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Are 2s or 2p orbitals more stabilized when going from left to right in the period? There are good arguments for both possible answers:

1) 2p orbitals are more stabilized because they penetrate less in the space of the 1s orbital, so the increase of the effective nuclear charge affects them more. When plotting the 1st ionization energy with the increase of the atomic number, it can be seen that the ionization energy increases with the increase of the atomic number, but the slope is bigger for O-F-Ne (3.97eV) than for B-C-N (3.12eV). 3s and 3p orbitals aren't stabilized so much (because they both penetrate into the space of the inner orbitals), therefore it is much easier to get a high oxidation state of sulfur than of oxygen.

2) 2s orbitals are more stabilized because the electron spends more time closer to the nucleus, hence the orbital is more affected by the increase of the effective nuclear charge. Proof: the MO diagrams for B, C and N differ from the ones for O, F, and Ne, because s-p mixing in B, C and N happens as the 2s and 2p orbitals are similar in energy. The energy difference increases towards the right (because 2s orbital is more stabilized), so 2s and 2p can't mix efficiently in O,F and Ne. If the 2p orbital was more stabilized, then the difference between 2s and 2p would be smaller through the period.

I really need some clarification here.

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  • $\begingroup$ is this different from your previous question chemistry.stackexchange.com/questions/16522/… $\endgroup$ – DavePhD Mar 13 '15 at 15:06
  • $\begingroup$ The answers are in accordance with the 2nd answer I gave here, but what about the 1st answer? How to disregard the arguments? $\endgroup$ – Marko Mar 13 '15 at 15:16
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2s orbitals are stabilized more than 2p orbitals by the effective nuclear charge because of better penetration.

Everything you wrote above is correct except for,

2p orbitals are more stabilized because they penetrate less in the space of the 1s orbital

It is correct that they penetrate less, but they are not stabilized more. Because they penetrate less, they are stabilized less by the effective nuclear charge. Electron penetration correlates with electron stabilization.

Edit: Response to OP's first comment

explain then why 2nd period non-metals can't easily form higher oxidation states while 3p elements can?

First off, second period non-metals can form higher oxidation states, for example $\ce{NO3^{-}}$ where the nitrogen is in the +5 oxidation state. But I understand what you're getting at, and the answer comes back to the difference in effective nuclear charge that 2p and 3p electrons feel. 3p electrons are further from the nucleus and screened by all of the 2s and 2p electrons. 3p electrons are therefore held more loosely than 2p electrons. This leads to lower electronegativities,

enter image description here

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lower ionization potentials

enter image description here

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and higher oxidation states being more common for third period non-metals compared to second period non-metals.

Edit: Response to OP's second comment

3p electrons penetrate well into the space near the nucleus (link) Therefore it should be easier for 2p elements to form higher oxidation states than 3p elements, but it's not, how?

Your link (Rose-Hulman) to the 2p-3p radial electron distribution is good, it shows that the 3p electrons are "on average" further from the nucleus than 2p electrons - which is the key point. Here is a 3-dimensional representation of the same thing. Look how much further away the 3p electrons are from the nucleus than the 2p electrons. Given that the 3p electrons are 1) better screened (more interior electrons) and 2) much further away from the nucleus than the 2p electrons, it will much easier to remove more 3p electrons than 2p electrons and achieve higher oxidation states. This is consistent with the electronegativity and ionization potential arguments presented in my first edit.

enter image description here

Edit: Response to OP's third comment

the effective nuclear charge has bigger increase for 2p than 2s (except N-O) according to this chart. How so?

The key here is that the s orbital is spatially symmetric, it has a spherical pattern. An electron in an s orbital can screen outer electrons from the nucleus equally in all directions. On the other hand, a p orbital is directional, it is not spherically symmetric. An electron in the 2px orbital will not screen outer electrons from the nucleus equally in all directions.

As we proceed from boron to carbon, that first 2s electron is effective in screening the second 2s electron because of its spherical orbit (2s effective nuclear charge increase=0.641). However, the first 2p electron is not as effective in screening the nucleus because of its non-spherical shape. So when we add the second 2p electron, it "feels" the nucleus more strongly than the second 2s electron did so the change in effective nuclear charge is greater for the second p electron than it was for the second 2s electron (2p effective nuclear charge increase=0.715).

You noticed the "anomaly" for the 2p electrons as we go from nitrogen to oxygen. This is because with 3 2p electrons, all of the nitrogen 2p orbitals (px, py, pz) have an electron and produce a spherical electron distribution pattern. The nucleus is better-screened by the 3 2p electrons in nitrogen, so when we add one more p electron to yield oxygen, the effective nuclear charge increase is noticeably smaller (N -> O 2p effective nuclear charge increase=0.619).

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  • $\begingroup$ Okay, but how to explain then why 2nd period non-metals can't easily form higher oxidation states while 3p elements can? $\endgroup$ – Marko Mar 15 '15 at 20:08
  • $\begingroup$ I've edited my answer to address this point. $\endgroup$ – ron Mar 15 '15 at 20:43
  • $\begingroup$ Thank you very much Ron. It's only I still have some doubts. 3p electrons penetrate well into the space near the nucleus rose-hulman.edu/~tilstra/Engineering%20Chemistry%20I/…. Therefore it should be easier for 2p elements to form higher oxidation states than 3p elements, but it's not, how? $\endgroup$ – Marko Mar 16 '15 at 16:43
  • $\begingroup$ See my second edit and let me know if you still have questions. $\endgroup$ – ron Mar 16 '15 at 22:00
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    $\begingroup$ "2p feel more the increase of the ENC (because they penetrate less)" and because of their shape they do not screen as well as s electrons. I agree with everything else you wrote. $\endgroup$ – ron Mar 17 '15 at 19:57

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