1.14: Slater's Rules
- Page ID
- 487371
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We have previously described the concepts of electron shielding, orbital penetration and effective nuclear charge, but we did so in a qualitative manner. In this section, we explore one model for quantitatively estimating electron shielding and calculating the effective nuclear charge experienced by an electron in an atom. The model we will use is known as Slater's Rules (J.C. Slater, Phys Rev 1930, 36, 57).
Effective Nuclear Charge
The effective nuclear charge \(Z_{eff}\) is less than the actual nuclear charge (\(Z\)) because of the repulsive interaction between core and valence electrons. We can quantitatively represent this difference between \(Z\) and \(Z_{eff}\) as follows:
Equation \(\PageIndex{1}\)
\[ S=Z-Z_{eff} \label{2.6.1} \]
Rearranging this formula to solve for \(Z_{eff}\) we obtain Equation \(\PageIndex{1}\)
\[ Z_{eff}=Z-S \label{2.6.2} \]
The shielding constant S is estimated using Slater's Rules. Then the shielding constant is substituted into Equation \(\PageIndex{2}\) to calculate an estimate of \(Z_{eff}\) for the corresponding atomic electron.
Slater's Rules
Shielding is calculated for an electron-of-interest. Shielding occurs when "other" electrons in lower energy shells (or the same energy shell) provide a repulsive force to an electron-of-interest in a higher energy shell. The repulsive force negates some of the attractive force from the positive nucleus on the electron-of-interest. The "other" electrons in the same shell as the electron-of-interest shield least effectively, while those closest to the nucleus shield the most effectively. Slater's rules specify the degree of shielding for each inner shell electron and are used to calculate the overall shielding experienced by an electron in an s-, p-, d-, or f- orbital.
The shielding numbers in Table \(\PageIndex{1}\) were derived semi-empirically (i.e., derived from experiments) as opposed to theoretical calculations. This is because quantum mechanics makes calculating shielding effects quite difficult and outside the scope of this text.
Slater's Rules estimate effective nuclear charge for an electron-of-interest based on how many other electrons exist in the atom and on the location of the electron-of-interest. In the calculation of the shielding constant S, it is assumed that electrons closer to the nucleus than the electron-of-interest cancel some of the nuclear charge; those farther from the nucleus have no effect. For an electron-of-interest in an s or p orbital, electrons in the same shell (with the same principle quantum number n)will shield the electron-of-interest 0.35 units. Conceptually, this means electrons in the same shell shield each other 35%. Electrons in the lower energy shell n-1 contribute 0.85 to S. All the electrons in even lower shells contribute 1.00 to S. Conceptually, this means that s and p electrons are shielded 85% by the electrons one shell lower, and 100% by all electrons in shells n - 2 or lower. For an electron-of-interest in a d or f orbital, all electrons in the same shell (with the same principle quantum number n) and subshell (with the same angular momentum quantum number l) will shield the electron-of-interest 0.35 units. All other electrons contribute 1.00 to S. Conceptually, this means that electrons in the same subshell shield each other 35%, and all other electrons shield 100%. Note: The shielding constant S of a 1s electron is just S=0.3, no matter the element.
Table \(\PageIndex{1}\): Shielding values for electrons in inner orbitals based on their principle and angular moment quantum numbers relative to the orbital-of-interest.
| Orbital of Electron-of-Interest | n and l | n and l-1 | n and l-2 | n-1 | n-2, n-3, etc. |
|---|---|---|---|---|---|
| 1s | 0.30 | - | - | - | - |
| ns, np | 0.35 | 0.35 | - | 0.85 | 1.0 |
| nd, nf | 0.35 | 1.0 | 1.0 | 1.0 | 1.0 |
Calculating Shielding and Effective Nuclear Charge
- Write the electron configuration of the atom with orbitals arranged by increasing principle quantum number n, as shown here:
(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . .
- Identify the electron of interest, and ignore all electrons that are higher in energy (to the right in the list). Higher energy electrons do not shield lower energy electrons.
- Identify the principle quantum number n and angular momentum quantum number l of the electron-of-interest.
- Assign shielding values to all other electrons that are lower in energy (to the right in the list).
- When the electron-of-interest is in a 1s orbital, the other electrons shield 0.30.
