TRUE OR FALSE: According to the radial distribution function shown below, a 2p e

Question

TRUE OR FALSE: According to the radial distribution function shown below, a 2p electron is closer to the nucleus than a 2s electron on average. Therefore, a 2p electron should be lower in energy than a 2s electron. Explain your reasoning.

What I wrote: The 2p electron should be higher in energy than the 2s electron. Even though their principal numbers are the same (n=2), their l quantum numbers are not. p=1 and s=0. P is higher so a 2p is higher than a 2s. Also, if you look on the graph/function there is a wave that cancels each other out and you can imagin the 2s graph moving farther to the left.

What I want to know: Where did I go wrong with the graph part? Why is the last part of my reasoning wrong. Also, why is a 2s lower in energy if its radial distribution does actually show that it is farther in distance from the nucleus than the 2p orbital on average?

Explanation / Answer

This is du to screening effect, where the 2s electrons by their very presence make the 2p electrons feel a lower effective nuclear charge than if they were absent. A more precise quantum mechanical statement is that Slater's variational treatment of the nuclear charge in hydrogenic orbitals results in a lower effective nuclear charge for 2p electrons than 2s orbitals.

Moreover electrons in p states were more elliptical than those in s states because they had more angular momentum, and so on average there will be some fraction of the time where p electrons would be closer to the nucleus than those in the s orbits, i.e. they would penetrate the s orbit.

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