Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kataklysmic on March 05, 2007, 10:44:36 PM
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would anyone be able to explain the electron affinity trend in the second period to me (Li to F)? i understand the basic concept of large amounts of energy being given off when flourine or oxygen gains an electron as a result of it's high electronegativity, but why does nitrogen have such a low electron affinity compared to carbon? if someone could explain the trend it would help me out so much.
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What's carbon's electron configuration? What's nitrogen's electron configuration? Why might adding an electron to carbon be favorable and why might adding an electron to nitrogen be unfavorable?
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i think i understand it now. it has to do with the number of paired electrons in the p orbital of nitrogen.
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It actually takes energy to attain a more stable configuration, which is low spin configurations for some atoms, high spin for others.
By high spin I mean number of unpaired electrons is high.
I can't remember what transition elements have high spin configurations. Can anyone remind me?
I'm thinking of copper for some reason.
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Electron configurations with completely full shells and half-full shells exhibit an increased stability over other electron configurations. Therefore, nitrogen (1s22s22p3), which already has a half full 2p shell, will not want to eliminate the stability of a half-full shell by gaining an electron. Therefore, it has a lower electron affinity than expected by electronegativity alone. On the other hand, carbon (1s22s22p2) is one electron away from the stability of a half-full shell; therefore, it has extra "incentive" to seek out an extra electron and obtain a half-full 2p shell. So, carbon exhibits a higher electron affinity than expected by electronegativity alone. In much the same way, oxygen (1s22s22p4) will want to get rid of an electron in order to obtain a half-full 2p shell, so its ionization potential will be lower than expected.
g english: High spin/low spin in transition metal complexes is dictated by the ligands as well as in which row the metal sits. But, what does this have to do with the question?
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Ahh I mixed things I meant high/low multiplicity.