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Lesson 7.2: Effective Nuclear Charge

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Electrons in higher energy shells are on average further from the nucleus. A crude way to visualize these shells, when we are not concerned about the details of the orbital shape studied in lesson 6, is to imagine the electrons as existing in increasingly distant layers from the nucleus. Given this characterization, it does not come as a surprise that lithium with two layers of electrons, two electrons in the first and 1 electron in the second, would be larger that hydrogen with just 1 electron in the first layer. Here is a picture of their relative sizes in pm.

The same trend continues as we move down the periodic table for group 1 elements. Notice that we have drawn the second shell as if it were a fixed size – the same size in lithium and sodium.

This way of looking at things would incline to say that shells have a fixed size, and that when we move across a period we would predict that the size of the atoms remains constant. However, this is not the case. Here is a picture with beryllium included.

Although beryllium contains more stuff – an extra electron, proton and neutron – it is smaller! How are we to make sense of this?

When trying to make sense of the size of atoms, along with other important properties, we need to consider three factors. (Keep in mind that the size of an atom is just the region of space where the electrons exist. A single electron far from the nucleus of an atom can make an atom very large, just like an outer planet can make a solar system very large.)

  1. The charge on the nucleus - how much positive charge is attracting a given electron. The greater nuclear charge, the greater the attraction the electron and the smaller the size contribution of that electron.
  2. The number of electrons between the nucleus and the electron. The nucleus attracts all of the electrons in an atom, but the electrons between the nucleus and a given electron will repel it.
  3. The distance from the nucleus to the given electron.

We keep track of the first two of these factors with the concept of effective nuclear charge. The effective nuclear charge on an electron is just the amount of charge it experiences if we keep track of the repulsions of inner electrons. We will say that inner electrons shield or screen the electron from the charge of the nucleus, so that the charge that is effective at attracting the electron is the total charge minus the charge of the electrons that shield it. As an equation we can write $$ Z_{eff} = Z - S$$

Zeff is the effective nuclear charge, Z the total nuclear charge, and the S the shielding - the repulsion caused by electrons that exist between the electron and the nucleus.

Using this concept, lets return to the comparison of beryllium and lithium. A crude method for finding the effective nuclear charge on the outer electron of lithium is to subtract the charge of the inner shell electrons from the charge on the nucleus. Z = 3, the charge of the nucleus, and S = 2 since the electrons in the first shell exist between it and the nucleus and repel it. So the effective nuclear charge on the outer electron is \begin{align*} Z_{eff} & = Z - S \\\\ Z_{eff} & = 3 - 2 \\\\ Z_{eff} & = 1 \end{align*}

For beryllium we again subtract the charge of the inner shell electrons from the charge on the nucleus. Z = 4 , the charge of the nucleus, and S = 2 since the electrons in the first shell exist between it and the nucleus and repel it. So the effective nuclear charge on the outer electron is \begin{align*} Z_{eff} & = Z - S \\\\ Z_{eff} & = 4 - 2 \\\\ Z_{eff} & = 2 \end{align*}

With a greater effective nuclear charge the electrons in beryllium are drawn closer to the nucleus resulting in a smaller atom.

This trend continues as we move from left to right across the periodic table. The electrons are drawn ever closer to the nucleus due to the increase in effective nuclear charge. The outer electrons nonetheless remain further out than those of hydrogen or helium since they occupy a higher energy level.

The process starts again once we have come to the end of a row on the periodic table. With neon the 2nd energy level is filled and the last electron of sodium is forced to occupy the third energy level. This last electron therefore has 10 electrons between it and the nucleus, giving it an effective nuclear charge of 1. \begin{align*} Z_{eff} & = Z - S \\\\ Z_{eff} & = 11 - 10 \\\\ Z_{eff} & = 1 \end{align*}

Moving from left to right across the periodic table we see the same trend as in the second period. The electrons are drawn ever closer to the nucleus due to the increase in effective nuclear charge.

