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.
Calculate Z* for a valence electron in fluorine.
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.