What is Effective Nuclear Charge?
Effective Nuclear Charge and Shielding (or Screening) Effect
Consider situation when only one electron is revolving around Nucleus in its own shell. Consider the attraction force between the Nucleus and Electron is F1.
Now imagine there are some more electrons are added into inner shell of the above atom which are also revolving around Nucleus but are present in inner shells . The attraction force between the Nucleus and outermost electron is F2.
Now answer which is stronger force F1 or F2 ?
Yes, F1 is stronger force. The inner electrons which are added repel the outermost electron and thus net attraction force over the outermost electron by Nucleus is reduced. This is called Shielding (or Screening) Effect. And due to this the Effective attraction of nucleus over electron is reduced. This effective attraction of Nucleus over electron is called Effective Nuclear Charge.
How to measure Effective Nuclear Charge
The effective nuclear charge is measured by
Zeff = Z − σ
or Zeff = Zactual − σ
where Zeff is effective nuclear charge
Z = number of protons inside the nucleus
σ = screening constant or shielding constant
The electrons residing in the shells between the nucleus and the valence-shell are called intervening electrons and these intervening electrons are affecting Effective Nuclear Charge.
σ is a measure of the extent to which the intervening electrons screen the out-most shell electrons from the nuclear pull on it.
Factors affecting the magnitude of σ and Zeff and variation in the periodic table
Following are the important factors which affect the magnitude of s and Zeff and predict their variation in the periodic table.
(i) Number of intervening electrons. Greater is the number of electrons intervening between the nucleus and the outer-most shell (i.e., intervening electrons), more will be the magnitude of s and hence the magnitude of Zeff will decrease (Zeff = Zactual − σ) to a greater extent.
When we move down a group, the number of intervening electrons increase and hence the magnitude of s also increases. The increase in the value of s decreases the value of Zeff.
Thus on going down a group, the magnitude of Zeff goes on decreasing.
(ii) Size of the atom: With the increase in the size of the atom, Zeff decreases. Thus:
(a) on moving down the group with the increase in the size of the atom, Zeff decreases in the same direction.
(b) Since the size of atoms decreases as we move along a period from left to right, Zeff increases in the same direction.
Slater’s rules
Slater’s rules are a guideline for determining shielding and, therefore, Zeff.
Effective Nuclear Charge (Zeff) is net attraction experienced by the electron(s) due to presence of nucleus.
Remember, when we wish to determine or conceptualize Zeff, we are considering only one of the outer most electrons. We are interested in the electron with the highest energy (n, l) as the electron that is being shielded by the other electrons in the atom. For example, a sodium atom has 11 electrons, but 10 of those electrons are shielding the outermost electron in the highest n, l. This means that for sodium atom all of the electrons in n= 1, and n=2 shield the electron in 3s from the nucleus, with the 1 s electrons being more effective at shielding than the 2s and 2p electrons.
- Electrons with the same n and l as the electron of interest are not as good at shielding that electron as inner electrons of n-1, or smaller n’s. Of course, electrons closest to the nucleus shield the best.
How can we use the Slater’s Rules to calculate Zeff?
Consider the lithium atom. Lithium has an IE1=520kJ/mol, while the IE1 for hydrogen is 1310kJ/mol. This means one needs more energy to remove an electron from the hydrogen atom then one needs for the same process for lithium. The average radius for lithium is larger than that of hydrogen. Therefore, the outer electron is further from the nucleus for lithium because that electron is a 2s orbital. Inner core electrons repel the electron of interest, the 2s1 electron, so it does not penetrate the nucleus effectively. Therefore, this electron is easily removed. Most of the 1s electrons density is very close to the nucleus. The electron of interest has a Zeff of 1.3 based on Slater’s Rules. The radial distribution of the 2s orbital shows that it can penetrate to the nucleus-just not very well.
Zeff = Z – Φ ; where Φ is Shielding constant.
how to calculate Φ (shielding constant) ?
Rules to determine shielding constant based on Slater’s values:
1. Write the electronic configuration for the atom
2. Group electronic configurations by n value as shown :
(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5d)(5d)(5f)
3.
Case 1: The value of s for an electron residing in an ns or in np orbitals
To calculate s assign the values
(a) 0.35 for each of the remaining electrons is the nth shell
(b) 0.85 for each of the electrons in (n−1)th shell
(c) 1.0 for each of the electrons present in the rest of inner shells
(d) There will be no contribution to the value of s by the electrons residing in orbitals/shell having higher value of their principal quantum number than the orbital containing the electrons for electrons for which s is being calculated.
