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CHEM 20A/1 - winter 2001 LECTURE NOTES - tenth set
SHIELDING.
In the last set of notes we studied systems consisting of many non-interacting electrons. Here we recognize that the electrons in an atom are compressed into tight quarters and that they therefore not only interact with each other but interact very strongly. ( Why do they interact strongly when crowded together? ) Although the effect of the electrostatic interactions among electrons is great, the effect of the exclusion principle on the electronic structure of atoms and molecules is even greater; this is why the periodic features outlined in the last section persist even when the electron-electron repulsions are considered.
Stability and Shielding. We have already discussed to some extent the effect of the electron-electron repulsions. The lower the energy the more stable the system , and since the electron-electron repulsions increase the energy of the atom they make the atom less stable than would be the case for fictitious non-interacting electrons. ( Why? ) Secondly, because any given electron is subject simultaneously to the attractive force of the nucleus and the repulsive forces due to all the other electrons, the net attractive force holding the electron within the atom is less than that due to the nucleus alone. This reduction in the net attractive force that is attributed to the nucleus is called " shielding ;" the other electrons can be envisaged as partially shielding any given electron from the full attractive force of the nucleus. ( Think this one over .)
We focus on shielding. There is a theorem in the study of electrostatics which states that if charges are distributed spherically and observed from a large distance, they act as though they were all situated at the center of the sphere. Thus in an atom the electrons are distributed spherically about the nucleus, and they therefore act as though they were all concentrated at the nucleus. Since the nucleus has a plus charge equal to the negative charge due to all the electrons, the atoms as a whole acts as though equal positive and negative charge were concentrated at the nucleus. Consequently the atom is neutral. The electrons have completely shielded the positive charge on the nucleus, and no electrical force is exerted at the distant observation point.
However, note that the analysis above holds only if observed from a large distance; if observed from inside the atom only those charges that lie inside the observation point can exert an electrical force. This means that outer or valence electrons are well shielded from the nucleus by the core electrons but that the core electrons are not well shielded from the nucleus by the outer electrons. This is the concept of " shielding ." In the simplest of pictures, each core electron is totally effective in shielding outer electrons; thus each electron in an inner shell has the effect of reducing the attractive force on an electron in an outer shell by reducing the effective nuclear charge from Ze to (Z-1)e. At this level of approximation , shielding of inner electrons by outer electrons is neglected as is shielding by electrons in the same shell. The following examples are worth considering:
Problem. Neglecting shielding we previously estimated the ionization energy of H, Li and Na to be RH, 9RH/4, and 121RH/9, respectively. With the rough shielding model given above one might expect the ionization energies to be RH, RH/4, and RH/9, respectively. Why? Problem. Neglecting shielding we estimated the ionization energy of F and Cl to be 81RH/4 and 289/9, respectively. With the rough model of shielding given above one might expect the ionization energies to be 49RH/4 and 49RH/9, respectively. ( Why? )
But this model of shielding is too extreme. The electrons are distributed as described by the probability densities given by the square of the orbitals. Outer electrons penetrate the core to some extent, and the core electrons leak out beyond the outer electrons. The major effect of this is that an electron in a given shell is somewhat effective, but not completely effective in shielding the other electrons in that shell. This has the effect of reducing most of the ionization energies estimated in the two problems above. ( Why? )
Problem. Compare the ionization energy of O (with Z = 8 and configuration (1s)^2 (2s)^2 (2p)^4 ) with that of F (with Z = 9 and configuration (1s)^2 (2s)^2 (2p)^5 ). The 1s electrons do quite an efficient shielding job, thereby reducing the effective charges to just slightly over 6 and 7 for O and F, respectively. ( Why? ) One of the outer n = 2 electrons, the one to be ionized, is then partially shielded by the other five n = 2 electrons in O and by the six in F. If one were to estimate the partial shielding as 0.7 per electron in the same shell , then the effective charge on the electron to be ionized would be about 2.5e for O and about 2.8e for F. It follows that the ionization energy for O and F would be approximately (2.5)^2 RH/4 and (2.8)^2 RH/4, respectively. Be sure you understand this. **** NOTE** : This model is slightly different than that proposed in class in that here the average shielding of one electron by another in the same orbital energy level is taken as about 0.7e, whereas in lecture it was taken as about 0.5e.
This model is closer to the "truth," but is still quite rough. We shall not try to do better here. There are, however, a few other aspects of shielding that we investigate.
