Research: Science and Education
Probing the Orbital Energy of an Electron in an Atom James L. Bills Department of Chemistry and Biochemistry, Brigham Young University, Provo, UT 84602-5700;
[email protected] This article answers an appeal (1) for simple theoretical interpretations of atomic properties. Some previous correlations of atomic ionization energies were based on the quantum-mechanical energy levels En of a one-electron atom (2–5), Z En = − 1 2 n
2
Eh
(1)
where Z is the nuclear charge, n is the principal quantum number, and Eh is the Hartree atomic unit, equal to 2625.50 kJ兾mol.1 For atoms with more than one electron, a shielding constant s is subtracted from Z in eq 1 (2–5). E n ′ = − 1 2
Z − s n
2
Eh
(2)
In some cases an empirical number n* is used for n. The virial theorem states that the total electronic energy of an atom equals the negative of its total electronic kinetic energy (6b). Slater (2) identified En´ in eq 2 as the negative of the average kinetic energy of each electron. He summed eq 2 to get an approximate total energy for both the atom and its ion, and he used their difference to estimate the ionization energy. Other authors (3–5) identified ᎑En´ of eq 2 with the ionization energy, I. They chose s to fit the I of a single atom (3), or (with optimized n*) of an atom and isoelectronic cations (4, 5). The fits are good, but the physical basis of eq 2 is obscure to most students. By contrast, Coulomb’s law is familiar to most students, and its meaning is more intuitive. A theoretical snapshot of an atom, showing the screened nuclear charge and the electron to be ionized at its radius of zero kinetic energy, enables anyone to approximate its ionization energy. The trends in the screened charges and radii are of interest in their own right. In this article, we examine these trends for the 20 atoms from H to Ca. Values are also given for the ions He+, Be+, Mg+, Ca+, H−, F−, and Cl−. The H Atom and One-Electron Atomic Ions The energy levels of a one-electron atom or ion are given in eq 1. Every value of En is an eigenvalue and therefore a constant energy, independent of the distance r of the electron from the nucleus (6c). Each eigenvalue is the constant sum of classical values of potential and kinetic energy (6d).
Potential and Kinetic Energies The classical potential and kinetic energies of an electron both depend on r. In atoms, r is often measured in units of the Bohr radius, a0 = 52.9177 pm. In this article, we measure r in picometers. The classical potential energy, V(r), is independent of n, V (r ) = −
Z E h a0 r
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(3)
•
but the classical kinetic energy, Tn(r), depends on n (6d). Tn (r ) = E n +
Z E h a0 r
(4)
Equation 4 appears in the quantum-mechanical operation Topψ = Tn(r)ψ. We add eqs 3 and 4 and get an obvious equation for En. E n = V ( r ) + Tn (r )
(5)
While eq 1 gives En directly, it tells us nothing about V(r) and Tn(r). Alternatively, eq 5 has us find V(r) and Tn(r) separately and get En as their sum. The latter approach is far more informative and is well worth the effort.
The Radius of Zero Kinetic Energy, r0 We shall take a shortcut. Since En is constant, we need not add the full functions V(r) and Tn(r) in eq 5. We can evaluate both V(r) in eq 3 and Tn(r) in eq 4 at any convenient single value of r and combine them in eq 5. Obviously, all other values of r must yield that same value of En in eq 5. We can simplify both the calculation of En and its interpretation by choosing that value of r called r0 that makes Tn(r) = 0 in eq 4. This r0 is analogous to the classical turning point of the harmonic oscillator. For any one-electron atom, eq 4 gives the quantum-mechanical formula for r0.2 r0 =
2n 2 a0 Z
(6)
With zero kinetic energy at r0, the potential energy at r0 must be the total energy En. Z E a r0 h 0
(7a)
pm kJ Z 138935 mol r0
(7b)
E n = V ( r0 ) = −
En = −
Recall that r0 depends on n in eq 6. When a one-electron atom or ion is ionized from its ground state, n goes from 1 to ∞. The electron starts at E1 and ends at E∞, where r0 is infinite and E is zero.