- When the electron of interest is in a ns or np orbital
- electrons in the same orbital shield 0.35
- electrons with n and l-1 shield 0.35
- electrons with n-1 shield 0.85
- all other electrons shield 1.00
- When the electron-of-interest is in a nd or nf orbital
- electrons in the same orbital shield 0.35
- all other electrons shield 1.00
- Sum the contributions of all electrons (except the electron-of-interest) to obtain the shielding constant S from the shielding values assigned to each electron, Si.
Equation \(\PageIndex{3}\)
\[ S = \sum n_i S_i \label{2.6.0}\]
- Subtract the shielding constant S from the nuclear charge Z to obtain the effective nuclear charge, as shown in Equation \(\PageIndex{2}\)
What is the shielding constant experienced by a 2p electron in the nitrogen atom? What is the effective nuclear charge experienced by this electron?
Strategy:
- Determine the electron configuration of nitrogen, then write it in the appropriate form.
- Use the appropriate Slater Rules to calculate the shielding constant for the electron.
- Subtract the shielding contant from the nuclear charge to find the effective nuclear charge.
Solution:
N: 1s2 2s2 2p3
N: (1s2)(2s2,2p3)
From Table \(\PageIndex{1}\):
- There are four 2s and 2p electrons that shield the 2p electron-of-interest equally at 0.35 units. The electron of interest does not shield itself!
- There are two 1s electrons that shield the 2p electron-of-interest at 0.85 units.
\[S[2p] = \underbrace{0.35(4)}_{\text{the 2s and 2p electrons}} +\underbrace{0.85(2)}_{\text{the 1s electrons}} = 3.10\nonumber\]
For N, Z = 7
\[ Z_{eff} = 7 - 3.1 = 3.9 \]
What is the shielding constant experienced by a 3d electron in the bromine atom? What is the effective nuclear charge experienced by this electron?
Strategy:
- Determine the electron configuration of nitrogen, then write it in the appropriate form.
- Use the appropriate Slater Rules to calculate the shielding constant for the electron.
- Subtract the shielding contant from the nuclear charge to find the effective nuclear charge.
Solution:
Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
Br: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p5)
Ignore the group to the right of the 3d electrons. These are in higher energy orbtials and do not contribute to the shielding constant.
From Table \(\PageIndex{1}\):
- There are nine 3d electrons that shield the 3d electron-of-interest equally at 0.35 units.
- There are eight 3s and 3p electrons that shield the 4d electron-of-interest equally at 1.00 units.
- There are eight 2s and 2p electrons that shield the 4d electron-of-interest at 1.00 units.
- There are two 1s electrons that shield the 4p electron-of-interest at 1.00 units.
\[S[3d] = \underbrace{0.35(9)}_{\text{the 3d electrons}} +\underbrace{1.00(8)}_{\text{the 3s and 3p electrons}} +\underbrace{1.00(8)}_{\text{the 2s and 2p electrons}} +\underbrace{1.00(2)}_{\text{the 1s electrons}} = 21.15\nonumber\]
For Br, Z = 35
\[ Z_{eff} = 35 - 21.15 = 13.85 \]
Problems
What is the shielding constant experienced by a valence p-electron in the boron atom? What is the effective nuclear charge experienced by this electron?
- Answer
-
B: 1s2 2s2 2p1
B: (1s2)(2s1)
From Table \(\PageIndex{1}\):
- There are two 2s electrons that shield the 2p electron-of-interest equally at 0.35 units. The electron of interest does not shield itself!
- There are two 1s electrons that shield the 2p electron-of-interest at 0.85 units.
\[S[2p] = \underbrace{0.35(2)}_{\text{the 2s electrons}} +\underbrace{0.85(2)}_{\text{the 1s electrons}} = 2.40\nonumber\]
For B, Z = 5
\[ Z_{eff} = 5 - 2.4 = 2.6 \]
What is the shielding constant experienced by a valence p-electron in the bromine atom? What is the effective nuclear charge experienced by this electron?
- Answer
-
Br: (1s2)(2s2,2p6)(3s2,3p6,3d10)(4s2,4p5)
From Table \(\PageIndex{1}\):
- There are six 4s and 4p electrons that shield the 4p electron-of-interest equally at 0.35 units.
- There are 18 3s, 3p, and 3d electrons that shield the 4p electron-of-interest equally at 0.85 units.
- There are eight 2s and 2p electrons that shield the 4p electron-of-interest at 1.00 units.
