1) Write the electron configuration for the atom using the
(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
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.
(4d10) (4f14) (5s2,5p6)
Rule 2 does not apply; 0.35 · 1 + 0.85 · 16 + 60 · 1.00 =
Z* = 78 73.95 = 4.15 for a valence
The first ionization energy for hydrogen is 1310 kJ·mol1
while the first ionization energy for lithium is 520 kJ·mol1.
The IE for lithium is lower for two reasons;
1) The average distance from the nucleus for a 2s
electron is greater than a 1s electron;
2) The 2s1 electron in lithium is repelled
by the inner core electrons, so the valence electron is easily
For reason #2 the inner core electrons shield the valence
electron from the nucleus so the outer most electron only
experiences an effective nuclear charge. In the case of the
lithium the bulk of the 1s electron density lies between the
nucleus and the 2s1 electron. So the valence
electron `sees' the sum of the charges or approximately +1. In
reality the charge the valence electron experiences is greater
than 1 because the radial distribution show their is some
probabilty of finding the 2s electron close to the