I = ∆E = 0 − E1 =
pm kJ Z 138935 r0 mol
(8)
We illustrate eqs 6 and 8 with two examples, H and He+. For H, r0 = (2兾1)(52.9177 pm) = 105.8 pm. Hence I = (1兾105.8 pm)(138935 pm kJ兾mol ) = 1313 kJ兾mol. For He+, r0 = (2兾2)(52.9177 pm) = 52.92 pm. Thus I = (2兾52.92 pm)(138935 pm kJ兾mol) = 5251 kJ兾mol. Experimental values of I are 1312.05 kJ兾mol for H and 5250.52 kJ兾mol for He+ (7).
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Research: Science and Education
A More General Way To Find r0 The energies in eq 1 and the values of r0 in eq 6 apply only to one-electron species. We need a way to find r0 that can be used in atoms with more than one electron. This way is not difficult, but it is more complex. (The desired values of r0 are tabulated below, so you may skip this section if you wish.) Radial wavefunctions, R(r), are readily available (8–11). Slater multiplied R(r) by r to get P(r), that is, P(r) = rR(r) (6d, 6f ). In effect, he found the kinetic energy from P(r), as −1 2
d2 dr
2
+
ᐉ ( ᐉ + 1)
P (r ) = T (r ) P (r )
2r 2
(9)
where ᐉ is the azimuthal quantum number: 0 for s orbitals, 1 for p, 2 for d, and so forth. We use P˝(r) for the second derivative of P(r), and solve eq 9 for T(r).
T (r ) = −
1 P ′ ′ ( r ) 2
+
P (r )
ᐉ ( ᐉ + 1) 2r
(10)
2
We find the distance r0 that makes T(r0) zero. Any nodes in P(r) occur far from r0 and cause no problem in finding r0. For s orbitals, where ᐉ = 0, r0 occurs at the point of inflection in the tail of P(r). We convert r0 from atomic units to picometers as in eq 6. We illustrate eqs 9 and 10 using R(r) for the one-electron 1s orbital at nuclear charge Z.
R (r ) =
P (r ) =
1
( 2 Z )3
2
(2 Z )
3
1
P ′ ′ ( r ) = Z 2 −
ε i = V (r 0 ) = −
−Z 2 Z + 2 r 2 a0 Z
(11)
Hence r0 is 105.8 pm in H and 52.92 pm in He , as found above. Polyelectronic Atoms and Atomic Ions In the Hartree–Fock model of a polyelectronic atom, each electron i has a spin-orbital energy, εi, that is somewhat analogous to the En of a lone electron. However, the potential energy of electron i is its nuclear attraction ᎑Z兾ri plus its repulsions and exchanges at ri with the other electrons j (each integrated over θi, φi, θj, φj, and rj) (6f ). Each electron i has a constant orbital energy εi, despite its variable potential energy, pi(ri). Slater defined for electron i its effective nuclear charge for potential energy, Zpi(ri) (6g). •
Z pi ( ri ) ri
E h a0
(12)
(13)
pm kJ Z0 138935 r0 mol
(14)
Z0
Z0 r0
r0 e 2 NA 4 π ε0
1−
=
Z0 E h a0 r0
(15)
pm kJ Z0 138935 mol r0
Here NA is Avogadro’s number and ε0 is the permittivity of a vacuum, 8.854188 × 10᎑12 C2兾(J m). Values of Z0 and r0 are given for atoms 1 to 20 (H to Ca) in Table 1 and for seven ions in Table 2. They were found from Hartree–Fock orbitals (8, 9, 11) via eqs 10 and 15. These values are of special interest. Each r0 measures the orbital size of the weakest-held electron, while Z0 measures the effective nuclear charge felt by it.
2Z P (r ) r
Journal of Chemical Education
E h a0
For a lone electron, eqs 12–14 reduce to eqs 3, 4, and 7, where ε = E and I = ᎑E accurately. From here on, we drop the subscript i for simplicity. Our theoretical snapshot of the atom shows only the screened nucleus plus the targeted electron with zero kinetic energy at r0.