(e) If s is being calculated for an electron of 1s orbital, there will be a contribution of 0.30 from other single electrons in 1s orbitals ( for H and He only)
Net σ = sum of all the above contribution made by electrons
Case 2 : The value of σ for an electron residing in (n – 1)d orbitals of (n – 1)th shell of an atom or an ion
(a) There will be no contribution to the value of s by the electrons residing in ns orbitals in ns orbital.
(b) Each of the remaining electrons present in (n – 1) d orbitals, makes a contribution of 0.35
(c) Each of the electrons present in the (n – 1)s, (n – 1)p orbitals and inner shells [i.e. 1st 2nd, 3rd,….(n – 2)th shells] makes a contribution of 1.0
Thus :
σ for a (n – 1)d electron = 0.35 × [No. of the electrons in (n – 1)d orbitals] + 1.0 × [No. of electrons in (n – 1)s, (n – 1)p orbitals and inner shells]
Electrons to the outer of the electron of interest do not shield.
EXAMPLE 1: Calculate the effective nuclear charge for outermost electron of Fluorine ?
Ans: Fluorine electronic configuration is : 1s22s22p5 or (1s)2(2s, 2p)7
There are 2 electrons in n=1. There are 7 e in n=2 but this includes the electron of interest. Since electron of interest would not shield itself so subtract 1 leaving 6 electrons that we count in n=2.
There aren’t any other electrons.
Φ = 2 × 0.85+6 × 0.35 = 3.8;
Zeff = Z – Φ = 9-3.8 =5.2
The Zeff on the outer most electron of fluorine is 5.2 . It means outer most electron of Fluorine experience charge of 5.2 units although its nucleus has 9 protons.
EXAMPLE 2: Calculate the effective nuclear charge for 4s electron in calcium ?
Ans: Ca: 1s2 2s22p63s2 3p6 4s2 or (1s)2(2s,2p)8(3s,3p)8 (4s,4p)2
count number of electrons in outermost shell = 1 ; because one electron is the one over which we are calculating the shielding constant.
Φ = 1.00nn-2+0.85 nn-1+0.35 n0
Φ = 2 × 1.00 + 8 × 1.00 + 8 × 0.85 + 1 × 0.35 =17.15;
Zeff = Z – Φ = 20-17.38 = 2.85
The Zeff of Ca is 2.85 .
EXAMPLE 3: Calculate the effective nuclear charge for 4s electron in scandium ?
Ans: Scandium : 1s2 2s2 2p6 3s2 3p6 4s2 3d1 or 1s2 2s2 2p6 3s2 3p6 3d1 4s2
or (1s)2(2s,2p)8(3s,3p)8(3d)1(4s,4p)2
Φ = 1.00nn-2 + 0.85 nn-1 + 0.35 n0
Φ =2 × 1.00 +8 × 1.00 +9 × 0.85
Zeff = Z – Φ = 21-18 = 3
Note : For transition metals, energy order nd > ns or np . In the Slater’s rules, nd electrons are considered inner core electrons. Sc is the odd ball. The nuclear “pull” is about the same for 3d as for 4s.
4s for Cu vs 3d for Cu
4s for Cu : 1s2 2s2 2p6 3s2 3p6 3d10 4s1 or 1s2 2s2 2p6 3s2 3p6 4s1 3d10
or (1s)2(2s, 2p)8(3s, 3p)8(3d)10(4s,4p)1-1
Φ = 1.00nn-2 + 0.85 nn-1 + 0.35 n0
Φ = 2 × 1.00 + 8 × 1.00 + 8 × 0.85 +10 × 0.85 + 0 × 0.35 = 25.3
Zeff = 29-25.3 = 3.70
3d for Cu : 1s2 2s2 2p6 3s2 3p6 3d10 4s1 or 1s2 2s2 2p6 3s2 3p6 4s1 3d10 or
(1s)2 (2s, 2p)8 (3s, 3p)8 (3d)10-1(4s, 4p)1
Φ = 2 × 1.00 + 8 × 1.00 + 8 × 1.00 + 9 × 0.35 + ignore 4s =21.15
Zeff = 29 – 21.15 = 7.85
Explanation of values: Copper has 10 electrons that are considered inner core, n-2 or lower in energy. These are composed of the 2 electrons from n=1, and 8 electrons from n=2 and are the same for any outer electron for copper.
We find differences in values when we look at the electron shielding in different higher energy orbitals on the copper atom.
