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Author Topic: electron affinity and shielding  (Read 277 times)
uglydaryl
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« on: January 07, 2010, 04:52:10 AM »

hello i have some problems here again and any help could do

1. how would you account for the decrese in electron affinity between carbon and nitrogen?

2. can someone explain the difference between the first ionization energy of calcium with that of zinc in terms of the balance between shielding with increasing numbers of d electrons and the effect of increasing nuclear charge

thanks for any reply
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kingchemist
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« Reply #1 on: January 07, 2010, 05:46:34 AM »

C = 1s2 2s2 2p1 2p1 while N = 1s2 2s2 2p1 2p1 2p1
First Ionisation energy of N is greater than C - you might say because of increasing nuclear charge without a change is screening effect. But it is more than this. N has a half-filled p orbital set which makes it much more stable than C (and O) and this makes it more difficult to remove an electron. This is likely to impact on electron affinity which will consequently be lower than expected.

C =  1086.1 kJ.mol -1; Zn = 904.5 kJ.mol -1
So to remove an electron from C means removing one close to the nucleus where attraction will be greater than in Zn. Also, although Zn has more protons, the atom is larger and there is much greater shielding effect on the outermost electons which are relatively easier to remove.
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uglydaryl
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« Reply #2 on: January 28, 2010, 03:19:48 AM »

hello.. are there different orbitals or sub orbitals other than the s p d f? on my status im not aware of any other than the spdf
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kingchemist
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« Reply #3 on: January 28, 2010, 07:55:28 AM »

There are only 4 quantum numbers needed to describe the position of an electron.
The first describes the energy level eg n=1, n=2
The second describes the orbital type eg s, p, d or f
The third describes the orientation of the orbital eg px, py, pz etc
The fourth describes the spin of the electron (+ 1/2 or -1/2)
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