=
+
474
ri
The smallest ᎑εi in the atom approximates the ionization energy, I (12).3 Values of ᎑ε and I are paired for comparison in Table 1. Individual values of ᎑ε often differ from I by several percent, but the trends in ᎑ε faithfully mirror the trends in I. Again defining r0 to make Ti(r0) = 0 and Z0 = Zpi(r0) (6g), eqs 12 and 13 are solved for εi.
2
We solve for the r where T(r) = 0. r0 =
Z pi (ri )
T (ri ) = εi +
r exp ( − Z r )
2
T (r ) =
V ( ri ) = −
I ≈ −ε =
exp ( − Z r )
2
The analogs of eqs 3 and 4 are:
Applications
The He Atom A He 1s orbital P(r) = rR(r) and its energies T(r), V(r), and ε(r) are shown in Figure 1 (8). The false rise in ε(r) at small r shows that this P(r) only approximates the eigenfunction.4 The point of inflection in P(r) locates r0 at 61.74 pm, where Z0 = Zpi(r0) = 1.070 (6g) and ε(r0) = ᎑2408 kJ兾mol. (For simplicity, eq 15 was used to find Z0 from ᎑ε = 2408 kJ兾mol.) The experimental ionization energy is 2372 kJ兾mol (7). Remember that our goal is to account for ᎑ε. We cannot control how closely ᎑ε approximates I. (However, we could fit I instead of ᎑ε.5 See the discussion of H−, F−, and Cl− below.) The Z0 of 1.070 in He is only 7% higher than the 1.000 in H. When He electron i reaches r0, much of the charge of electron j is distributed within r0. In effect, this inner nega-
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tive charge cancels an equal quantity of charge from Z. Since we use the orbital energy (as ᎑εi) to estimate the ionization energy, we must include the outer shielding by electron j (6g). Thus Z0 is lowered from 2.000 in He+ to 1.070 in He. The screening of the nuclear charge has much less effect on the distance r0, raising it from 52.92 pm in He+ to 61.74 pm in He. Each ionization energy is best understood from its value of Z0兾r0, and the ratio of two ionization energies is best understood from the ratio of the two values of Z0兾r0. The experimental He and H values of I1 have the He兾H ratio of 1.81. The He and H values of Z0兾r0 have the ratio of 1.84. Moreover, the changes in the Z0 and r0 values enable us to see the physical reasons for the higher value of Z0兾r0 in He. Table 1 shows that the He兾H ratio of Z0 values is 1.07, while their ratio of r0 values is only 0.59. Both factors contribute to the higher value of Z0兾r0 in He, but the major factor is the decrease in r0.
Figure 1. Properties of a Hartree–Fock 1s orbital of He, P(r) = rR(r) (8). Its kinetic energy is T(r) = {᎑1/2[d2P(r)/dr2]/P(r)}Eh. Its potential energy is V(r) = {᎑2/r + [0∫r[P2(r)dr]/r + r∫∞[P2(r)dr]/r}Eh (6g). Its orbital energy is ε(r) = T(r) + V(r). At r0 = 61.74 pm, T(r0) = 0, so ε = V(r0) = ᎑(Z0/r0)Eha0, with Z0 = 1.070.
Trends in Z0 and r0 The trends in Table I are interesting. On going from left to right through a period, Z0 generally increases and r0 decreases. Both trends are caused by each successive atom having an additional proton and an additional electron that incompletely screens the proton from the electron at r0. These trends are commonly taught in general chemistry. The quantitative data in Table 1 are compelling evidence. Exceptions to the trend in Z0 occur at O and S, where px↓ has no exchange avoidance of px↑, py↑, and pz↑, and px↓ is in the same orbital with px↑. The resulting repulsions
cause the drops in Z0 at O and S that in turn cause the unexpected drops in ᎑ε and I, despite continued decreases in r0. The smaller I gives relative instability to the p4, one-beyond-half-filled, subshell. Ionization disrupts a p3 half-filled subshell in both N and P. The relative increase in I is 29.1% from C to N and 28.6% from Si to P. Neither increase is disproportionately large. There is no special stability (13, 14) to a p3 half-filled subshell.