Energy level diagram for atomic orbitals.
For hydrogen-like atoms the 2s and 2p orbitals are degenerate. Because of shielding ( i.e ., because of electron-electron repulsions) this is not true for other atoms. In fact the orbital energies for the ns state lie slightly lower than those for the np states. This can be understood by examining the shapes of the wave functions, and hence of the probability densities. Specifically the ns functions have finite probability densities at the nucleus whereas the np functions have nodes at the nucleus. This implies that ns electrons tend to get "inside" the np electrons; consequently, the ns electrons shield the np electrons more than the np electrons shield the ns electrons. Since shielding cuts down the attractive forces of the nucleus, it raises the potential energy. This explains why the ns levels lie below the np levels.
Problem. It does explain it, doesn't it?
Similar reasoning leads to the conclusion that nd levels lie above np levels , and nf levels lie above nd levels.
With the information above one can draw an energy level diagram for atomic orbitals.
- At the bottom one finds the 1s level which can accommodate two electrons.
- Then, with a very large gap comes the 2s level which can accommodate two electrons. The gap is large for two reasons. If there were no shielding the gap would be a
large Z^2 RH(3/4); furthermore, whereas there is little shielding of the 1s electrons, the 1s electrons provide considerable shielding for the 2s electrons, thereby increasing the energy of these.
incorporate the property of spin in which case one can describe a spin-orbital which describes the state of an electron including its spin, and only one electron can occupy any given spin orbital. In describing the energy of the atom by means of the "occupancy" of orbitals, one must take account of the degeneracy of the orbital energy levels.
** The electron-electron repulsive interactions increase the energy and reduce the stability of the atom. This effect can be represented as shielding where each electron is envisaged as partially shielded from the attractive force of the nucleus by the repulsive forces of all the other electrons. This shielding effect can be described in terms of reduction in the effective charge Ze of the nucleus. Shielding removes the degeneracy between the ns, np, nd, and nf levels so that εns < εnp < εnd < εnf.
** Because of the nature of the degeneracies, the atoms develop a shell structure with inner (core) and outer (valence) electrons. The core electrons are less shielded, are held very tightly by and very closely to the nucleus, and they are not altered much in chemical reactions. The outer electrons are more shielded, held less tightly and less closely to the nucleus, and it is these that participate in chemical reactions.
** Because of the shell structure of atoms, similar chemical properties are expected (and found) associated with electronic structures outside each shell core. This gives rise to the periodicity of properties summarized by the periodic table.
The Periodic Table.
Tacking account of the exclusion principle and the concept of shielding one can get a good understanding of the observed periodicity in the properties of the atoms as well as an understanding of their chemical (reactive) properties. I shall not detail these here, but will discuss them in lecture. Your text also gives information on these properties. In particular we can examine ground state electronic structure (configurations), ionization energies, electron affinities, atomic sizes, and reactivity.
The alkali metals {Li, Na, K, Rb, Cs, Fr} all lie in the same column of the periodic table. H also lies in this column. The ground state configurations of all these elements are similar in that they have a single outer electron, only a single ns electron outside a fully occupied core. (H is slightly different in that it has no core.) The outer ns electron is thus heavily shielded and the effective nuclear charge holding it is not much greater than 1e, i.e .,
similar to that for H. Since the orbital energy for this outer ns electron should vary as n-2, as one moves down the column one might expect the ionization energy to decrease, but because the shielding is not perfect, one might expect the effective nuclear charge holding the electron within the atom to increase slightly and counteract the decrease in ionization energy. As a consequence the ionization energies for the alkali metals should be rather low and decrease, but not much as one goes down the column. And this means that these elements (and H as well) can readily lose an electron and form positive ions or cations
{H+,Li+, Na+, K+, Rb+, Cs+, Fr+}. In chemical reactions it is usually the ease with which these elements give up an electron that controls their roles.
