﻿ Ionization Energy and Charge Shielding of Electrons in Atoms and Ions
San José State University

applet-magic.com
Thayer Watkins
Silicon Valley

Ionization Energy and Charge
Shielding of Electrons
in Atoms and Ions

## Ionization Energies

The ionization energy IE, or as it is usually called the ionization potential, for an electron in an atom or ion is the amount of energy required to dislodge it. The Bohr model of a hydrogen-like atom or ion indicates that the energy required to remove an electron, called the ionization potential, should follow the for

#### IE = RZ²/n²

where R is the Rydberg constant (approximately 13.6 electron Volts (eV), Z is the net charge experienced by the electron and n is the principal quantum number, effectively the shell number.

Here is the graph of the ionization potential of the innermost electron of the first five elements. Clearly the relationship is very regular and quadratic. Let #p denote the number of protons in the nucleus of the atom or ion, If IE is regressed upon (#p)², as the above equation indicates, the result is

#### IE = 13.60828 (#p)²

The Rydberg constant is 13.60569 eV.

The coefficient of determination for this equation is 0.999999993 and the standard error of the estimate is 0.01794 eV. The statistical fit is excellent, but it will be shown later that it can be improved upon.

## Charge Shielding

The Bohr model is strictly for a hydrogen-like atom or ion; i.e., one in which there is a single electron in the outermost shell. However the regression equation also fits very well the cases of multiple electrons in the outer shells if charge shielding is taken into account. That shielding is by the electrons in the inner shells and may also be by electrons in the same shell. But the shielding by electrons in the same shell is likely only for a fraction of their charge. As it turns out, shielding even for electrons in the inner shells the shielding is less than the full value of their charges.

Here is the rationale for the shieldings. If inner shell electrons execute trajectories that take them over a spherical shell it is as though their charges are smeared over a spherical shell and their effect on outer shell electrons is the same as though the charges of the inner shell electrons are concentrated at the center of the atom or ion and thus cancel out an equal number of positive charges.

The shielding by electrons in the same shell is a bit more complicated. The effect of a charge distributed over a spherical shell on an electron entirely within that spherical shell is zero. If the electron is entirely outside of the spherical shell the effect is the same as if the charge were concentrated at the center of the spherical shell. But if the center of the electron is located exactly on the shell then roughly half of the electron is inside of the spherical shell and is unaffected by its charge. Thus an electron is shielded by an amount approximately equal to one half of the charges in the same shell. That is the rough theory. It needs to be tested empirically. This partial shielding by electrons in the same shell explains how there can be negative ions. In a negative ion such as O= there are outer electronsclinging to a structure with overall negative charge. That appears to be a puzzle. The oxygen nucleus has eight protons. There are two electrons in the first shell and six in the second shell of the oxygen atom. Overall that is electrostatically neutral. But for a seventh electron in the second shell the two electrons in the inner shell and other six electrons in the second shell shield only a portion of the positive charge of the nucleus. Thus there is a positive attraction for that seventh electron. And since the seventh electron shields only a fraction of a unit charge for the eighth there is a positive attraction for the eighth electron. However another electron would go into a third shell and the ten (2+8) other electrons would be inner shell electrons and could more than shield the eight positive charges of the nucleus. Thus there would be a repulsion of an eleventh electron.

## Empirical Analysis

The value of Z in the above Bohr formula is the number of protons in the nucleus #p less the shielding ε by the electrons in inner shells or in the same shell. Thus the ionization energy would be

#### IE = (R/n²)(#p−ε)² which can be put in the form IE = (R/n²)((#p)² − 2(#p)ε + ε²)

This is equivalent to a regression equation of the form

#### IE = c2(#p)² + c1(#p) + c0

Such a form does give a very good fit to the data. The value of ε is found as

#### ε = −½c1/c2

However, according to the equation, it also should be that c0/c2 is equal to ε² and thus equal to the square of the value found from c1 and c2. The regression coefficients are not constrained to achieve that equality. Thus effectively the form assumed for the relationship for ionization potential is

#### IE = (R/n²)[(#p−ε)² + ζ]

where R is an empirical value, rather than necessarily being the Rydberg constant, and ζ is a constant. However the values found for R by regression analysis are notably close to the Rydberg constant. A nonzero value of ζ could arise from other factors not taken into account by the Bohr equation, such as the energy involved in the spin pairing of electrons.