- There are two 1s electrons that shield the 4p electron-of-interest at 1.00 units.
\[S[4p] = \underbrace{0.35(6)}_{\text{the 4s and 4p electrons}} +\underbrace{0.85(18)}_{\text{the 3s, 3p, and 3d electrons}} +\underbrace{1.00(8)}_{\text{the 2s and 2p electrons}} +\underbrace{1.00(2)}_{\text{the 1s electrons}}= 27.4\nonumber\]
For Br, Z = 35
\[ Z_{eff} = 35 - 27.4 = 7.6 \]
What is the shielding constant experienced by the valence p-electrons in the three isoelectronic species: fluoride (F-), neon (Ne), and sodium cation (Na+)? What is the effective nuclear charge experienced by these electrons?
- Answer
-
F-, Ne, Na+: 1s2 2s2 2p6
F-, Ne, Na+:: (1s2)(2s2,2p6)
From Table \(\PageIndex{1}\):
- There are five 2s and 2p electrons that shield the 2p electron-of-interest equally at 0.35 units.
- The electron of interest does not shield itself! There are two 1s electrons that shield the 2p electron-of-interest at 0.85 units.
\[S[2p] = \underbrace{0.35(6)}_{\text{the 2s and 2p electrons}} +\underbrace{0.85(2)}_{\text{the 1s electrons}} = 3.80\nonumber\]
For F, Z = 9
\[ Z_{eff} = 9 - 3.8 = 5.2 \]
For Ne, Z = 10
\[ Z_{eff} = 10 - 3.8 = 6.2 \]
For Na, Z = 11
\[ Z_{eff} = 11 - 3.8 = 7.2 \]
What is the shielding constant experienced by a valence d-electron in the copper atom?What is the effective nuclear charge experienced by this electron?
- Answer
-
Cu: 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Cu: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s1)
Ignore the group to the right of the 3d electrons. These are in higher energy orbtials and do not contribute to the shielding constant.
From Table \(\PageIndex{1}\)
- There are nine 3d electrons that shield the 3d electron-of-interest equally at 0.35 units.
- There are eight 3s and 3p electrons that shield the 4d electron-of-interest equally at 1.00 units.
- There are eight 2s and 2p electrons that shield the 4d electron-of-interest at 1.00 units.
- There are two 1s electrons that shield the 4p electron-of-interest at 1.00 units.
\[S[3d] = \underbrace{0.35(9)}_{\text{the 3d electrons}} +\underbrace{1.00(8)}_{\text{the 3s and 3p electrons}} +\underbrace{1.00(8)}_{\text{the 2s and 2p electrons}} +\underbrace{1.00(2)}_{\text{the 1s electrons}} = 21.15\nonumber\]
For Cu, Z = 29
\[ Z_{eff} = 29 - 21.15 = 7.85 \]
What is the shielding constant experienced by a valence d-electron in the zinc atom? What is the effective nuclear charge experienced by this electron?
- Answer
-
Zn: 1s2 2s2 2p6 3s2 3p6 4s2 3d10
Zn: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2)
Ignore the group to the right of the 3d electrons. These are in higher energy orbtials and do not contribute to the shielding constant.
From Table \(\PageIndex{1}\)
- There are nine 3d electrons that shield the 3d electron-of-interest equally at 0.35 units.
- There are eight 3s and 3p electrons that shield the 4d electron-of-interest equally at 1.00 units.
- There are eight 2s and 2p electrons that shield the 4d electron-of-interest at 1.00 units.
- There are two 1s electrons that shield the 4p electron-of-interest at 1.00 units.
\[S[3d] = \underbrace{0.35(9)}_{\text{the 3d electrons}} +\underbrace{1.00(8)}_{\text{the 3s and 3p electrons}} +\underbrace{1.00(8)}_{\text{the 2s and 2p electrons}} +\underbrace{1.00(2)}_{\text{the 1s electrons}} = 21.15\nonumber\]
For Cu, Z = 30
\[ Z_{eff} = 30 - 21.15 = 8.85 \]
References
- James L. Reed, "The Genius of Slater's Rules" , J. Chem. Educ., 1999, 76 (6), p 802
- David Tudela, "Slater's rules and electron configurations", J. Chem. Educ., 1993, 70 (11), p 956
- Kimberley A. Waldron, Erin M. Fehringer, Amy E. Streeb, Jennifer E. Trosky and Joshua J. Pearson, "Screening Percentages Based on Slater Effective Nuclear Charge as a Versatile Tool for Teaching Periodic Trends", J. Chem. Educ., 2001, 78 (5), p 635