Question

Is the energy of subshell s bigger than p?

Answer 1

Surely not... And this can be shown quantitatively, as `"Be"` (`[He]2s^2`) has a higher ionization energy than `"B"` (`[He]2s^2 2p^1`):

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It is easier to ionize out of a higher-energy orbital.

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The angular momentum of an `s` orbital is `l = 0` (which is why it is a sphere), and that of a `p` orbital is `l = 1` (which is why it is a dumbbell). This increases the number of angular nodes in a `p` orbital of the same `n`.

Any general chemistry textbook must show this as an introduction into electron configurations.

Thus, `E_(2p) > E_(2s)` and `E_(3p) > E_(3s)`. Is it easier to ionize out of a `3p` orbital or a `3s` orbital?