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The Ionization Potentials and Shielding
of the First Five Electrons in the Fourth Shell

The Bohr model of a hydrogen-like ion predicts that the total energy E of an electron is given by

E = −Z²R/n²

where Z is the net charge experienced by the electron, n is the principal quantum number and R is a constant equal to approximately 13.6 electron volts (eV). This formula is the result of the total energy being equal to

E = − Ze²/(2rn)

where e is the charge of the electron and rn is the orbit radius when the principal quantum number is n. The orbit radius is given by

rn = n²h/(Zmee²)

where h is Planck's constant divided by 2π and me is the mass of the electron.

Shell Structure

Electrons in atoms are organized in shells whose capacities are equal to 2m², where m is an integer. Thus there can be at most 2 electrons in the first shell, 8 in the second shell, 8 in the third shell and 18 in each of the fourth and fifth shells. Here only the fourth shell is being considered.

Here are all the ionization potentials for such ions. The values are for the elements for which the data is available in the CRC Handbook of Physics and Chemistry 82nd Edition (2001-2002).

Ionization Potential of First Five Electrons in the
Fourth Shell for Elements from Vanadium to Gallium
(electron volts)
23 65.2817 46.709 29.311 14.66 6.7463
24 90.6349 69.46 49.16 30.96 16.4857
25 119.203 95.6 72.4 51.2 33.668
26 151.06 124.98 99.1 75 54.8
27 186.13 157.8 128.9 102 79.5
28 224.6 193 162 133 108
29 265.3 232 199 166 139
30 310.8 274 238 203 174
31 391 350 308 268.2 230.85

The Ionization Potential of the First Electron
as a Function of Proton Number

An ion with only one electron in a shell is equivalent to the hydrogen atom but having a positive charge of Z instead of one, where Z is the proton number #p of the nucleus less the amount of shielding by the electrons in the inner shells. The Bohr theory applies to such system. According to the Bohr theory the ionization potential should be

IE = (R/n²)(#p−ε)²

Where R is constant known as the Rydberg constant and is equal to about 13.6 electron volts (eV), n is the principal quantum number which here is the same as the shell number. The quantity #p is the proton number of the nucleus and ε is the amount of shielding by electrons in inner shells or the same shell. For the first electron in the third shell it is usually presumed that the ten electrons in the first and second shell shield exactly ten units of charge. It is immediately discovered that this not the case. Here is the plot of the relationship for the first electron.

This appears to be a quadratic relationship but shifted; i.e. something proportional to (#p−ε)². Thus equation is then

IE = (R/n²)(#p−ε)²
which can be expressed as
IE = (R/n²)(#p² − 2#p*ε + ε²)

The appropriate regression equation would be

IE = c0 + c1#p + c2(#p)²
in which

The regression results are

IE = 643.299224 − 73.06773382#p + 2.079254775(#p)²
[6.9] [-9.6] [13.7]
R² =0.997488104

The numbers in the square brackets are the t-ratios for the regression coefficients. For a regression coefficient to be statistically significantly different its magnitude must be greater than 2.0. As can be seen the regression coefficients for are highly significant.

The value of ε can be found as

ε = ½(−c1/c2) = 17.57065432

Thus the shielding of the first electron in the fourth shell by the eighteen electrons in the first, second and third shells is not exactly 18. Instead it is about 97.6 percent of that value. This could be due to the distributions of the charges of the two inner electrons, either their radial dispersion or their asymmetry.

The regression results for the first four of the electrons in the fourth shell are

Number of
in Shell
Coefficient of
117.57065432 0.997488104
218.53150712 0.997193112
319.54117024 0.99726095
420.6161118 0.9969434
521.40504006 0.997849703

The graph of the shielding versus the number of electrons in the shell is:

The slope of the relationship is found by regressing the shielding S on the number of electrons #e. This gives

S = 16.60688386 + 0.975337616 #e
[200.6] [39.1]
R² = 0.9804

Thus, on average, each additional electron in the second shell shields about 0.975 units of positive charge of the nucleus.


The values of the ionization potentials IE are accurately explained by a function of the form

IE = (R/n²)(#p−ε)²

where R is an empirical constant approximately equal to the Rydberg constant, n is the shell number, #p is the proton number of the nucleus, and ε is an empirical value which is a function of the number of electrons in the shell and the number which are in inner shells.

(To be continued.)

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