For the second electron the quadratic regression that fits the data is

#### IE = 13.61003(#p)² − 17.00677(#p) + 4.16385        [7774.7]     [-1201.3]     [159.9]

The coefficient of determination for this equation is 0.999999999 and the standard error of the estimate is 0.00574 eV.

The estimate of ε which comes from this equation

#### ε = −½(-17.00677)/13.61003 = 0.62479 electron charges.

Thus the shielding by another electron in the same shell is real and is approximately 0.5 electron charges.

For the third electron (the first in the second shell) the regression equation is

#### IE = 3.42839(#p)² − -11.15594(#p) + 7.99664

The coefficient of determination for this equation is 0.999999913 and the standard error of the estimate is 0.01547 eV.

The estimate of ε which comes from this equation

#### ε = −½(-11.15594)/3.42839 = 1.62699 electron charges.

Full shielding by the two electrons in the first shell would give ε=2.0.

The coefficient of (#p)² is R/n² where n for the second shell is 2. Thus the R value for this case is 2²(3.42839)=13.71357, close to the .Rydberg constant.

Now it is worthwhile to apply the above methodology to the case of the first electron. There is no shielding in this case so ε should be zero.

The regression equation for the first electron is

#### IE = 13.61453(#p)² − 0.03161(#p) + 0.01798        [8874.2]     [-3.4]     [1.5]

The coefficient of determination for this equation is 0.999999999 and the standard error of the estimate is 0.00574 eV.

The estimate of ε which comes from this equation

#### ε = −½(-0.0316067)/13.61453 = 0.00116

This is essentially zero, thus confirming the methodology.

For the fourth electron ε is equal to 2.19001 rather than an "expected" 2.5 from the two electrons in the first shell and the one electron in the same shell.. Comparing this value with the 1.62699 shielding for the third electron indicates that the electron in the same shell as the fourth electron contributed 0.56302 charges to the shielding.

For the fifth electron ε is 3.16743 rather an "expected" 3.0. For the sixth electron it is 3.83669 rather an "expected" 3.5. For the seventh electron (the fifth electron in the second shell) it is 4.50280 rather than the "expected" 4.0. For the eighth it is 5.54623 rather than 4.5.

What the above indicates is that the shielding ratio for electrons in the same shell is approaching 1.0 rather than 0.5. This can be explained by the existence of subshells within a shell. The electrons in a lower subshell may be entirely interior to an outer subshell. Thus the analysis should be in terms of shielding by electrons in inner shells and subshells versus shielding by electrons in the same subshell.

The s-subshell of the second shell can contain at most two electrons. Therefore the third and fourth electrons are in that subshell. The "expected" shieldings for the third and fourth electrons are 2.0 and 2.5, respectively. But the fifth through tenth electrons are in the p-subshell of the second shell. Therefore their "expected" shieldings are 4.0, 4.5, 5.0, 5.5, 6.0 and 6.5, respectively. The shielding values found for the fifth through eighth electrons are 3.16743, 3.83669, 4.50280 and 5.54623. Thus there is a near perfect match for the eighth electron. The measured shielding for the ninth electron is 6.10890, reasonably close to 6.0. . The shielding for the tenth electron is 6.80810.

The eleventh electron is the first for the third shell so its "expected" shielding is 10.0. The value which comes from the analysis is 9.62699. Not bad as a confirmation of the theory and methodology.

(To be continued.)

## Conclusions

The energy required to dislodge an electron from a position in an atom is determined by the net positive charge it experiences from the nucleus. The net experienced charge is the charge of the nucleus less the amount that it is shielded from by the electrons in the inner shells and subshells and also by the electrons in the same subshell. The shielding by the electrons in the same subshell is a fraction of their charge, roughly one half. The shielding by electrons in inner shells or shells is generally less than one for one, say 0.8.

Fractional shielding accounts for the existence of negative ions.