Table 1. Electrostatic View of Hartree–Fock Orbital Energy Parame t e r
Group 1
2
13
14
15
16
17
18
A t om
H
He
Z0
1.000
1.070
r0 a ᎑εb
105.8
61.74
1313
2408
Ic
1312
2372
A t om
Li
Be
B
C
N
O
F
Ne
Z0
1.000
1.044
1.148
1.250
1.346
1.110
1.262
1.396
r0 ᎑ε
269.6
178.6
196.1
152.6
125.5
110.6
97.58
86.87
515
812
813
1138
1490
1394
1797
2233
I
520
899
801
1086
1402
1314
1681
2081
A t om
Na
Mg
Al
Si
P
S
Cl
Ar
Z0
1.000
1.045
1.098
1.227
1.350
1.133
1.295
1.443
r0 ᎑ε
290.6
218.5
276.8
218.5
182.4
161.5
143.7
129.2
478
664
551
780
1028
975
1252
1552
I
496
738
578
787
1012
1000
1251
1521
A t om
K
Ca
Z0
1.000
1.039
r0 ᎑ε
358.8
281.2
387
513
I
419
590
a
r0 in pm and ᎑ε and I in k J /mol (8, 9, 11). b᎑ε approx imat e s t he ionizat ion e ne rgy I of t he at om and Z0 is t he e f f e ct iv e nucle ar charge at r0, radius of ze ro k ine t ic e ne rgy . ᎑ε = (Z0/r0)(138935 pm k J /mol). cRe f 7.
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Research: Science and Education Table 2. Electrostatic View of Hartree–Fock Orbital Energy He +
Be+
M g+
C a+
H−
F−
C l−
Z0
2.000
2.000
2.000
2.000
0.120
0.384
0.443
r0 / p m
52.92
158.9
195.5
254.2
137.5
112.3
156.0
Parame t e r
᎑ε/(k J /mol)a
5251
1749
1421
1093
121
475
395
I/(kJ /mol)b
5251
1757
1451
1145
72.8
328.2
348.6
---
---
---
---
0.072
0.2652
0.3914
Z0 ´
c
a
᎑ε approx imat e s t he ionizat ion e ne rgy I of H e +, B e +, M g+, and C a+; and H −, F−, C l− f or t he e le ct ron af f init y A of H , F, and C l. ᎑ε = (Z0/r0)(138935 pm k J /mol); Z0´ w as chos e n t o f it I (8,11). bRe f 7. cC alculat e d E(H −) = ᎑1280 k J /mol but E(H ) = ᎑1312 k J /mol, s o H − is not s t able in t he H art re e –Fock mode l.
Uncommon increases in r0 occur at B and Al, where the last electron is a relatively nonpenetrating p electron compared to the s electron at Be and Mg. This leads to the pronounced dip in ionization energy for B and Al. On going from top to bottom down a family in Table 1, r0 always increases. Although the n shell of an upper family member has shrunk in the member below, its added n+1 shell is larger than the n shell above it. There is very little change in Z0, because both family members have the same number of screening electrons in their outer shell that incompletely screen their respective protons. Both shell structure and family structure are evident here. Teachers of general chemistry can illustrate these points with quantitative data from Table 1. In Table 2 each cation r0 is smaller, and each anion r0 is larger, than the r0 of the neutral atom. The change in r0 helps to raise the ionization energy of each cation and to lower the ionization energy of each anion. But the major factor is the much larger Z0 in each cation, and the much smaller Z0 in each anion, than in the neutral atom. The agreement of ᎑ε with I is poor for H᎑, F᎑, and Cl᎑. Wavefunctions that allow electron correlation give good results. Correlation affects the repulsion between electrons much more than it does the properties of a single electron (15). In terms of eq 15, correlation would alter Z0 much more than r0. Instead of choosing Z0 so eq 16 fits ᎑ε, one can choose Z0´ so it fits I (i.e., the electron affinity of H, F, and Cl). This was done in the last line of Table 2. If agreement with I is required, the same tactic can be used throughout Tables 1 and 2.5 Acknowledgment The author thanks a reviewer for extensive assistance. Notes 1. Strictly speaking, the energy levels depend slightly on the relative masses of the electron and the atom (6a). Let m and M be masses of electron and atom. An excellent approximation for µ兾m, where µ is the electron’s reduced mass, is µ兾m = 1 − m兾M. The factor µ兾m belongs on the right side of eqs 1 and 2 and as a divisor of a0 in eqs 6 and 11. Because m兾M is small, µ兾m ≈ 1, and we can use 1 with little error. 2. We substitute ᎑1/2(Z兾n)2 Eh for En in eq 4, set Tn(r) = 0 at r = r0, and solve for r0.
