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Topic: Slaters rules for effective nuclear charge  (Read 30596 times)

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Offline BeepoGirl

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Slaters rules for effective nuclear charge
« on: May 02, 2009, 11:45:22 AM »
I understand most of the rules. But in all the examples I look at in my notes, my textbook and online seem to be inconsistent.

It says to add 0.85 for each electtron in sets with quantum number n-1 and add 1 for each electron in sets with quantum number <n-1.

For the example in my notes, it looks at the Mn 3d electrons. So, [1s]2 [2s2p]8 [3s3p]8 [3d]5 [4s]2 is the set up.

We ignore the 4s electrons, add 0.35 for the electron in the 3d set. Then I get confused because the [3s3p] electrons have the same quantum number as 3d but the examples state that it 5 x 1 for that set.

Then the [2s2p] electrons I am told contributes 8 x 1 even though in the rules it clearly says that it should be 0.85..?

Any help would be appreciated.


Offline kimyacı

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Re: Slaters rules for effective nuclear charge
« Reply #1 on: May 02, 2009, 04:43:14 PM »
 
 Slater's Rules:
1) Write the electron configuration for the atom using the following design;
(1s)(2s,2p)(3s,3p) (3d) (4s,4p) (4d) (4f) (5s,5p)
2) Any electrons to the right of the electron of interest contributes no shielding. (Approximately correct statement.)
3) All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units
4) If the electron of interest is an s or p electron: All electrons with one less value of the principal quantum number shield to an extent of 0.85 units of nuclear charge. All electrons with two less values of the principal quantum number shield to an extent of 1.00 units.
5) If the electron of interest is an d or f electron: All electrons to the left shield to an extent of 1.00 units of nuclear charge. 6) Sum the shielding amounts from steps 2 through 5 and subtract from the nuclear charge value to obtain the effective nuclear charge.
Examples:
Calculate Z* for a valence electron in fluorine.
(1s2)(2s2,2p5)
Rule 2 does not apply; 0.35 · 6 + 0.85 · 2 = 3.8
Z* = 9 – 3.8 = 5.2 for a valence electron.
Calculate Z* for a 6s electron in Platinum.
(1s2)(2s2,2p6)(3s2,3p6) (3d10) (4s2,4p6) (4d10) (4f14) (5s2,5p6) (5d8) (6s2)
Rule 2 does not apply; 0.35 · 1 + 0.85 · 16 + 60 · 1.00 = 73.95
Z* = 78 – 73.95 = 4.15 for a valence electron.

 you  careful  the 5.th rule. Mn ends d 5.

 
 


Offline BeepoGirl

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Re: Slaters rules for effective nuclear charge
« Reply #2 on: May 03, 2009, 08:46:09 AM »
Ah.. the way it was worded here made me think that rule was for when a d or f electron is shielding, not when it's the electron being considered. Thanks a lot!

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