EPTSo my question is, how do I "prove" electro negativity? I know the trend, I just don't really understand why it's like that.
[H]Basic Chemistry
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Demoninja
United States1190 Posts
So my question is, how do I "prove" electro negativity? I know the trend, I just don't really understand why it's like that. | ||
LosingID8
CA10828 Posts
http://en.wikipedia.org/wiki/Electronegativity#Methods_of_calculations even your teacher will have no clue wtf you did lol but really, do you understand the theory behind EN other than the simple periodic trends? | ||
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Demoninja
United States1190 Posts
Edit: another stupid question, are metalloids considered non metals or are they just metalloids? | ||
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LosingID8
CA10828 Posts
as you move to the right, the nucleus becomes larger and has a greater attraction to the electrons. as you move up, the overall radius decreases so the nucleus has a greater attraction to the electrons. example: HCl: Cl has greater EN so it pulls the electrons closer to it. the H atom becomes partially electron deficient, making it partial positive. the Cl atom becomes partially electron rich and becomes partially negative. (this leads into the idea of dipoles and dipole moments, etc) | ||
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LosingID8
CA10828 Posts
http://ibchem.com/ use the quick jump menu at the top right. sl = standard level hl = higher level if you're taking an AP class the hl material is probably right for you. | ||
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KlaCkoN
Sweden1661 Posts
On January 20 2009 19:25 Demoninja wrote: I need help on this question. I have no idea what I'm supposed to do. All my teacher said was "prove ionization energy and electro negativity. Provide 2 examples of each." Going from what we did in class, I assume we just provide an example like "The IE of Li is less than the IE of Be because the radius of Be is smaller than the radius of Li." So my question is, how do I "prove" electro negativity? I know the trend, I just don't really understand why it's like that. As you go from Li to Be you increase the nuclear charge, but at the same time you add a second electron to a 2s orbital. While this electron will experience electrostatic repulsion from beeing forced to adopt a quantum state almost identical (differing only in spin) to the electron already present it is far from enough to offset the increased electrostatic attraction between the heavier nucleus and the outer electrons. As you continue to boron you again increase the nuclear charge but this time you add the electron to a 2p orbital. This means the electron suddenly has angular momentum and thus it has zero probability to be found at the nucleus. Net effect is that this electron will actually experience a net attractive force _less_ than the electron we wanted to remove in the Be case. This is due to the repulsive effect effect of the two 2s electrons already present that sort of occupy the space between our 2p electron and the nucleus. Thus we go down in IE. Etc etc. Electronegativity is a bit of a weird concept since it has many different definitions. To be honest I have no idea at all how you would go about "proving it". X-ray studies comparing say the electron denisty around a chlorine atom bonded to a hydrogen to the electron density of a chlorine atom bonded to another chlorine atom? :s Anyway in general you can say that electronegativity is the ability of an atom to attract electrons to itself in a bond. It is closely related to both electron affinity and ionization energy. (Higher of these two means higher EN, lower means lower EN) | ||
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