Lets look at the electron in the 4s orbital. It is shield by everything in n=3. The 8 electrons from 3s and 3p and the 10 electrons from 3d comprise the 18 electrons that are shielding at 0.85. There is one electron in 4s, so it does no shielding (it can’t shield itself). Contrast this with an electron in 3d. 9 other electrons in the same sub shell shield the 3d electron. The other electrons shield these extremely well, and are give a value of 1.00. The 4s electron does not shield the 3d electrons; it is higher in energy.
The 3 electron has a stronger attraction to the nucleus than the 4s electron. The Zeff is higher; a stronger attraction means it takes more energy to remove that electron. The electron that is removed for ionization must come from 4s5.
Cations and anions shielding work in a similar manner.
EXAMPLE 4 : 6s electron in platinum 1s22s22p63s23p64s23d104p65s24d105p66s14f145d9
(1s)2 (2s,2p)8 (3s,3p)8 (3d)10(4s, 4p)8 (4d)10 (4f)14 (5s,5p)8 (5d)9 (6s)2-1
1.00 × 2 + 1.00 × 8 + 1.00 × 8 + 1.00 × 10 + 1.00 × 8 + 1.00 × 10 + 1.00 × 14 + 0.85 × 17 = 74.45
Zeff = 78-74.45 = 3.55
Explanation of values: Platinum has a lot of electrons. Consider the outer electrons first. The electron of interest is in the 6s orbital. The energy shell below that is n = 5. There are 16 electrons in n = 5; the electrons in n=4 and lower shield by 1.00.
EXAMPLE 5: 1s electron in Os
Os: (1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)10(4f)14(5s5p)8(5d)6(5f)0(6s6p)2
For the 1s valence electron: Φ = (1)(0.30) + the rest don’t matter since they don’t shield) so Zeff = 76 – 0.3 = 75.7
The 1s electrons are poorly shielded. It will be very difficult to remove an electron from the 1s orbital.
Applications of Slater’s rules and concept of effective nuclear charge
1. Why is 4s orbital filled earlier than 3d orbitals in potassium atom (Z =19)?
We known that the configuration of Ar (Z = 18) which is the last element of 3rd period of the periodic table is 1s2, 2s2p6, 3s2p6. Thus 3rd shell is not completely filled in Ar atom, since 3d orbitals remain vacant in it.
Why 4s orbitals is filled in preference to 3d orbitals can be explained as follow :
The two configuration that are theoretically possible for K-atom are:
(a) K(Z = 19) = 1s2, 2s2p6, 3s2p6 4s1
(b) K(Z = 19) = 
In example 1 we have calculate the value of Zeff experienced by 4s1 electrons of K-atom [configuration (a)] equal to 2.20. The value of Zeff experienced by 3d electron of K-atom [configuration (b) can be calculated as follows :
σ for 3d electron in structure (b) = 0.35 × 0 + 1.0 × 10 = 18
Zeff experienced by 3d electron = 19 – 18 = 1.0
Since Zeff experienced by 3d electron = 19 – 18 = 1.0
Since Zeff for 4s1 electron is greater than that for 3d1 electron, the attraction between 4s1 electron and the nucleus is greater than that between the 3d1 electron and nucleus of K-atom. Thus 1s2, 2s2p6, 3s2p6, 4s1 configuration would be more stable than 1s2, 2s2p6, 3s2p6d1 configuration. In other words 4s orbitals is filled earlier than 3d orbital.
2. When transition metals are converted into cations, it is ns electrons, and not (n – 1)d elements, which are removed first from the isolated gaseous atoms of transition metals. Explain?
In order to explain this fact, let us consider the formation of Mn+ ion from Mn atom (Z = 25).
The configuration of Mn is 1s2 2s2p6, 3s2p6d5, 4s2.
Mn+ possible configuration :
a) 1s2 2s2p6, 3s2p6 3d5, 4s1
b) 1s2 2s2p6, 3s2p6 3d4, 4s2
for option a configuration : 4s electron is removed . the Zeff of 4s electron is 3.60
for option a configuration : 3d electron is removed . the Zeff of 3d electron is 5.60
Since Zeff for 3d1 electron is greater than that for 4s1 electrons, the attraction between 3d electron and the nucleus of Mn atom is greater than the attraction between 4s electron and the nucleus. Therefore, the removal of a 4d electron from an isolated gaseous atom of manganese is more difficult than that removal of an electron from 4s orbitals, i.e. it is ns electron, that is lost first by a transition metal to convert itself into a cation.
