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S117Lecture62011

# S117Lecture62011 - Polyelectronic atoms 3 energy...

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Polyelectronic atoms 3 energy contributions K.E. of electrons P.E. of attraction between electrons and nucleus P.E. of repulsion between electrons The repulsion between electrons leads to the electron correlation problem. In a particle view you can imagine how their motions are correlated. The second electron tries to stay as far away as possible from the first so that if the first electron moves, so does the second. They are not independent motions. The same effect occurs in a wave treatment. This means that the Schrodinger equation CANNOT be separated rigorously into equations which involve only one electron.

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Experimentally: He He + + e - I.E. 1 = 2372 kJ He + He 2+ + e - I.E. 2 = 5248 kJ If the two electrons were uncorrelated you would expect that the first I.E and second I.E. would be equal. Since I.E.1 < I.E.2 we can conclude that somehow the first electron in the atom is making it easier to remove the second electron in the atom (the first one removed). One can view this as the first electron “screening” the nucleus from the second electron and thus reducing the nuclear charge that the second electron experiences. Z effective = Z actual x (effect of electron-electron repulsions) This only accounts for the effect of the electron-electron repulsions in a crude way.
Ion Electron Configurations Atoms from Groups 1A, 2A and 3A give up (“lose”) 1, 2, or 3 valence electrons respectively to become isoelectronic with the preceding noble gas forming the +1, +2, and +3 ions. e.g. Ca Group 2A Ca 2+ Atoms from Groups 7A, 6A, and some 5A pick up (gain ) 1, 2, or 3 valence electrons to become isoelectronic with the following noble gas of the same period (main shell). They thus form 1 , 2 , or 3 ions. Metal atoms lose (give up) electrons to form cations with a positive charge equal to the group number. Non-metals gain (pickup) electrons to form anions with a negative charge equal to the A group number 8. e.g., 0 Group 6A 6-8 = 2

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