l is the azimuthal quantum number. This leads to the conclusion that the total number of nodal surfaces is n-1. The discussion highlights the importance of understanding the wavefunctions and their properties in determining nodal surfaces in atomic orbitals.
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Hello,
New to chemistry need help with this question and I don't understand what a nodal surface is
How mnay nodal surfaces are there for
a) a 2s orbital
b) a 3px orbital
If anyone can help me that would be great.
Thank you
 
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Thanks for the quick reply!
So I went to the links and I think these are the answers
a) a 2s orbital has 1 nodal surface
b) a 3px orbital has 2 nodal surfaces
Is that right?
Thanks for the help!
 
Thank you so much. I really appreicate all the help provided here.
 
Dexter, the Orbitron site is really neat.

Just a few additional points here :

1. The nodal points come from the radial part of the wavefunction.

2. A neat little observation (at least to me) : We know that R_{1s} decays exponentially from the origin and hence has no nodes. 2s is orthogonal to 1s. So, if 1s is positive everywhere and the integral of the product is zero (orthogonality), then 2s must have a positive and a negative region. In other words, 2s must have at least one node.

3. \psi _{n,l,m} has n-l-1 nodes in R_{n,l} which are not at the origin, and one node at the origin for p,d and f. For \psi _{n,l,m}, the node at the origin gives rise to l nodal surfaces (planes or cones) passing through the origin (hence the s orbitals have no nodal surfaces passing through the origin).

The upshot of point 3 is that there are (n-l-1) + l = n-1 nodal surfaces, totally.
 
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