![]() 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. You noticed the "anomaly" for the 2p electrons as we go from nitrogen to oxygen. 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). However, the first 2p electron is not as effective in screening the nucleus because of its non-spherical shape. ![]() An electron in the 2px orbital will not screen outer electrons from the nucleus equally in all directions.Īs 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). On the other hand, a p orbital is directional, it is not spherically symmetric. An electron in an s orbital can screen outer electrons from the nucleus equally in all directions. The key here is that the s orbital is spatially symmetric, it has a spherical pattern. The effective nuclear charge has bigger increase for 2p than 2s This is consistent with the electronegativity and ionization potential arguments presented in my first edit. 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. Look how much further away the 3p electrons are from the nucleus than the 2p electrons. Here is a 3-dimensional representation of the same thing. 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. Should be easier for 2p elements to form higher oxidation states than This leads to lower electronegativities,Īnd higher oxidation states being more common for third period non-metals compared to second period non-metals.ģp electrons penetrate well into the space near the nucleus ( link) 3p electrons are therefore held more loosely than 2p electrons. 3p electrons are further from the nucleus and screened by all of the 2s and 2p electrons. 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. Electron penetration correlates with electron stabilization.Įxplain then why 2nd period non-metals can't easily form higherįirst off, second period non-metals can form higher oxidation states, for example $\ce$ where the nitrogen is in the +5 oxidation state. Because they penetrate less, they are stabilized less by the effective nuclear charge. It is correct that they penetrate less, but they are not stabilized more. If the 2p orbital was more stabilized, then the difference between 2s and 2p would be smaller through the period.Ģs orbitals are stabilized more than 2p orbitals by the effective nuclear charge because of better penetration.Įverything you wrote above is correct except for,Ģp orbitals are more stabilized because they penetrate less in the 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. 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. 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.Ģ) 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. 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). Are 2s or 2p orbitals more stabilized when going from left to right in the period? There are good arguments for both possible answers:ġ) 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.
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