Tn (r ) = E n +
Tn (r0 ) = −
476
1 Z 2 n
Z E h a0 r
2
Eh +
Z E a = 0 r0 h 0
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r0 =
2n 2 a0 Z
3. When the nᐉ subshell has unequal numbers of electrons with opposite spins, the smallest ᎑εi is the ᎑εnᐉs where s is the spin of fewer electrons. A published value of εnᐉ is the weighted average of both values of εnᐉs. 4. The published ε is the integral of ε(r)P 2(r) from r = 0 to ∞. The false values of ε(r) at small r hardly affect the integral because P 2(r) is very small there. 5. The ᎑ε and I for an atom would agree if the correlation energy of the atom equaled the relaxation energy of the ion (the true energy of the ion minus its energy calculated with atomic orbitals). That equality can be imitated by changing ᎑ε in eq 15 to I and using the old r0 to find the new Z0.
Literature Cited 1. Gillespie, R. J.; Moog, R. S.; Spencer, J. N. J. Chem. Educ. 1998, 75, 539–540. 2. Slater, J. C. Phys. Rev. 1930, 36, 57–64. 3. Pimentel, G. C.; Spratley, R. D. Chemical Bonding Clarified Through Quantum Mechanics; Holden–Day: San Francisco, CA, 1969; pp 53–58. 4. Agmon, N. J. Chem. Educ. 1988, 65, 42–44. 1 5. Mirone, P. J. Chem. Educ. 1991, 68, 132–133. Mirone fit (En´) /2 1/ to I 2. 6. Slater, J. C. Quantum Theory of Atomic Structure; McGraw– Hill: New York, 1960; Vol. 1, (a) pp 395–401. (b) p 351. (c) p 95. Every eigenvalue of the Schrodinger equation exists at all coordinates as the constant sum of kinetic and potential energies. (d) pp 166–182. (e) pp 35–37. (f ) pp 306–311. (g) pp 227–229. Slater defined Zpi(r) for the Hartree model, but its equivalent for the Hartree–Fock model appears in his eqs 13–18 to 13–25. 7. CRC Handbook of Chemistry and Physics, 84th ed.; Lide, D. R., Ed.; CRC Press: New York, 2003; pp 10-178, 10-147. 8. Clementi, E.; Roetti, C. Atomic Data and Nuclear Data Tables 1974, 14, 177–478. 9. Sekiya, M.; Tatewaki, H. Theor. Chim. Acta 1987, 71, 149–167. 10. Bunge, C. F.; Barrientos, J. A.; Bunge, A. V. Atomic Data and Nuclear Data Tables 1993, 53, 113–162. 11. Koga, T.; Kanayama, K.; Watanabe, S.; Thakkar, A. J. Int. J. Quant. Chem. 1999, 71, 491–497. 12. Koopmans, T. A. Physica 1933, 1, 104–113. 13. Lowe, J. P. J. Chem. Educ. 2000, 77, 155–156. 14. Rioux, F.; DeKock, R. L. J. Chem. Educ. 2002, 79, 429–430. 15. Tsien, T. P.; Pack, R. T. J. Chem. Phys. 1968, 49, 4247–4248.
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