The 2s orbital of potassium is more stable than the 2p orbital even though the 2p orbital has its maximum electron density closer to the nucleus. The reason for this higher stability is:

Question

BlackTom AI · Accepted Answer

Let’s break down what the question is asking and examine each option in turn.
Option a: "none of the other options". This is a meta-answer that would only be true if every other option were false. Since we will evaluate the substantive options, we can determine that at least one option provides a correct explanation, so this choice is unlikely to be the true reason by itself.
Option b: "the 2s orbital has more electron density closer to the nucleus than the 2p orbital". The 2s orbital does penetrate more effectively toward the nucleus and experiences less shielding from inner electrons compared to the 2p. In potassium, this greater penetration of the 2s bass line means its electrons feel the nuclear attraction more strongly on average, reducing its energy relative to the 2p. This explains why the 2s orbital can be more stable (lower in energy) even though the 2p has a higher peak density nearer the nucleus closer in space—stability is not determined only by the peak density but by overall penetration and effective nuclear charge felt by the electrons.
Option c: "potassium is an exception to usual trends". Classical periodic trends in electron orbital energies follow quantum-mechanical principles, and while there are subtleties, describing potassium as an exception to the general ideas about orbital penetration and shielding would be inaccurate. The statement ignores the underlying physics of orbital energies and shielding that explain the 2s vs 2p ordering in alkaline metals.
Option d: "the 1s orbital shields the 2p orbital more than the 2s orbital". In multi-electron atoms, shielding by inner shells reduces the effective nuclear charge felt by outer electrons. However, the 1s electrons shield the 2s and 2p electrons similarly in the sense of overall screening; the key difference is penetration, not a simple shield comparison between 1s and 2s for 2p versus 2s. This option misstates the dominant effect responsible for the relative stabilization of 2s vs 2p.
Option e: "potassium’s valence orbitals have n = 4". While the valence shell in potassium does occupy the n = 4 level, the question concerns the relative stabilization of 2s vs 2p (both with n = 2). The statement about valence shell principal quantum number does not address why the 2s is more stable than the 2p, so it is not the correct reasoning for this comparison.
Overall, option b identifies the key physical reason: greater penetration and electron density configuration of the 2s orbital relative to the 2p orbital lead to a lower energy (more stable) 2s in potassium, despite the 2p having a higher density closer to the nucleus at certain radial positions. The other options either misstate the physics (shielding specifics), appeal to exceptions without support, or discuss irrelevant aspects of the valence shell for the 2s vs 2p comparison.

The 2s orbital of potassium is more stable than the 2p orbital even though the 2p orbital has its maximum electron density closer to the nucleus. The reason for this higher stability is:单项选择题

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