While this is an easy way to approximate S (the number of core electrons) it does fully make sense. For example, it does not make sense to say that core electrons are 100% effective at shielding valence electrons, and that valence electrons do not shield each other at all. To make sense of these more subtle distinctions we need a more sophisticated method for finding S - a more sophisticated method for thinking about electron-electron repulsion in atoms.

Consider the diagram for beryllium shown below. First, the drawing makes it look like the 1s electrons are restricted to the small volume near the nucleus which is not the case, but even as drawn, it is clear that the two core 1s electrons are not in the same location and thus not able to equally screen a valence electron in the 2s shell. Only if the two electrons were in the same location as the proton would they screen with 100 % efficiency. This condition is only approximated if the valence electrons are sufficiently far from the core electrons existing in a relatively small space around the nucleus. Second, it is clear that the valence electrons can shield each other – will exert repulsive forces on each other that will vary as they move about the 2s orbital.


Slater's Rules

How can we capture these considerations in an effort to generate a better approximation of S and thus Zeff? A set of rules has been devised called Slater's Rules which are easy to apply, captures these insights, and yields Zeff values close to those found with the most sophisticated methods. Here are the rules.

Rule 1: Electrons in the same energy level of s and p orbitals are 35% effective at screening electrons.

For example, how much is an electron in a 3p electron of Cl , 1s22s22p63s23p5, repelled or screened by the other electrons in the 3s or 3p orbitals? There are 7 electrons in the 3rd energy level of chlorine. Any one of the 7 electrons in the 3s and 3p orbitals is repelled by 6 electrons with 0.35 screening efficiency. This makes a total screening of 0.35 (6) = 2.1 for an electron of chlorine in the 3rd energy level. Notice that this screening of 2.1 would be ignored by the crude method of using the core electrons to find S.

Rule 2: Electrons in the next lower energy level, n-1, of a s or p electron, screen with 85% efficiency.

To find their contribution to S simply add up the number of electrons in the next lowest energy level as the one under consideration and multiply by 0.85. For example, a 3p electron in Cl, 1s22s22p63s23p5, is screened by the 8 electrons in the 2nd energy level. The contribution to S is 0.85 x 8 = 6.8 units of charge. Notice that in the crude method we would assume that these screen by 8.

Rule 3: Electrons in the next lower level, n-2, and below, are considered 100% screening electrons.

To find their contribution to S simply add up their number. For example, a 3p electron in Cl , 1s22s22p63s23p5, is then screened by the 2 electrons in the 1st energy level by 2 units of charge, the same amount as we would calculate with the crude method.

That gives us a final value of \begin{align*} Z_{eff} & = Z - S \\\\ S & = 2.1 + 6.8 + 2 = 10.9 \\\\ Z & = 17\\\\ Z_{eff} &= 17 – 10.9 = 6.1. \end{align*} The crude method gave us Zeff = 17 – 10 = 7.

Rule 4: If the electron under consideration is an f or d orbital electron, the electrons that occupy that subshell are assumed to screen by 35% and all the rest by 100%.

For example, Fe has 6 electrons in the 3d orbital. How much is one of these screened by other 5 d electrons? 5 x (0.35). Note that the 4s electrons do not screen at all and that the rest of the electrons in the 3rd, 2nd , and 1st shield 100%. For Fe this gives us a final value of \begin{align*} Z_{eff} & = Z - S \\\\ S & = 5(0.35) + 18 = 19.75 \\\\ Z & = 26 \\\\ Z_{eff} &= 26 - 19.75\\\\ Z_{eff} &= 6.25 \end{align*}

Rule 5: A 1s electron is assumed to screen another 1s electron by 30%.

For example, Fe, like all atoms past hydrogen, has 2 electrons in the 1s orbital. How much does the one electron in that 1s orbital screen the other 1s electron? With 30% efficiency or with S = 1 x (0.30). Note that none of the other electrons in hgher energy levels screen the electron in the 1s orbital. For a 1s electron in Fe this gives us an effective nuclear charge of \begin{align*} Z_{eff} & = Z - S \\\\ S & = 1(0.30) = 0.30 \\\\ Z & = 26 \\\\ Z_{eff} &= 26 - 0.30\\\\ Z_{eff} &= 25.7 \end{align*}

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