• anonymous
Half filled is more stable or full filled and why???
  • Stacey Warren - Expert
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  • jamiebookeater
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  • JFraser
fully filled is more stable than half-filled, but half-filled is better than nothing. If you look at the electron configuration for nitrogen, you get\[1s^{2}2s^{2}2p^{3}\] the p-orbitals each have 1 electron, so the p-level is half-filled. This makes nitrogen a little more stable than you might think, because half-filled is "ok". carbon doesn't have this stability, because carbon's electron configuration is\[1s^{2}2s^{2}2p^{2}\]
  • anonymous
Keep in mind stability is the difference between nuclear-electron attraction (which lowers the energy) and electron-electron repulsion (which raises it). As you add electrons to an atom AND ADD PROTONS TO THE NUCLEI, you get a little more of both. Whether the net result is greater or lesser stability depends on the relative magnitudes. Now, if you are adding electrons to an empty subshell, e.g. the 2p subshell going from B to Ne, then initially you get to add each new electron to a different orbital (let us say the px, py and pz, although it would be more correct to say with m = -1, 0 and 1). Because these electrons are in different orbitals, they tend to be in different regions around the nuclei. They don't come near each others as often, so the added electron-electron repulsion is not very high. However, the added nuclear-electron attraction is the same each time, because all these electrons are at the same distance from the nuclei. With plenty of extra nuclear-electron attraction and not so much electron-electron repulsion, the net result is a reduction in energy, and increase in stability, as we go from B ([He]2s2,2p1) to N ([He]2s2,2p3). But now we run out of orbitals, and to add one more electron to get to O, we have to "double up" and put the next electron in an orbital already occupied by another electron (albeit with spin opposite). Since these electrons will spend more time closer to each other, the additional electron-electron repulsion is significantly greater. We still have the same nuclear-electron attraction, so the net in this case is an increase in energy, and O ([He]2s2,2p4) is less stable than N. Adding the 5th and 6th electron is no worse, so F and Ne are no worse. But now we have filled the subshell entirely, and the next electron, to move from Ne to Na, has to go into an entirely new subshell, in this case the 3s. The new subshell is entirely empty, so the added electron-electron repulsion is quite low. However, it is also further from the nucleus, which reduces the added nuclear-electron attraction, and, to make matters worse, there is now an entire new layer of electrons (the filled 2p subshell) between it and the nucleus, reducing the amount of nuclear charge it sees. Both effects substantially lower the nuclear-electron attraction, so the net result is, again, a reduction in energy and reduction in stability. To sum up: we see a gap in stability at the half-filled subshell because adding electrons to a half-filled subshell necessarily means "double up" electrons in the same orbital, which increases electron-electron repulsion sharply. We see a gap in stability at the filled subshell because the next electrons must go into a higher subshell, which is further from the nucleus and experiences greater inner-electron shielding, which reduces the nuclear-electron attraction sharply.

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