3. Variation in the values of successive ionization energies of a given element. The successive ionization energies (IE1, IE2, IE3 etc.) of a given element (M) increases in the order IE1 < IE2 < IE3 <….This order has been explained on the basis of the increase in the effective nuclear charge experienced by the outermost electron in M, M+, M2+ etc.
PROBLEMS
1. Write the electron configuration for P, F, Cl, Mg, and C, then group the s and p electrons together, by n values.
| Element | Configuration |
| P | 1s22s22p63s23p3; (1s)2(2s2p)8(3s3p)5 |
| F | |
| Cl | |
| Mg | |
| C |
2. Write the electron configuration by grouping using the Slater equation.
| Element | Configuration |
| P | 1s22s22p63s23p3; (1s)2(2s2p)8(3s3p)5-1 |
| F | |
| Cl | |
| Mg | |
| C |
3. Determine Slater’s shielding value for each of the elements and the Zeff.
| Element | Configuration |
| P | Φ = 2 ×1.00 + 8 × 0.85 + 4 × 0.35 = 10.2; 15-10.2 = 4.8 |
| F | |
| Cl | |
| Mg | |
| C |
4. We are interested in the behavior of outer core electrons, but we can use Slater’s Rules for any electron. As seen in the rules, electrons that are in orbitals that are after the electron of interest do not contribute to the shielding of that electron. Determine the Φ and Zefffor each of the following.
| Element | Configuration |
| 2s electron in P | 1s22s22p63s23p3; (1s)2(2s2p)8-1(3s3p)5-1
Φ = 2 × 0.35 +7 × 0.85 =4.15. 15 -4.15 = 10.85 |
| 2p electron in Mg | |
| 2s electron in Cl | |
| 1s electron in He | |
| 1s electron in Mg |
5. for cations and anions ; the same rules are valid .
| Element | Configuration |
| 2s electron in Na+ | (1s)2(2s2p)8-1(3s3p)0
Φ = 2 × 0.35 +7 × 0.85 =4.15 Zeff = 11 -4.15 = 6.85 |
| 3p electron in Cl– | |
| 2p electron in Mg2+ | |
| 1s electron in Li+ |
6. Transition elements: electrons in the same (nd) group contribute 0.35 towards the overall shielding. Elements in all groups to the left in the Slater configuration format contribute 1.00 towards overall shielding. Electrons in all groups to the right in the configuration contribute nothing towards overall shielding.
| Element | Configuration |
| A 3d electron in Ni | Ni = 1s22s22p63s23p63d84p2;
(1s)2(2s2p)8-(3s3p)8(3d)8-1 (4s)2; Φ = 18×1.00+7×0.35=20.45; Zeff = 28 – 20.45 =7.55 |
| 3d electron in Mn | |
| 4d electron in Ru |
Using the techniques above, give a reason why the following occurs:
7. The radius of Cd2+is smaller than that of Sr2+. Calculate the Zeff for the outer most orbital for each atom and each ion. Calculate the Zeff for a 3s orbital for each atom. Give a rational explanation for your findings.
8. The ionization potentials increase subtly for Ru and Rh and have a large increase for Pd, but decrease in the series Fe, Co, Ni. The electron of interest is in the 5s orbital for Ru, Rh, and 4d orbital for Pd (4d because 5s is empty), and 4s for Fe, Co, and Ni.
710.2 kJ × mol-1 Ru
719.7 kJ × mol-1 Rh
804.4 kJ × mol-1 Pd
762.5 kJ × mol-1 Fe
760.4 kJ × mol-1 Co
737.1 kJ × mol-1 Ni
9. Consider the melting point data for the natural forms of the second row elements. Calculate Zeff using Slater’s rules for the outermost electron for each of these elements. Does Zeff give a reasonable accounting of the trend in the melting point? Why or why not? Should melting point be considered a periodic property of atoms? Why or why not?
| Element | Melting Point (°C) |
| Na | 98 |
| Mg | 650 |
| Al | 660 |
| Si | 1414 |
| P | 44 |
| S | 115 |
| Cl | –101.5 |
| Ar | –189 |
10. Calculate Zefffor the highest energy electron in the following ions: K+, Ca+, Sc+, Ti+, V+. Does Zeff account for the observed values of the second ionization potential of these elements? Why or why not? Some have argued that Ca+ should be considered a transition metal. Give an argument to support this contention.
11. Look at a plot for the electron affinity as a function of atomic number for the atoms from Li to Ne. In general, does Zeffaccount for the variation in electron affinity? Why or why not? Are there any specific anomalies in the trend? If so, give an explanation.
