Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: MIBOO on October 19, 2017, 08:07:36 PM
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In my general chemistry class in college we are taught that the first ionization energy of group 3 elements is smaller than elements in group 2 because the added electron in the p orbital is shielded by the electrons in the s orbital, so their effective nuclear charge and force of attraction decreases. But since radius decreases when force of attraction increases, why is it that group 3 elements still decrease in size when their p electron has lower force of attraction? Shouldn't it go further away from the nucleus since its force of attraction is smaller and increase the atomic radius?
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Hi guys :D I have been thinking about some concepts behind periodic trends and came up with some questions. In Gen Chem we were taught that the first ionization energy decreases for group 6 elements compared to group 5 elements because in group 6 electrons are paired in the p orbitals. Therefore, that added electron's potential energy increases, decreasing the first ionization energy. But does electron electron repulsion actually decrease force of attraction (according to Coulomb's law: (effective nuclear charge x valence electron charge (1-)/distance from nucleus)) or does it just counter it (so force of attraction stays constant but potential energy just went up?)? Also, since its potential energy is higher, shouldn't atomic radius increase from group 5 to group 6?
Many thanks to anyone who replies!
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why is it that group 3 elements still decrease in size
Please elaborate what do you mean by "group 3"? Which orbital is filled in the group 3?
Are you sure you are not mistaking group 3 and group III?
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And don't forget that (neutral) atoms with one electron more have also one proton more, so the nucleus attracts all the electrons more strongly hence closer to it.
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Yea, I meant group III. According to my professor, the electron added from Group II to Group III in the p orbital actually has a lower force of attraction to the nucleus because the previous s orbital electrons (which have higher penetration based on radial probability graphs) shield that electron in the p orbital. I just thought since that one electron in the p orbital is not as attracted to the nucleus than the electrons in the s orbitals in the previous element in Group II, it would move away from the nucleus and increase the atomic radius.
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Huh, I dunno...
At least, I've checked that from Be to B and Mg to Al, ionization gets easier and the "atomic radius" smaller
https://www.webelements.com/beryllium/atom_sizes.html
Though, the "atomic radius" (for which half a dozen definitions are common, give very different and incompatible figures) relates usually with a distance between nuclei in a molecule. Useful since molecules are more common than lone atoms, but then the "atomic radius" relates to a kind of binding capability of the outer shell.
And since B and Al are trivalent, their "atomic radius" would say that
- their distance of B and Al to other atoms once three bonds are established
is smaller than
- the distance of Be and Mg to other atoms once two bonds are established.
I too would be happy to read better informed answers.