The halogens {F, Cl, Br, I, At } all lie in the same column of the periodic table. The ground state configurations of these elements are similar in that they consist of five np
electrons outside a core and outside a two member (ns)^2 subshell. Even the (ns)^2 subshell does a pretty good shielding job, so the inner electrons reduce the effective attractive charge of the nucleus on the np electrons to about 5e. ( Why? ) Additionally, the np electrons shield each other to some extent, let us say about 70% each. This means that any single np electron feels an effective nuclear charge of about 2.2e. ( Why? ) This is much greater than
the effective nuclear charge felt by the lone ns outer electron in the alkali metals; furthermore, the orbital (binding) energy depends upon the square of the effective nuclear charge. Consequently, we expect the outer electron on a halogen (and it can be any one of the five np electrons) to be much more tightly bound than on its alkali metal brethren. Thus the halogens have very high ionization energies and do not loose electrons very easily. On the other hand they can gain an electron quite easily. If an extra electron is added it interacts repulsively with the other electrons, thereby increasing the energy and destabilizing the atom. However, since the shielding is not perfect the added electron still feels an attraction to the nucleus. To understand this note that there are now six np electrons, and that any one of these is shielded by the core electrons and by the other five np electrons; the effective nuclear charge acting on any one of the six np electrons is then approximately 1.5e. ( Why? ) This is less than the effective nuclear charge acting on an outer electron in the atom but it is still sufficient to give rise to a considerable attractive force. Thus the extra np electron is readily accommodated. This means that the halogens readily form negative ions
or anions: {F-, Cl-, Br-, I-, At-}. In chemical reactions it is usually the ability of these elements to attract an additional electron that determines their reactivity and chemistry. The energy of attraction of an extra electron is known as the electron affinity.
The noble gases {He, Ne, Ar, Kr, Xe, Rn} are all placed in the same column. They are all very inert chemically. This means that
- they do not lose electrons easily (have high ionization energies),
- they do not have electron affinities (they do not add electrons easily), and
- it is difficult to change the state of their electrons (no low energy excitations). They all have complete np subshells, i.e. , they have six outer np electrons (except for He which has a complete n = 1 shell). If an extra electron is added without violating the exclusion principle, it must be placed outside the completed shells, and so the attractive force due to the nucleus is almost completely shielded out. Consequently, there is nothing holding the extra electron to the atom, i.e. , there is no significant electron affinity, and it is difficult to form negative ions. At the same time they have large ionization energies and it is not easy to remove an electron and form positive ions. To understand this recognize that although the inner electrons shield the nuclear charge almost completely, the outer np electrons only shield each other partially, so that one might expect an effective charge of about 2.5e on any given outer np electron. ( Why ?) This is a very large effective charge and so the electrons are tightly bound, and it takes much energy to ionize them. The reasoning for why He has a high ionization differs slightly but is quite analogous. ( Explain .).
It is now left to you to explain the following:
Problem. Why on the whole do you expect ionization energies to increase as one moves from left to right on the periodic table? Problem. Why on the whole do you expect electron affinities to increase as one moves from left to right on the periodic table? Problem. Why do you expect elements on the left to form positive ions and those on the right to form negative ions? Problem. Why are second ionization energies higher than first ionization energies? Problem. Why might you expect the alkali earth elements {Be, Mg, Ca, Sr, Ba, Ra} to form doubly positive ions {Be2+, Mg2+, Ca2+, Sr2+, Ba2+, Ra2+}, whereas the alkali metals form only singly charged positive ions? Problem. Why might you expect {O, S, } to add two electrons without too much difficulty whereas the halogens can only add one? Problem. Discuss the relative sizes of the atoms as one moves across a row of the periodic table and as one moves down a column.
Note that the ground state configuration of the carbon atom is (1s)^2 (2s)^2 (2p)^2. The spins of the 1s electrons must be paired, i.e ., one "up" and one "down." (Why?) Thus as a pair they have no net spin and cannot interact with a magnetic field. The same is true of the two 2s electrons. ( Why? ) But the 2p electrons are different. There are many ways in which they could arrange themselves: if the orbitals occupied are identified by (n,l,ml), then the 2p electrons could occupy the {(2,1,1),(2,1,1)}, {(2,1,-1),(2,1,-1)}, {(2,1,0),(2,1,0)}, {(2,1,1),(2,1,-1)}, {(2,1,1),(2,1,0)}, {(2,1,-1),(2,1,0)} orbitals. The first three choices lead to zero net spin, i.e ., paired electrons with Ms = 0. ( Why? Be sure you understand why. ) The last three could lead either to Ms = 0 or Ms = 1. ( Why ?) All things being either, the system is most stable in the higher spin state, i.e ., the Ms = 1 state in this case. This is one of Hund's rules. Just what constitutes "all things being equal" will not be of concern to us here.
