Shielded nuclear charge

Since the same orbitals are used for the ground state and the hole state, no relaxation effects due to the change in the shielded nuclear charges are taken into account, and this model is called the frozen atomic structure approximation.) Due to the interpretation of els as the binding energy of one ls-electron, the differential equation for the Pls(r) orbital, equ. (7.66b), can be interpreted as a one-particle Schrodinger equation for the Is orbital (see equs. (1.3) and (1.4)) ... 

Iron cloud lies closer to the nucleus than the periphery of the completed 4d subshell. The valence electron is shielded from the positive nucleus only incompletely, thus being held more firmly (ionization potential 7.57 ev), than is the valence electron in the rubidium atom (ionization potential 4.19 ev), for which the shielding is more nearly complete. Likewise, there is attraction between the incompletely shielded nuclear charge of one atom in silver metal and the peripheries of the electron clouds of adjacent atoms and breakup of the metal structure to the individual atoms is far more difficult for silver (heat of sublimation 67 kcal per gram-atom) than for any of the alkali metals (heats of sublimation ranging from 20 to 36 kcal). If any one factor may be said to explain the nobility of the coinage metals, it would thus be the incomplete shielding of the valence electron by the inner d orbitals. 

Between La (Z = 57) and Lu (Z = 71) the 4/ orbitals are sequentially filled. The very diffuse /-orbitals are relatively ineffective at shielding nuclear charge, even more so than d-electrons. The increased effective nuclear charge in... 

In more detail, since the proton in Equation (1.14) bears a positive charge, the energy increase results from the repulsion of the proton by the shielded nuclear charges of B. These are largest for a short bond to the proton, as we have in H3O+, and are lower for H3S+ and H3Se+. Also the nuclear repulsions decrease as charge is removed from the nucleus to the outer parts of the molecule, as in going from H2F+ to NH. Since Equation (1.15) depends inversely on R, the overall effect is to make small molecules reluctant to add protons in the gas phase. 

In any column in the periodic table, the IP s decrease as the atomic number increases. Let us examine, for example, the Li and Cs atoms. Lithium, which has IPi = 5.390 eV, has the electronic configuration [He]2j-. Cesium, with IPi = 3.893 eV, has the structure [Xe]6j. The 2s electron in Li spends much more time near the nucleus than the 6s electron does in Cs. This means that the net attraction between the electron and Zeii, the shielded nuclear charge, is substantially larger for the Li 2s electron than for the Cs 6s electron, a fact that is illustrated in Fig. 1-9. 

The decrease in atomic radius moving across the periodic table can be explained in a similar manner. Consider, for example, the third period, where electrons are being added to the third principal energy level. The added electrons should be relatively poor shields for each other because they are all at about the same distance from the nucleus. Only the ten core electrons in inner, filled levels (n = 1, n = 2) are expected to shield the outer electrons from the nucleus. This means that the charge felt by an outer electron, called the effective nuclear charge, should increase steadily with atomic number as we move across the period. As effective nuclear charge increases, the outermost electrons are pulled in more tightly, and atomic radius decreases. 

As well as being attracted to the nucleus, each electron in a many-electron atom is repelled by the other electrons present. As a result, it is less tightly bound to the nucleus than it would be if those other electrons were absent. We say that each electron is shielded from the full attraction of the nucleus by the other electrons in the atom. The shielding effectively reduces the pull of the nucleus on an electron. The effective nuclear charge, Z lle, experienced by the electron is always less than the actual nuclear charge, Ze, because the electron-electron repulsions work against the pull of the nucleus. A very approximate form of the energy of an electron in a many-electron atom is a version of Eq. 14b in which the true atomic number is replaced by the effective atomic number ... 

The electron density in d-orbitals is low near the nucleus, and so they are not very effective at shielding other electrons from the positive nuclear charge. 

The atomic radii of the second row of d-metals (Period 5) are typically greater than those in the first row (Period 4). The atomic radii in the third row (Period 6), however, are about the same as those in the second row and smaller than expected. This effect is due to the lanthanide contraction, the decrease in radius along the first row of the / block (Fig. 16.4). This decrease is due to the increasing nuclear charge along the period coupled with the poor shielding ability of /-electrons. When the d block resumes (at lutetium), the atomic radius has fallen from 217 pm for barium to 173 pm for lutetium. 

Metals in Groups 11 and 12 are easily reduced from their compounds and have low reactivity as a result of poor shielding of the nuclear charge by the d-electrons. Copper is extracted from its ores by either pyrometallurgical or bydrometallurgical processes. 

As the value of n increases, d- and /-electrons become less effective at shielding the outermost, highest-energy elec-tron(s) from the attractive charge of the nucleus. This higher effective nuclear charge makes it more difficult to oxidize the metal atom or ion. 

The orbital distortions due to relativity can be pictured in a qualitative way as follows the relativistic potential appears as a short-range attractive term that, above all, stabilizes s-orbitals and makes them more compact as well as steep close to the nucleus. The same happens to a lesser extent for the p-orbitals. Since the low-lying s- and /7-orbitals are now closer to the nucleus they do shield the nuclear charge better than in the nonrelativistic case. Consequently, d- and /-orbitals are... 

As the atom radii increase down group 1, the outermost electron is further from the nucleus (1) and more shielded from the nuclear charge, so is held less tightly (despite the increasing number of protons in the nucleus of each atom) (1). Always answer in terms of factors such as distance, shielding and nuclear charge. 

As we move across the period, atomic radius decreases as electrons are added to the same shell ( = 3) (1) and, therefore, the outermost electrons experience the same shielding (1) from the nuclear charge. The increase in the number of protons (1) causes the decrease in radius from Na to Ar. These factors also explain the general increase in first ionisation energy across the period. 

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