In the Classroom802 Journal of Chemical Education • Vol. 76 No. 6 June 1999 • JChemEd.chem.wisc.eduMore than 60 years ago Slater proposed a set of verysimple rules for the computation of the effective nuclearcharge experienced by an electron in an atom (1). Since thattime these rules have been used in applications ranging fromevaluating atomic overlap integrals (2) to establishing tablesof atomic radii (3). Recently they were used to provide newinsights into the often elusive nature of chemical hardness(4). Because Slater’s procedure provides for the computationof one-electron energies rather than orbital energies, it is es-pecially well suited for instructional purposes and has beenincluded in a number of textbooks (5). The need for simpleactivities that illustrate and reinforce an intuitive understandingof atoms and their properties is needed throughout theundergraduate curriculum. However, one difficulty withSlater’s original procedure has been that it fails to yield thelevel of detail needed to illustrate important concepts. Withrelatively minor modifications, Slater’s method would provideto both students and teachers a model of the atom that isinstructive as well as surprisingly simple and intuitive.The Basic ConceptsSlater’s method and its implementation require only areasonable mastery of three concepts, which are already part ofthe first-year curriculum. The first of these is that the electron–electron interactions can be reasonably described as shieldingof the nuclear charge by each of the other electrons, or Zi*= Z – ci= Z – cijΣi≠ jall electrons(1)where Zi* is the effective nuclear charge experienced by elec-tron i and cij is the shielding of electron i by electron j. Oncethis is determined, the energy of the electron may be com-puted, using the relationship ei= {1312kJmol{1Zi*ni2(2)This relationship most often appears in first-year texts inits more complicated form as the Rydberg equation. Finally,since Slater’s procedure yields one-electron energies, the totalelectronic energy is the sum of the one-electron energies:E = e1 + e2 + e3 + … + eN + epairing(3)The pairing energy was not explicitly included in Slater’soriginal formulation, but its explicit consideration is necessaryfor a number of applications.ShieldingThere are potentially a very large number of shieldingconstants. However, Slater found that, because many werevirtually equal, the number could be reduced to only three.We have refined the scheme to use only three for the lighterelements (no d electrons) and a total of eight for the elementsthrough barium (no f electrons). These rules are applicableto positive, negative, and fractionally charged atoms. The rulesgiven below are summarized in Figure 1.The Rules for Light ElementsLarger Shell. An electron is not shielded by any electronin a larger shell (cij = 0.00).Same Shell. Both s and p electrons in the same shell shieldeach other: 0.3228.Next Smaller Shell. The shielding of an s electron by an sor p electron in the next smallest shell is 0.8366. Theshielding of a p electron by an s or p electron in the nextsmallest shell is 0.9155.Inner Shells. An electron is completely shielded by elec-trons in shells that are smaller than the next smallest (cij= 1.00).The pairing energy for p electrons is 1360.1/n2 kJ/mol.The Rules for Heavier ElementsSame Shell. In the same shell the shielding of a d electronby s and p electrons is 0.8933, but the shielding of an sor p electron by a d electron in 0.0352. The shielding ofd electrons by d electrons is 0.3228.Smaller Shells. The shielding of both s and p electrons by ad electron in the next smaller shell is 0.9143. A d electronis completely shielded by any electron in a smaller shell.The pairing energy for d electrons is 9610.7/n2 kJ/mol.The Genius of Slater’s RulesJames L. ReedDepartment of Chemistry, Clark Atlanta University, Atlanta, GA 30314; [email protected] 1. A diagrammatic presentation of the modified Slater’s Rules.In the ClassroomJChemEd.chem.wisc.edu • Vol. 76 No. 6 June 1999 • Journal of Chemical Education 803Evaluation of the Shielding ConstantsThese shielding constants are the ones which yield thebest least squares (5) agreement between the experimental (6 )and computed first ionization energies for elements hydro-gen through xenon. Thus for atom M the ionization energywas computed viaIE(M) = E(M+) – E(M) (4)where the energy is evaluated using eqs 1–3 (see the samplecalculation below for details). The first ionization energiesare known very accurately for hydrogen through xenon, whichare the elements having only s, p, and d electrons. For mostelements the ionization of d electrons occurs only for the thirdand subsequent ionizations of the transition metals. These wereused to determine the shielding constants for d electrons. Thevalidity of both the model and the shielding constants hasbeen evidenced by their ability to reproduce the experimentalfirst ionization energies and successive ionizations of the sameatom (7).Sample CalculationA potassium atom has been selected for the sample cal-culation. The configuration we have selected is1s22s22p63s23p63d1For processes involving only valence electrons the core electronenergies do not change. Thus for most exercises the numberof computations is greatly reduced, and students need concernthemselves with only one shell. Of course to illustrate theshell structure of an atom or core ionizations (XPS or ESCA), itwould be necessary to do computations for core electrons also.The determination of the shielding of a 3s electron isillustrated diagrammatically below along with the computa-tions for all the third-shell electrons.1s23s23p63d11s22p61.00 0.03520.32280.8366Z*3s = 19 – [(1)(0.0352) + (7)(0.3228) + (8)(0.8366) + (2)(1.00)]Z*3s = 8.0124 e3s = {9,358.7 kJ/molZ*3p = 19 - [(1)(0.0352) + (7)(0.3228) + (8)(0.9155) + (2)(1.00)]Z*3p = 7.3812 e3p = {7,942.3 kJ/molZ*3d = 19 - [(8)(0.8933) + (8)(1.00) + (2)(1.00)]Z*3d = 1.8536 e3d = {500.9 kJ/molepairing = 1360.1/9 = 151.1 kJ/molThe electronic energy (valence electrons only) is thusE[3s23p63d1]=(2)({9358.7)+(6)({7942.3)+(1)({500.9)+(3)(151.1)E[3s23p63d1]={66,418.8 kJ/molTo allow for comparison, the effective nuclear charges andone-electron energies for the configuration1s22s22p63s23p64s1have been computed to yielde3s = {9,441.1 Z*3s = 8.0476e3p = {8,018.2 Z*3p = 7.4164e4s = {436.5 Z*4s = 2.3072E[3s23p64s1] = {66,974.9This is one of the more involved computations,