D4.1 Effective Nuclear Charge
Periodic trends in atomic properties can be predicted by applying these ideas about electron-nucleus attraction and electron-electron repulsion:
- Electron-density distributions are in shells that increase in size as the principal quantum number, n, increases. Electrons in larger shells are, on average, farther from the nucleus and less strongly attracted.
- Electrons repel other electrons, raising Coulombic potential energy. This partly counteracts the attractive force between an electron and the nucleus. Electrons are said to screen or shield other electrons from nuclear charge.
Exercise: Coulombic Attraction and Repulsion
The electron density of a core electron (an electron in an inner shell) is, on average, closer to the nucleus than the electron density of a valence electron. Thus, core electrons can significantly counteract the effect of nuclear attraction. Consider a lithium atom (Li, 1s22s1), which has three protons in the nucleus. Because the 2s orbital is larger than the 1s orbital, the 1s electron density is mostly located between the nucleus and the 2s electron density. (Move the slider in the middle of the figure below to see how much of the 1s electron density lies between the nucleus and 2s electron density.) Thus, the two 1s electrons repel the 2s electron away from the nucleus, counteracting part of the 3+ charge of the nucleus.
To account for such electron-electron repulsions, we use effective nuclear charge, Zeff, the positive nuclear charge (given by the atomic number) reduced by the repulsion of a specific electron by all the other electrons. In the case of the Li 2s electron, quantum mechanics calculate that the repulsions from the two 1s electrons reduce the nuclear charge by 1.72; that is, Zeff for the 2s electron is 3 − 1.72 = 1.28. If all the electron density of the 1s electrons were between the nucleus and the 2s electron, Zeff would be reduced to 1.
The effective nuclear charge for the 2s electron in Li is 1.28. The effective nuclear charge for one of the 2s electrons in Be is 1.91.
In your class notebook, write a clear and concise explanation for the relative sizes of these values. Use your explanation to predict which experiences a larger effective nuclear charge: a 2p electron in B or a 2p electron in O.
Write in your notebook, then left-click here for an explanation.
In a Li atom, the 2s electron is screened from the nucleus by two 1s electrons. Because the 1s electron density is mostly closer to the nucleus than the 2s electron density, the two 1s electrons are between the nucleus and the 2s electron most of the time. Screening is strong; this reduces the effective nuclear charge significantly from the actual nuclear charge of 3 to the effective nuclear charge of 1.28.
A Be atom differs from a Li atom in two important ways: first, there is one more proton in the nucleus, so the actual nuclear charge is now 4, greater by one; second, there is one more electron, which occupies the 2s orbital. Because both valence electrons in Be are in the same shell (n = 2), both 2s electron-density distributions are about the same distance from the nucleus. One 2s electron is between the nucleus and the other 2s electron less than half the time. Thus, a 2s electron is much less effective than a 1s electron in screening the other 2s electron from nuclear charge. Most of the additional actual nuclear charge is not well screened and the effective nuclear charge is larger than for Li.
Applying this argument to B and O, O has three more protons in its nucleus and three more electrons (all in 2p orbitals), than does B. This additional nuclear charge is incompletely screened by the three additional electrons. Therefore, the effective nuclear charge is larger for O than for B and the valence electrons in O are more strongly attracted to the nucleus.
This is a general rule: effective nuclear charge increases from one element to the next moving across a row in the periodic table.
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